Today we're going to try to cover a lot of topics in algebra including solving equations, linear quadratic equations, factoring, working with exponents, graphing equations, and a lot of other stuff that you've learned in algebra. We're going to try to cover that today in this video. So let's begin, let's get right into it. So, let's begin our discussion with numbers. A natural number.
is a number that's a whole number but greater than zero so 1 2 3 4 5 these are natural numbers a whole number is a natural number that includes 0 an integer are whole numbers that can also be negative All of these are real numbers, and they're all rational. A rational number is a number that can be put into a fraction using integers. So 2 over 3 is rational. 4 is rational. The square root of 9 is rational because that's equal to 3. So all of those numbers are rational.
An irrational number is one that cannot be put into a fraction using integers. So an example would be the square root of 5. If you type in the square root of 5 using a calculator, you're going to get a number that looks like this. 2.236067977 and it keeps on going.
So that's an example of an irrational number. It doesn't end. Now, let's say if you have a number like Point 2 repeating.
This number goes on forever. It doesn't end, but there's a pattern. Point 2 repeating is 2 over 9. Since it could be put into a fraction using integers, this is a rational number. So, if this number goes on forever and you don't see a pattern, it's irrational. So the examples that we've considered are real numbers.
Imaginary numbers are numbers that look like this. You might see an i. i is equal to the square root of negative 1. And whenever there's no number here, it's a 2. It's an even number.
So anytime you have the square root of a negative number, this is equal to an imaginary number. But if you have the cube root of a negative number, that's equal to a real number. So anytime the index number is an odd number, it's going to produce a real number.
If it's an even number, you're going to have an imaginary number if there's a negative sign inside. So now let's go over some basic rules of addition and subtraction. So, let's say if you want to add 5 plus 4. Now, this is pretty much an easy question, but for some of you who don't like to deal with negative numbers, using a number line can help.
So, if we start with 5, whenever you're adding... move to the right on the number line so if you're adding 4 to 5 go 4 units to the right if you count it this is 6 7 8 9 so now let's say if you want to find the value of 6 actually let's change it let's say 2 minus 5 Now you know it's negative 3, but using the number line, you can see why. If we start at 2, and if we subtract it by 5, anytime you need to subtract, go to the left.
Go 5 units to the left. 1, 2, 3. three, four, five. So this is negative one, zero. I take that back. That is not negative one.
This is one, zero, negative one, negative two, negative three. Now what about this one? Let's say if you have negative five minus six, what is the answer to that question? What is the value of negative 5 minus 6?
So let's start with negative 5 on the number line. And if we're going to subtract it by 6, we need to move 6 units to the left. So 1, 2, 3, 4, 5, 6. So this is negative 6, negative 7, negative 8, negative 9, negative 10, negative 11. So negative 5 minus 6 is equal to negative 11. Now what about this one? Negative 7 minus negative 4. Whenever you have two negative numbers next to each other, they will change into a positive number.
So this is the same as negative 7 plus 4. So if you start at negative 7, add 4 to it. Go 4 units to the right. 1, 2, 3, 4. So this is negative 6, negative 5, negative 4, negative 3. So that's a simple way you can do addition and subtraction. Now you want to get to a point where you don't have to write it down. You can do it mentally in your head.
But I just want to go over the basics for those of you who may want to refresh on this concept. So now, let's talk about add-in and subtracted numbers, like bigger numbers. So let's say if you wish to add 126 plus 97. So first you add 6 and 7. 6 and 7 is 13. So you need to write the 3 and carry over the 1. What's 1 plus 2 plus 9?
9 plus 1 is 10, plus 2 is 12. So we're going to put a 2 here, and carry a 1. And then 1 and 1 is 2, so it's 223. That's 126 plus 97. So that's a simple way in which you can add. Now let's go over subtraction. Let's say if we want to subtract...
94 actually let's use 294 and Actually let's change it let's say 253 minus 167 So how would you do this one? So, let's subtract the top number by the bottom number. 3 minus 7, that's a negative number, so we're not going to do that right now. We're going to borrow a 1 from the 5. So the 5 is going to become a 4, and the 3 will now be...
a 13. So 13 minus 7 is 6. Now if we subtract 4 by 6, we're going to get a negative number. So we need to borrow another 1 from a 2, and that's going to turn into 1. And a 4 will be a 14. So 14 minus 6 is 8. And 1 minus 1 is 0, so it's simply 86. If you have your calculator, you can check it. 253 minus 167 is indeed equal to 86. Now, let's try another example. Let's say if we were to subtract 124 from 268. How would you do it? So notice that the 268 is bigger, so you're going to get a negative number.
so instead of subtracting this way let's reverse the numbers let's subtract 268 by 124 now this is going to give us a positive number whatever answer we get yet we're going to just put it here and add a negative sign and the answer will be the same so 8 minus 4 is 4 6 minus 2 is 4 2 minus 1 is 1 so it's 144 so therefore the answer is negative 144 since we reversed it and that's how you can subtract two numbers if you're going to get a negative number using that technique now let's talk about multiplication so let's say if we want to apply 74 by 26 how can we do it 6 times 4 is 24 add the 2 6 times 7 that's 42 plus 2 so that's 44 next add a 0 2 times 4 is 8 let's get rid of this 2 we don't need anymore and then 2 times 7 that's 14 and then let's add the 2 numbers 4 and 0 is 4 4 and 8 is 12 Carry the 1 over. 4 plus 4 is 8 plus 1. That's 9. And then the 1 dropped down. So, 74 times 26 is indeed 1924, if you check with the calculator. So that's how you can multiply two numbers by hand.
So let's say if we wish to divide 299 by 13. How can we do it? Whenever you have a fraction, it basically represents division. You're dividing the numerator by the denominator. So to divide 13, or 299 by 13, let's write it this way. Let's put the 13 on the outside and 299 on the inside.
So how many times does 13 go into 29? Let's make a list. If we keep adding 13, 13 plus 13 is 26, 26 plus 13 is 39. So if you add 13 three times, you get 39, which means 13 times 3 is 39. 39 is too much, it exceeds 29, so we need to use the 26. 13 times 2 is 26. Next we need to subtract, so 29 minus 26 is 3, and then we need to bring down a 9. So how many times does 13 go into 39? We know it's 3 times, because 3 times 13 is 39. And so we have a remainder of 0. Therefore, 299 divided by 13 is equal to 23. So that's how you can divide two numbers using long division.
So now let's move on into fractions. So let's say if you want to add two fractions. 3 over 5. plus 2 over 7. How would you do it?
Here's one technique that can work. 5 times 2 is 10, and 3 times 7, that's 21, and 5 times 7 is 35. So, 21 plus 10, that's 31, so it's 31 over 35. Now another way in which you can get the same answer is by getting common denominators. So for the second fraction, multiply the top and bottom by 5. And for the first one, multiply the top and bottom by 7. So 7 times 3 is 21, and 7 times 5 is 35. 2 times 5 is 10. So 21 plus 10 is 31. As soon as the two denominators are the same, you may add the numerators. So now, what about this one? Let's say we have 4 over 7 minus 5 over 9. So we can use this technique.
4 times 9 is 36, and 7 times 5 is 35. 7 times 9 is 63. By the way, the 36 should come first because the 4 comes before the 5. So, it shouldn't be 35 minus 36. It's 36 minus 35. So, when you subtract, you have to be careful with the order in which you write it because it is important. When you're adding, it doesn't matter. So, the final answer is 36 minus 35, which is 1. So, it's 1 over 63. Now, what about this one? Let's say if we have 2 thirds plus 4 over 3. Well, actually...
let's say 4 over 5 minus 1 over 6. So if you're adding and subtracting three fractions, what do we need to do? So I really wouldn't use the criss-cross method because it works well for two fractions. Instead, what I would use is the common denominator.
We need to find that. So what number can 3, 5, and 6 go into? 5 times 6 is 30, so 5 and 6 can go into 30. 3 can go into 6, so 3 can also go into 30. Let's multiply top and bottom by...
Let's attempt to get a common denominator of 30. So, in order to get a 30 for the third fraction, we need to multiply the top and the bottom by 5, because 6 times 5 is 30. In order to get a 30 for the first fraction, we need to multiply top and bottom by 10. And to get a 30 for the second fraction, we need to multiply top and bottom by 6. So, each fraction will have a denominator of 30. 10 times 2 is 20, 6 times 4 is 24, and 1 times 5, well, that's just 5. 20 plus 24 is 44. 44 minus 5 is 39. So the final answer for this problem is 39 over 30. But it turns out that we can reduce it. 30 and 39 are divisible by 3. 39 divided by 3, that's 13. And 30 divided by 3 is 10. So this is the answer in its reduced form. Now, how can we multiply two fractions? So let's say if we have 7 over 6. multiplied by 5 over 9. What can we do? Whenever you need to multiply two fractions, multiply across.
7 times 5 is 35, and 6 times 9 is 54. And that's all you got to do when multiplying two fractions. But now let's say if you have two fractions that contain very large numbers. Let's say You have 60 over, hmm, 60 over 28 times 35 over 48. What would you do in this case? Would you multiply across? Now when multiplying fractions with very large numbers, it's going to be very helpful if you can break those numbers into smaller numbers.
For example, 60 is 12 times 5. 28 is 7 times 5. 4. 35 is 7 times 5. And 48 is 12 times 4. So notice that we can cancel a 12, we can cancel a 7, and that's about it. So what we have left over on top is 5 times 5, and that's 25. And on the bottom, 4 times 4, which is 16. And this is the answer. It's 25 over 16. Now, what if you need to divide two fractions?
Let's say if you have 12 over 5 divided by 4 over 15. If you need to divide two fractions, use the expression keep change flip. Keep the first fraction the same, change division to multiplication, and flip the second fraction. so now we can multiply across 12 is 4 times 3 15 is 5 times 3 and so we can cancel a 4 and we can cancel a 5 so the final answer is simply 3 times 3 which is 9 Now, let's say if you get a fraction that looks like this, 7 over 8 divided by 16 over 21. If you get a fraction that looks like this, rewrite it as 7 over 8 divided by 16 over 21. And then using the keep change flip, this is the same as 7 over 8 times 21 over 16. So there's nothing we can cancel here.
21 can be broken into 7 and 3, and there's no 3 in 16 or 8. So therefore, we have to multiply across. So what's 21 times 7? 20 times 7 is 140, because 2 times 7 is 14. If you add a 0, you get 140. And 7 times 1 is 7, so this is going to be 147. Now what about 8 times 16?
8 times 10 is 80. 8 times 6 is 48. If you add 80 and 48, you're going to get 128. And so this is the answer, 147 over 128. Now what should you do if you have a complex fraction? Let's say you have 5 over 1 plus 1 third. How can you simplify this expression?
So what you want to do is get rid of the small fraction in the large fraction. So notice that the denominator of the small fraction is 3, therefore multiply top and bottom by 3. So 5 times 3 is 15. On the bottom we're going to distribute the 3. So 3 times 1 is simply 3 and 3 times 1 3rd the 3's cancel and you're just left with 1. 3 plus 1 is 4 so it's 15 over 4. Let's try another example like that. So let's say if we have 7 minus 1 half divided by 5 plus 1 3rd.
So notice that we have too many fractions within the larger fraction. So you want to multiply by the common denominator which is 6. So, let's distribute the 6. 7 times 6, that's 42. And then 6 times 1 half is the same as 6 divided by 2, which is 3. On the bottom, we have 5 times 6, which is 30. And 6 times the third, or 6 divided by 3, that's 2. 42 minus 3 is 39, and 30 plus 2, that's 32. So this is the answer. Now the next thing we're going to talk about is how to convert an improper fraction to a mixed number.
So let's say if we have 11 over 4. So how many times does 4 go into 11? 4 goes into 11 two times. 4 times 2 is 8. So we're going to break the fraction into 8 over 4 and 3 over 4. Notice that 8 over 4 plus 3 over 4 is 11 over 4. 8 divided by 4 is 2. 2 plus 3 fourths is the same as 2 and 3 fourths. So that's a simple way in which you can convert an improper fraction into a mixed number. Let's try another example.
So 19 over 5. 5 goes into 19 three times. 3 times 5 is 15, so we're going to break it into two improper fractions. 15 over 5 and 4 over 5. 15 divided by 5 is 3. 3 plus 4 fifths is the same as 3 and 4 fifths.
Now, how can we convert a mixed number into an improper fraction? So, let's say if we have 2 and 4 fifths. What do we need to do?
So, the 5 is going to remain on the bottom. Multiply 2 and 5 and add the numerator, 4. 2 times 5 is 10. 10 plus 4 is 14, so it's 14 over 5. Let's try another example. 5 and 3 sevenths.
So the 7 is going to stay the same. 5 times 7 is 35 plus 3. That's 38. Now the next thing we need to go over is converting fractions into decimals. How can you convert 1 over 7 into a decimal?
The simplest way to do this is to use long division. Whatever is in the denominator or the bottom of the fraction is going to go on the outside. The 1 is going to go on the inside.
Now 7 doesn't go into 1, so we need to add a 0. 7 goes into 10 one time. And 7 times 1 is 7. 10 minus 7 is 3. So let's add another 0. How many times does 7 go into 30? 7 goes into 30 4 times. 7 times 4 is 28. And so we have 2 remaining.
And so let's add another 0. 7 goes into 20 2 times. And we can stop here because this number is going to go on forever. So we can say 1 over 7 is approximately 0.14.
So that's how you can turn a fraction into a decimal. Now what about the reverse? How can we turn a decimal into a fraction?
So 0.3, what is that as a fraction? You can put it over 1, and if there's only one digit after the decimal place, multiply the top and bottom by 10. 0.3 times 10 is 3. 1 times 10 is 10. So 0.3 is basically 3 over 10 as a fraction. So now let's say if we have.24, if there's two digits after the decimal.
Multiply top and bottom by 100..24 times 100 is 24. 1 times 100 is 100. And then reduce it. Half of 24 is 12. Half of 100 is 50. So we can divide it by 2 again. Half of 12 is 6. Half of 50 is 25. And so.24... is 6 over 25 as a fraction.
So now let's go over percentages. If you have 57%, how can you convert this number into a decimal? To convert a percentage into a decimal divided by 100, 57% is the same as 0.57. 90% is 0.90. 8% is 0.08, 0.5% is 0.05.
I mean not 0.05 but 0.005. There we go. Now sometimes you may get a question that asks you, what is 20% of 50?
So how would you figure this out? There's an equation that goes like this. is basically equal to the percentage multiplied by of. So we're looking for what is 20% of 50. We're looking for the is part. So we need to multiply 20% by 50. 20% as a decimal is 0.20.
So what's 0.20 times 50? 0.20 times 10 is 2. 2 times 5 is 10. This is going to be 10. If you take a calculator and multiply it, you should get 10. Another way you could see it, what's 10% of 50? 10% of 50 is 5. So 20% is twice the value of 5. 20% must be 10. That's another way you could see it. So try this one. What is 35%?
of 80. So you can either multiply 0.35 times 80, or you can do it mentally. So let's find out what 10% of 80 is. 10% of 80 is simply 8. Just move the decimal one unit to the left. So 5% is half of that. 5% is half of 8, so therefore 5% is 4. 30% of 80 is 3 times the value of...
10% because 10 times 3 is 30. So then 30% is going to be 8 times 3 which is 24. So if we want to find 35% simply add these two numbers. So 24 plus 4 is 28. And so 35% of 80 is 28. So now let's move on into variables and exponents. x to the third multiplied by x to the fifth power. What is that equal to?
When you multiply common variables, you should add the exponents. 3 plus 5 is 8, so this is x to the 8. When you divide, you need to subtract. So 7 minus 3 is 4, this is x to the 4. And when you raise one exponent to another exponent, you need to multiply. 4 times 5 is 20. And anything raised to the 0 power is always equal to 1. So if you have 4xy squared all raised to the 0 power, this entire expression is equal to 1. Now you need to know what to do when you have a negative exponent.
For example, x to the negative 4 is the same as 1 over x to the 4. And 3 over x to the negative 2 is the same as 3x to the positive 2. If you move x from bottom to the top, or the top to the bottom, the exponent simply switches sign. So let's say if you have 5x squared, y to the third, raised to the second power. What is this equal to?
We need to multiply each exponent by 2. So 1 times 2 is 2, 2 times 2 is 4, 3 times 2 is 6. Anytime you raise one exponent to another, you need to multiply. 5 squared, that's 5 times 5, you're multiplying two 5's together. So it's 25x to the 4 or y to the 6. Let's try another example.
Let's say if we have 2 squared x to the 4th. y to the fifth, z to the seventh, raised to the third power. So we're going to multiply every exponent by 3. So 2 times 3 is 6, 4 times 3 is 12, 5 times 3 is 15, 7 times 3 is 21. Now what's 2 to the 6th? 2 to the 6th, you can break that down into 2 to the 3rd times 2 to the 3rd, because 3 plus 3 is 6. 2 to the 3rd, you're multiplying 3 2's together.
It's 2 times 2 times 2, which is 8. So this is 8 times 8, which is 64. So 2 to the 6th power is 64. So the final answer is 64, x to the 12th power, y to the 15th power, and z raised to the 21st power. Now let's say if we have 35. x to the third power, y to the negative fourth. z to the sixth power divided by 14 x to the 8 y to the negative 2 and z to the negative 7 all raised to the second power. So how would you simplify this expression? So we can either divide 35 and 14 and everything else, or we can raise everything to the second power.
Now, it's better if we cancel, if we simplify before we raise it to the second power, because 35 squared, that's a big number. So let's focus on the inside part first. 35 is 7 times 5, and 14 is 7 times 2. Now what's x to the third divided by x to the 8? So it's 3 minus 8, it's x to the negative 5. What's y to the negative 4 divided by y to negative 2 if you subtract it it's negative 4?
Minus negative 2 which is the same as negative 4 plus 2 so that's negative 2 So we're gonna have y to the negative 2 on top Initially after you subtract it you should place it on top. And then for the other one it's 6 minus negative 7. Since we have two negatives right next to each other this is the same as 6 plus 7. And so what we have is. z to the 13th and all of it is squared so we can cancel a 7 and any negative exponent that we have we want to make it positive so we have a 5 on top we're going to keep the z to the 13 on top the 2 is in the bottom and x and y we're going to move we're going to move those variables to the bottom so that the negative 5 will become positive 5 and the same will happen to the negative 2 it's going to change to positive 2 And this is still raised to the second power. So now we can square everything.
So let's multiply every exponent by 2. So we have 5 squared. 13 times 2 is 26. And then 2 squared. X to the 10th power, y to the 4th.
5 squared is 25. And 2 squared is 4. So this is the final answer. By the way, when you divide an x cubed by x to the 8th, or let's say x to the 5th power. So, as we did before, you can subtract them, and you're going to get negative 2, initially, which is on top. And then you can move it to the bottom and get x squared. That's one way of seeing it.
Another way you can see this problem, you can see it this way. On the bottom, you have 5 x's. And on top you have 3-axis.
So, 3 are going to cancel with each other, so you have 2 left over on the bottom. So, it's simply 1 over x squared. That's another way that can quickly help you to see what the appropriate answer is going to be. Now, let's say if you have this expression, 5x squared y to the negative 3 over 9x to the negative 4 y to the fifth power.
So, how can we simplify this expression? So, the first thing I'm going to do is I'm going to move the negative exponents into a different, into the other side to make them positive. So, I'm going to keep the x squared on top, but the x to the negative...
I'm going to move it to the top and the y to the negative 3 I'm going to move it to the bottom so this is another way you can solve and so now I can multiply these two and you can clearly see that it's going to be x to the 6th power and Multiplying y to the third by y to the fifth I can add 3 and 5 and that's going to give me 8 So this is another way you can do it, too Now let's briefly go over exponential fractions. So here's one, x to the 3 halves. You need to be able to convert this into a radical. This is the same as the square root of x cubed, where the index number is a 2. So let's say if you have x to the 5 thirds. How can we convert that into a radical expression?
And try this one, x raised to the 7 over 4 power. So this is going to be the cube root of x to the 5th, and this is the 4th root of x to the 7th. Now you also need to be able to simplify radical expressions.
What is the square root of x to the 4? So the index number is a 2 if you don't see it there. So how many times does 2 go into 4?
4 divided by 2 is 2, so this is simply x squared. Likewise, the square root of y to the 6 is 6 divided by 2, it's y to the third. And the square root of z to the 10th power 10 divided by 2 is 5, so it's z to the 5th power.
Now what about the cube root of x to the 12th power? And the cube root of x to the 17th? And let's say the 4th root of x to the 19th. So starting with the cube root of x to the 12th, we can just divide it.
12 divided by 3, that's 4. So it's x to the 4th. But in this case, 3 doesn't go into 17 evenly. How many times does 3 go into 17?
3 goes into 17 5 times, because 5 times 3 is 15. So now what's remaining? 17 minus 15 is 2, so we have 2x's remaining. And so this is the answer.
It's x to the 5th times the cube root of x squared. Now for this one, how many times does 4 go into 19? 4 goes into 19 4 times.
because 4 times 4 is 16, and 19 minus 16 will give us a remainder of 3. So we're going to have the fourth root of x to the third. So this is going to be the remainder. Now, what about this problem? What is the cube root of x to the 6th, y to the 9th, z to the 14th? So, 6 divided by 3 is 2, 9 divided by 3 is 3, and 3 goes into 14 four times.
So, we're going to have z to the 4 on the outside times the cube root of 14. Well, we said 3 times 4 is 12. 3 goes into 14 four times, so 4 times 3 is 12, and 14 minus 12 is 2. So, there's two z's remaining. So this is the answer. x squared y cubed z to the 4 times the cube root of z squared. Try this example. The square root of 25x to the 8, y to the 9, z to the 10. So we know the square root of 25 is simply 5. And we need to divide every number by 2. 8 divided by 2 is 4. 10 divided by 2 is 5. Now, 2 goes into 9 four times.
2 times 4 is 8, so 9 minus 8 is 1. We have 1 remaining. So, this is just going to be the square root of y, or y to the first power. So, this is the answer for that particular example.
Now, what about this one? The cube root of 8x to the 12th, y to the 17th, and z to the 21st power. Feel free to pause the video and try this example.
So the cube root of 8 is 2. 12 divided by 3, that's 4. 3 goes into 17 5 times. And 17 minus 15 is 2, so we have 2 remaining. And let's keep the 3 on the outside.
21 divided by 3 is 7, and so that's how you can simplify that example. Now sometimes you might have to simplify an expression that's within a radical and that's associated with division. The cube root of 27 is 3. And the cube root of 125. What times what times what is 125? It's 5. 5 times 5 times 5, 3 times, is 125. So the cube root of 125, that's 5. Now notice that each of these numbers are divisible by 3. So, as best we apply the exponent. Let's divide everything by 3. x to the 9, or 9 divided by 3 is 3. 15 divided by 3, that's 5. And 12 divided by 3 is 4. And 30 divided by 3 is 10. So now we can reduce it.
So, x to the third divided by x to the fourth. We need to subtract it. If we subtract it backwards, 4 minus 3 is 1, so that means that we're going to have one more x on the bottom.
If you cancel three x's on top and three on the bottom, you have one left over on the bottom. And 10 minus 5 is 5 on the bottom, if you subtract it backwards. So this is the answer.
It's 3 over 5x y to the fifth power. Now let's go over equations. So let's say if we have the equation x plus 8 is equal to 20. How can we solve for x? X is simply a number. It's a number that you currently don't know the value of.
So what number plus 8 is 20? That's what we're looking for. It turns out that 12 plus 8 is 20, so X is 12. To get 12, subtract 8 from both sides. Whenever you're solving for X, your goal is to get X on one side of the equation by itself. If you can do that, then you can find the value of X.
So here we can see that the 8s will cancel. negative 8 is 0 and on the right side 20 minus 8 is 12 and that's the value of x that's how you solve for it. So let's try another example 3x minus 5 is equal to 9. So how can we solve for x? Notice that there's a negative sign in front of the 5. The opposite of subtraction is addition so let's add 5 to both sides. 9 plus 5 is 14. Now, how can we separate the 3 from the x?
Notice that the 3 is multiplied by the x. The opposite of multiplication is division, so we need to divide both sides by 3. And so now we have the value of x. x is equal to 14 divided by 3. Now, what is 14 over 3? What is that as a decimal?
So, we can separate 14 into 12 and 2. 12 divided by 3 is 4. And 1 third is.33. 2 thirds is.66 repeated. Or if you round it,.67.
So this is about 4.67 as a decimal. Now, let's say if we have the equation 5 times x plus 2 plus 7 is equal to 37. How can we solve for x? Feel free to try this problem and pause the video, and then unpause it when you're ready to see the solution. So, let's go ahead and subtract both sides by 7, since the opposite of addition is subtraction. 37 minus 7 is equal to 30. So notice that the 5 is multiplied to the x plus 2. If we divide both sides by 5...
We don't need to have the parentheses anymore because there's no number in front of it. So all we have now is x plus 2 on the left side. And 30 divided by 5 is 6. So to separate the x from the 2, they're separated by addition.
So let's subtract both sides by 2. So therefore, x is equal to 6 minus 2, which is 4. And so that's the answer for this particular problem. So now let's try an example where we have variables on both sides with parentheses. So feel free to solve this equation. In this particular example, I'm going to distribute first before combining like terms. So let's distribute the 3 to the x and the negative 3. So 3 times x is simply 3x.
And 3 times negative 3, that's negative 9. we can rewrite the 4 and now let's go ahead and distribute the negative 2 to x and 1 so it's going to be negative 2x and negative 2 so now let's combine like terms on the left side we can combine negative 9 and 4 Negative 9 plus 4 is negative 5. And 22 minus 2 is equal to 20. So we have 20 minus 2x. So let's add 2x to both sides. And simultaneously, let's add 5 to both sides. So we can cancel the 2x's on the right side and the 5's on the left. So 3x plus 2x is 5x, and 20 plus 5 is 25. So now the last thing we need to do is divide both sides by 5. So the 5's will cancel on the left.
25 divided by 5 is 5. So this is the value of x. That's how you can solve this particular problem. Now let's say if we have a fraction.
within the equation. How can we solve for x in this particular instance? So before we deal with the fraction, we could subtract both sides by 7, and that should make it a lot easier.
So what we now have is 2 thirds x is equal to 11 minus 7, which is 4. Now notice that we have a 3 in the bottom. The best thing to do in this particular case is to multiply both sides by 3, because on the left side, the 3's will cancel. So on the left side, all we have is 2x, and on the right side, 4 times 3, which is 12. So now we could just divide both sides by 2, and so therefore, x is equal to 12 divided by 2, which is positive 6. Let's try another example like that.
So let's say if we have 3 over 7x is equal to, let's say, 21. So instead of multiplying just by 7, another thing that you could do is you can multiply both sides by the reciprocal of the fraction. That's by 7 over 3. And so, at the same time, the 3's will cancel, and the 7's will cancel as well. So on the left side, you just have x.
And on the right side, you have 21 times 7 thirds. You can put this over 1. 21 is basically 7 times 3. And then we're going to multiply that by 7 thirds. So the 3's cancel.
All we have left over is just 7 times 7. So therefore, 21 times 7 thirds is 49. Try this example. 5 over 2x minus 1 fourth is equal to 2 thirds x plus 1. So here we have a lot of fractions. And in this particular problem, I'm going to multiply both sides of the equation by the least common denominator.
those three fractions. So what number goes into 2, 3, and 4? The answer is 12. 4 can go into 12, and so can 3 and 2. So I'm going to multiply everything by 12. So 12 12 times 5 over 2x.
What is that equal to? Well it's easier if you divide first before you multiply. 12 divided by 2 is 6 and 6 times 5 is 30 and there's an x in front of it so the first term is 30x.
And 12 times 1 fourth. 12 divided by 4 is 3, so this is simply negative 3. Next, we have 12 times 2 thirds. 12 divided by 3 is 4. 4 times 2 is 8. And we're going to carry over the x. And 12 times 1 is 12. So now, we don't have any more fractions.
So now, it's a lot easier to solve it. So, let's add 3 to both sides. And simultaneously, let's subtract 8x from both sides. sides. So these will cancel.
30x minus 8x is 22x and 12 plus 3 is 15. So now let's divide both sides by 22 to isolate x. So now we have our final answer. x is equal to 15 divided by 22. Now sometimes, you might have an equation that contains decimals. Let's say if we have 0.32x minus 0.15, and that's equal to 0.23x plus 0.21.
How can we solve for x under these circumstances? So what you want to do, you want to multiply both sides by 100 because we have two decimals after, I mean two digits after the decimal place. place. So let's multiply everything by 100. Whenever you multiply a number by 100, simply move the decimal two units to the right.
So 0.32x times 100 is 32x. 0.15 times 100 is 15. And so we're going to have 23x and 21 on the other side. And so now it's much easier to solve. So let's subtract 23x from both sides, and let's add 15. So these will cancel. 32x minus 23x, that's 9x.
And 21 plus 15 is 36. So now let's divide both sides by 9. So 36 divided by 9 is 4. So x is 4. Now, how would you solve for x if you have this equation, x squared is equal to 25? What would you do? So, in order to solve for x, you want the exponent to be 1. Right now, it's a 2. So, what you want to do is you want to take the square root of both sides.
When you take the square root, what you're really doing is you're dividing the exponent by 2. So, 2 divided by 2 is 1, so it becomes x to the 1. The square root of 25 is basically the value of a number when you multiply it by itself that will give you 25. 5 times 5 is 25, thus the square root of 25 is 5. Now, it can be positive 5 or it can be negative 5, because positive 5 times positive 5 is positive 25, and negative 5 times negative 5 is also positive 25, so that's why we have two answers in this situation. Now, what about this one? Let's say if we have x to the third is equal to 27. What can we do? So in this particular case, we need to take the cube root of both sides.
So 3 over 3 is 1, so we just have x, or x to the first power. So what number times itself 3 times is equal to 27? 3 times 3 times 3 is 27, so the cube root... of 27 is only positive 3. It's not negative 3 because negative 3 times negative 3 times negative 3 is negative 27, not positive 27. So if you have an odd index, you're only going to get one answer and not two answers.
Now let's say if we have this equation x to the three halves is equal to 8. How can we solve for x in this problem? So, keep in mind, whenever you solve for x, you want the exponent to be 1. Right now, it's 3 over 2. So, in order to change it to 1, we need to raise both sides to the reciprocal of 3 over 2. The reciprocal is 2 thirds. Whatever you do to the left side, you have to do it to the right side, so that the value of both sides of the equation remain the same. So, whenever you raise one exponent to another exponent, you're multiplying.
So we're multiplying 3 over 2 times 2 thirds. The 2's will cancel, and the 3's will cancel. So on the left side, what we now have is x to the first power.
But on the right side, we have 8 to the 2 thirds. How can we simplify that? What is 8 to the 2 thirds? So we could change it to a radical expression.
This is the same as the cube root of 8 squared, which I'm going to put the 2 on the outside. You want to take the cube root first and get a small number and then square it, rather than doing it the other way. So what is the cube root of 8? So what number times itself 3 times is 8?
2 times 2 times 2 is 8. Therefore, the cube root of 8 is 2. And 2 squared is 4. So this is the value of x. x is equal to 4. And we can check it. If we plug in 4... into this equation 4 to the 3 halves this is the same as the square root of 4 to the third the square root of 4 is 2 and 2 to the third is 8 so the answer works so let's try another one try this one x to the 3 fourths is equal to 8 so go ahead and practice problem and see if you can get the answer. So just like before, we're going to raise both sides to the 4 3rds.
So we know 3 4ths times 4 3rds is going to give us 1. So we just have x on the left side. So 8 to the 4 3rds is the same as the cube root of 8 raised to the 4th power. We know that the cube root of 8 is 2, and 2 to the 4th, that's 2 times 2 times 2 times 2, 4 times, and so that's equal to 16. So x... is equal to 16 Sometimes you might get a similar problem, but with more steps. So let's say if we have this problem, x plus 2 raised to the 4 fifths plus 5, and let's say it's equal to 21. Go ahead and solve for x.
Now the first thing I would do is subtract both sides by 5. So 21 minus 5, that's equal to 16. Next, we need to get rid of the 4 fifths, so I'm going to raise both sides to the reciprocal, which is 5 fourths. So, once the exponents cancel on the left, there's no need for the parentheses, so we just have x plus 2. So, now what's 16 raised to the 5 over 4? So, this is the 4th root of 16 raised to the 5th power. The 4th root of 16, we know it's 2. And 2 to the 5th power. 2 to the 4th is 16, so 2 to the 5th, you just multiply 16 by 2, and you get 32. So what we now have is x plus 2 is equal to 32. And so we just got to subtract both sides by 2. And so now we have the answer x is equal to 30. Let's try this one.
x plus 3. to the second power is equal to 5. Now this one might seem like a simple problem. If we take the square root of both sides, we can get rid of the square on the left side. And so it's x plus 3. But we can't simplify root 5, so we have to leave it this way. And whenever you take the square root of both sides, you're going to get two answers.
So this is going to be plus or minus root 5. Now all we need to do at this point is subtract 3 from both sides. Now we can't combine the negative 3 and the root 5, so For this particular problem, you have to leave the answer like this. You can convert it to a decimal, but this is what it is.
There's two answers. It's negative 3 plus root 5, and the other answer is negative 3 minus root 5. Now let's try some other problems that involves radicals. So let's say if you have the square root of x is equal to 3. What do you need to do to solve for x? So this time, before... When we had x squared was equal to 25, to solve for x, we need to take the square root of both sides.
Now that we have the square root, we need to take the square of both sides. We've got to do the opposite. So, once you square a square root, the 2's will cancel. And so, on the left side, all you have is x.
3 squared is 3 times 3, which is equal to 9. And so, that's it for that problem. So let's try another similar problem but with more steps. So let's say if we have the square root of 2x minus 5 and it's equal to 4. So just like before we're going to square both sides. So we don't have the radical on the left side anymore so we just have 2x minus 5. On the right side we have 4 squared which is 4 times 4 and that's 16. So now let's add 5 to both sides. So, 16 plus 5 is 21. Now, the last thing we need to do is divide both sides by 2. So, now we have our answer.
It's 21 over 2. And if you want to turn 21 over 2 into a decimal, you can rewrite it as 20 over 2 plus 1 over 2, since 20 plus 1 is 21. 2 goes into 20 10 times, and a half is basically 0.5. So, 21 over 2 is 10.5 as a decimal. Try this one. The cube root of 3x plus 1 is equal to negative 2. So go ahead and solve it.
So for this problem, we need to raise both sides to the third power in order to solve for x. So the radical is gone. So we just have 3x plus 1 on the left side.
So negative 2 to the third power. That's negative 2 times negative 2 times another negative 2, which is negative 8. So now we can just solve for x, as we've been doing before. Negative 8 minus 1, that's negative 9. And if we divide both sides by 3, we can see that x is equal to negative 3. Here's one. Try this one. The square root of x plus 1 plus 3 is equal to the square root of 3x plus 2. So, you don't want to take the square root of both sides right now.
You want to try to simplify if you have two radicals. And you want one radical to remain on the left and the other to remain on the right. So, what we're going to do is we're going to subtract both sides by 3. So what we now have is the square root of x plus 1 that's equal to root 3x minus 1. So now, we have no choice, but we need to square both sides.
If we square it on the left, the 2 on the radical will cancel. So on the left, we'll just get x plus 1. On the right side, we need to FOIL. We need to multiply root 3 minus 1 times another root 3 minus 1, because we have two of them, based on the exponent.
So we're going to multiply root 3x times root 3x, which will become 3x. For example, the square root of 4 times the square root of 4 is the square root of 16, which is 4. So when you multiply two radicals that have the same inside, the radicals cancel. and this will be multiplied by negative 1 so this is negative root 3x and root 3x times another negative 1 and finally negative 1 times negative 1 that's positive 1 so let's make some space so we can combine like terms on the right side That's these two.
Now keep in mind there's a negative 1 in front of it. Negative 1 minus 1 is 2. So we have 3x minus 2 root 3x plus 1. Now we still have another radical that we need to get rid of. That's this one.
And so in order to do that, we need to get it by itself. So let's subtract 1 from both sides first. So these will cancel. and let's subtract 3x so X minus 3x this is the same as 1x 1 minus 3 is negative 2 and so we have this at this moment let's divide both sides by negative 2 so X is equal to root 3x So now, let's go ahead and square both sides.
So on the left side, we have x squared. On the right side, we simply have 3x. So what can we do in this particular problem?
So, if we divide both sides by x, x squared divided by x is x. We'll get the answer x equals 3. But notice that if we plug in 0, 0 squared equals 3 times 0. So 0 is also an answer. So there's two answers for this problem. x can be 0 or it can be 3. So now let's take a break from solving equations and let's move into functions.
Let's say if f of x is equal to x squared plus 4x minus 5. Let's say we have this function. What is the value of f of 3? Keep in mind y is equal to f of x, so we're really looking for the y value in this problem.
We have the x value, x is 3. So let's substitute x with 3. so 3 squared plus 4 times 3 minus 5 everywhere we see an x we're going to replace it with a 3 3 squared is 9 4 times 3 is 12 12 minus 5 that's 7 and 9 plus 7 to 16 so we have the coordinate 3 comma 16 from x is 3 y is 16 But what if you have a question that looks like this? Let's say f of x is 7x plus 3. So if f of x is equal to 24, what is the value of x? Now this is a typical SAT question that you'll see.
Sometimes you'll have functions with two variables. You'll have like x and y within the function. And so you need to understand functions well.
By the way, if you're going to take the SAT soon, check out my video on YouTube. It's SAT Math Prep Course Reviews, something like that, Algebra and Geometry. If you do a YouTube search, you should find it. But it's going to have a lot of questions like this on algebra, geometry, and word problems. A lot of stuff that you'll see on the SAT math section, so feel free to check that out when you get a chance.
But let's go back to this problem. So, we said that f of x is equal to y. So therefore, y is 24. So we're looking for the value of x. So let's replace... f of x with 24, and let's solve for x.
So, if we subtract both sides by 3, we'll get 21 is equal to 7x. So, all we need to do at this point is divide by 7. 21 divided by 7 is 3, so that's the value of x. So, we have the coordinate 3, 24. Now, let's say if we have two functions. f of x is equal to x squared plus 3 and g of x is 2x minus 5. How can we find f of g of 3? So this is a composite function, a function within another function.
So g is inside of f. So what we need to do is evaluate g of 3 first. So let's replace x with 3 in a g equation. So 2 times 3 minus 5. 2 times 3 is 6. 6 minus 5 is 1. So g of 3 is 1. So we can replace g of 3 with 1. So we're looking for f of 1. So now let's plug in 1 into x in the f equation. So f of 1 is 1 squared plus 3. 1 squared is 1. 1 plus 3 is 4. So therefore f of g of 3...
has a value of 4. So now it's your turn. Try this problem. Evaluate g of f of 2. So first, let's calculate f of 2. So let's substitute 2 for x in the f of x equation. 2 squared plus 3. 2 squared is 4. 4 plus 3 is 7. So f of 2 is 7. So now we're looking for g of 7. So let's substitute x with 7 in a g equation. So g of 7 is going to be 2 times 7 minus 5. And 2 times 7 is 14. 14 minus 5 is 9. So therefore, g of f of 2 is equal to 9. So sometimes, you may not have a number.
You might just have x. Let's say if you want to find f of g of x. So our answer should be in terms of x.
g of x is equal to 2x minus 5. So we can replace g of x with 2x minus 5. So now we've got to take 2x minus 5 and substitute for x in the f of x equation. So f of g of x is equal to 2x minus 5 squared plus 3. So we replace x with 2x minus 5. And so this is the answer. Now granted, we could FOIL 2x-5 if we want to, because sometimes the answer may not be in this form. If it's not, you may need to take the next step further.
So let's go ahead and FOIL it. So 2x times 2x, that's 4x squared, and 2x times negative 5, that's negative 10x. Negative 5 times 2x is also negative 10x. And negative 5 times negative 5, that's positive 25. Plus we have the 3 on the outside.
So now let's combine like terms. Negative 10x and negative 10x is negative 20x. And 25 plus 3, that's 28. So this is the answer.
4x squared minus 20x plus 28. So now let's get back into solving equations. Now, let's go back to that last problem that we have, where it was x squared is equal to 3x, and we got 0 and 3. Another way you can solve it is you can move the 3x to the other side, and you can factor. So if you have two terms, typically you want to look for the GCF.
That's usually the first thing you want to look for if you need to factor an expression. The GCF is the greatest common factor. The greatest common factor that we can take from both terms is an x. So, if we remove x, what goes inside?
To find out what goes inside, divide. x squared divided by the GCF, x, is equal to x. And negative 3x divided by x is simply negative 3. So now, once you have the expression in factored form, you need to set each factor equal to 0. So the first one is just 0. But the second one, if we add 3 to both sides, we could see that x is equal to 3. So let's try another similar problem. Let's say if we have 3x squared plus 18x is equal to 0. So let's factor the GCF.
What is the common term that can go into 3x squared and 18x? 3 can go into itself, and it can go into 18. And we can at least take out 1x from both terms. So the GCF is 3x. 3x squared divided by 3x is simply x.
And 18x divided by 3x is 6. So we can set 3x equal to 0 and x plus 6 equal to 0. So if we divide both sides by 3, we can see that x is 0 on the left side. And here, we just need to subtract both sides by 6. So x is going to be negative 6. So anytime you see an x on the outside, it's going to be 0. And for the x plus 6, if you see it's plus 6, reverse the sign. x is going to equal negative 6. Because when you move the 6 from the left to the right, it's going to change sign. Now what if you have an equation that looks like this? How can you factor it in order to solve for x?
You need to use the difference of squares method. So what you need to do is take the square root of x squared, which is just x, and the square root of 25, which is 5. One is going to be positive and the other is going to be negative. And so x is going to equal negative 5 and positive 5. So let's try a few more examples using that technique.
So x squared minus 64. The square root. of x squared is x, and the square root of 64 is 8. We're going to have a positive and a negative sign, so therefore, x will equal negative 8 and positive 8. So, let's say if we have 4x squared minus 49 is equal to 0. What's the square root of 4x squared? The square root of 4x squared is 2x.
And the square root of 49 is 7. And then let's add the two signs. So let's solve for x. So 2x plus 7 is equal to 0. And 2x minus 7 is equal to 0. So if we subtract both sides by 7, 2x is equal to negative 7. Here we need to add 7 to both sides, so 2x is equal to positive 7. And then divide by 2. So x is negative 7 over 2 and positive 7 over 2. How about this equation? 3x squared minus 48 is equal to 0. Go ahead and solve for x. Right now, we can't use the difference of squares method, because if we take the square root of 3 or 48, we're going to get a decimal.
However, we could factor out the GCF, the greatest common factor. So if we take out a 3, 3x squared divided by 3 is x squared, and negative 48 divided by 3 is negative 16. So now we can use the difference of squares method. So the square root of x squared is x, and the square root of 16 is 4, and one is plus, the other is minus, and so we can see that x is equal to plus and minus 4. Now what about this problem? x squared plus 8x plus 15. So here we have a trinomial where the leading coefficient is 1. How can we factor this expression in order to solve for x?
So look at the constant term the 15. You want to find two numbers that multiply to 15 but that add to 8. So two factors of 15 are 1 and 15 but These two, they add up to 16. 3 times 5 is also... 15, but 3 plus 5 adds to 8. So you can simply write it like this. It's going to be x plus 3 times x plus 5. And so now we can solve for x.
So x is going to equal negative 3 and negative 5 if you set each factor equal to 0. Now how about this one? x squared plus... 4x minus 21. What two numbers multiply to negative 21 but add to the middle term 4? So let's make a list. Let's start with 1. Negative 21 divided by 1 is negative 21. 2 doesn't go into 21, but 3 does.
Negative 21 divided by 3, that's negative 7. Now, these two numbers, if we add them, we're going to get negative 4. 3 plus negative 7 is negative 4. But if we change the sign, negative 3 plus positive 7, that's positive 4. So the factor is going to be x minus 3 times x plus 7. So now let's set each factor equal to 0. And let's solve for x. So here we can add 3 to both sides. And on the other one, we're going to subtract 7 from both sides. So x is equal to positive 3. And x is equal to negative 7. Try this one.
x squared minus 9x plus 18. So what two numbers multiply to 18 but add to negative 9? So let's go ahead and make a list. So we have 1 and 18. Now they both have to be negative so that they can add up to negative 9. 2 and 9 can go into 18, but these two they add up to negative 11. Negative 3 and negative 6 multiply to positive 18, and those two they add up to negative 9. So this is the pair that we need. So the factor is going to be x minus 3 times x minus 6. So therefore x will be equal to 3. and plus 6. Let's try this one. 2x squared minus x minus 1. So this time we have a trinomial where the leading coefficient is not 1, it's 2. So the way in which to factor is going to be a little different from the last method.
We need to multiply the first coefficient by the constant term. So 2 times negative 1 is negative 2. And then look for two numbers that multiply to negative 2, but that add to the middle term, negative 1. So this is going to be negative 2 and 1. Negative 2 times positive 1 is negative 2, but negative 2 plus 1 is negative 1. Now what we need to do next, we need to replace the middle term with negative 2x and plus 1x. So, negative 2x plus 1x. Those two, they add up to negative 1x. So, the value of the equation remains the same.
Now, what we need to do at this point, we need to factor by grouping. So, in the first two terms, we need to remove or factor out the GCF. So, let's take out a 2x.
2x squared divided by 2x, that's x. Negative 2x divided by 2x is negative 1. In the last two terms, there's nothing we can take out. If you have nothing to take out, just take out a 1. And so what's inside is going to be the same as what we have here, and that is x minus 1. Now, since you see two of these, if they're the same, that means you're on the right track. In the next step, factor out x minus 1, and you can do that by writing it once. And then what goes inside the next parentheses is what's on the outside.
the 2x and the plus 1. So it's just going to be 2x plus 1. And so that's how you can factor a trinomial where the leading coefficient is anything but 1. So now let's solve for x. Let's set each factor equal to 0. So for the first one, all we need to do is add 1 to both sides. And we can see that x is equal to positive 1. Now for the second...
factor. Let's subtract both sides by 1 and then let's divide by 2. So x is equal to negative 1 half and x is equal to 1. So those are the two answers for this particular problem. Now, if you ever get to a point where...
You find it difficult to factor, you can always solve for x using the quadratic formula. So let's go back to this equation. The quadratic formula is x is equal to negative b plus or minus the square root of b squared minus 4ac. all divided by 2a. So 2 is a, negative 1 is b, and negative 1 is also c in this particular equation.
It's in the form ax squared plus bx plus c. So, let's plug it in into the formula. So, b is negative 1, so it's negative negative 1 plus or minus the square root of b squared.
Negative 1 squared is just positive 1 minus 4. a is 2 and c is negative 1. And divided by 2a, or 2 times 2. So, negative times negative 1 is just 1. On the inside... Negative 4 times 2, that's negative 8, times negative 1, that's positive 8, plus the 1 over here, so that's 9 on the inside of the radical, divided by 4. Now the square root of 9 is 3. So this is what we have. 1 plus or minus 3 over 4. So once you get to this part, you can separate it into two values.
The first one is 1 plus 3 over 4, and the second one is 1 minus 3 over 4. 1 plus 3 is 4. 4 divided by 4 is 1. 1 minus 3 is negative 2, and negative 2 over 4 can be reduced to negative 1 over 2, which is the two answers that we got when we factor it. So both techniques can work. You can either factor the expression or use the quadratic equation to solve it. So let's try another example. Let's say if we have the equation 6x squared minus 9x.
Actually, instead of 9x, let's make it negative 7x. minus 3. So feel free to pause the video, solve by factoring, and by using the quadratic equation. Make sure you get both answers. Make sure both answers are the same.
So let's factor it first. Let's multiply the leading coefficient by the constant term. So 6 times negative 3 is negative 18. And what two numbers multiply to negative 18 but add to negative 7?
So, we have 1 and 18, 2 and 9, and 3 and negative 6. But notice that 2 plus negative 9 adds up to negative 7, so we're going to use that. So let's replace the middle term, the negative 7, with 2 and negative 9. So it's going to be plus 2x and negative 9x. If you wrote negative 9x first and then plus 2x second, you're going to get the same answer. It doesn't matter the order in which you write it. It works out to be the same.
So let's factor out the GCF in the first two terms. We can take out a 2x from 6x squared and 2x. 6x squared divided by 2x is 3x, and 2x divided by 2x is simply 1. in the last two terms, let's take out a negative 3. Negative 9x divided by negative 3 is 3x, and negative 3 divided by itself, negative 3, that's plus 1. So notice that these two are the same, which means we're on the right track.
So let's write it in the next line, 3x plus 1, and then the stuff on the outside, that's the 2x and the negative 3, we're going to put it inside the second parentheses. And so now we have our answer in factored form. So now let's set each factor equal to 0. So 3x plus 1 is equal to 0, and 2x minus 3 is equal to 0. So let's subtract 1 from both sides, and let's add 3. So 3x is equal to negative 1, and on this side, 2x is equal to 3. So, x is equal to negative 1 third, and for the other one, x is equal to plus 3 over 2. So now let's see if we can get the same answer using the quadratic equation.
So x is equal to negative b plus or minus square root b squared minus 4ac divided by 2a. And the original problem was 6x squared. minus 7x minus 3 So a is 6 B is negative 7 and C is negative 3 so negative negative 7 plus or minus the square root B squared 7 squared is 49 or negative 7 times negative 7 is also 49 minus 4 a is equal to 6 and C is negative 3 divided by 2a or 2 times 6 so negative times negative 7 is positive 7 and 4 times 3 is 12 12 times 6 is 72 so we have 49 plus 72 the reason why it's positive 72 is because we have two negative numbers which become positive 2 times 6 is 12 And 49 plus 72. 40 plus 70 is 110. 9 plus 2 is 11. 110 plus 11, that's 121. So now let's make some space.
So the square root of 121 is 11. So we have two answers, 7 plus 11 divided by 12, and 7 minus 11 divided by 12. So what's 7 plus 11? 7 plus 11 is 18 so we have 18 over 12 and 7 minus 11 is negative 4 over 12 18 over 12 we can reduce it if we divide top and bottom by 6 18 divided by 6 is 3, 12 divided by 6 is 2, so we have 3 over 2, which is what we had in the last example. Negative 4 over 12, let's divide both numbers by negative 4, so it's going to be negative 1 over 3. So, as you can see, we have the same answers as we did when we factored. Now, let's say if we have an exponential equation.
9 raised to the x plus 2 is equal to 27 raised to the 2x minus 3. How would you solve for x if you have a problem that looks like this? Now, notice that 9 and 27 share a common base. 3 squared is equal to 9. 3 times 3 is 9. 3 to the third power is 27. So we can replace 9 with 3 squared, and we can replace 27 with 3 cubed.
Now we still have... the same exponents and keep in mind whenever you raise one exponent to another exponent you need to multiply the two exponents so we need to multiply 2 times X plus 2 and if we distribute the 2 it's going to be 2x plus 4 And for the other one, 3 times 2x minus 3, that's 6x minus 9. So now that we have the same basis, therefore the exponents must be the same in value. So 2x plus 4 is equal to 6x minus 9. So now we can solve for x. So, let's add 9 to both sides, and let's subtract 2x from both sides.
so 6x minus 2x is 4x 4 plus 9 is 13 so all we got to do is divide by 4 we can see that x is equal to 30 over 4 and if you want the decimal value for that 13 over 4 can be broken into 12 over 4 and 1 over 4 so 4 goes into 13 three times with one remaining so as a mixed number it's 3 and 1 fourths now granted 1 fourth is 0.25 so as a decimal this is 3.25 Let's try another example like that. So, let's say 8 raised to the 4x plus 1 is equal to 16 to the 2x minus 3. So, pause the video and try this example. Go ahead and solve for x.
Look for a common base. So, how many 2's do we need to multiply to get to 8? Because 2 can go into 8 and 16. We need to multiply 3 2's to get to 8. 2 times 2 times 2, 3 times is 8. And 2 to the 4th power is 16. So let's replace 8 with 2 cubed and 16 with 2 to the 4th. And just like before, the next thing that we need to do is multiply the two exponents. So 2 to the 3rd raised to the 4x plus 1, that's going to be the same as 2 to the 12x plus 3. And 4 times 2x plus 1, that's going to be 8x plus 4. So make sure to distribute the 4 to the 2x and to the 1. So now that we have a common base, that means that the exponents are equal to each other.
12x plus 3 is equal to 8x plus 4. So let's subtract 8x from both sides, and let's subtract 3 from both sides. So 12x minus 8x is 4x, 4 minus 3 is 1, so x is equal to 1 fourth. or 0.25. Now let's say if you can't change both numbers into a common base. So let's say if you get a problem that looks like this.
We can't really change this equation much. So at this point you want to use logs. You can take the log of both sides or the natural log of both sides. So let's go ahead and take the natural log.
So we have ln 5 to the x is equal to ln 8. Now property of natural logs allows you to move the x to the front. So we have x ln 5 is equal to ln 8. So now we can divide both sides by ln 5. So x is equal to ln 8 divided by ln 5. If you divide these two numbers, if you type this in your calculator, ln 8 divided by ln 5, you should get... 1.292. And you can test to see if it works out. What's 5 raised to the 1.292 power?
If you type that in your calculator, it's about 7.9996, which is approximately 8, since this answer is a rounded answer. And so it works. So if you need to solve for x whenever it's in the exponent position, if you can't change the base into a common base, Take the log or the natural log of both sides.
Now sometimes you might see a similar problem, but with an e instead of a number. If you see an e, don't take the log, take the natural log. The reason for that is because the natural log of e is always equal to 1. So let's take the log of both sides, or the natural log of both sides.
So we can take this exponent and move it to the front. 2x plus 1 times ln e is equal to ln 3. And the natural log of e is 1. So we no longer need the parentheses. So 2x plus 1 is equal to ln 3. And so we can subtract both sides by 1. So 2x is ln 3 minus 1. These two are unlike terms, so you can't combine them. You have to write them separately.
So now all we need to do at this point is divide both sides by 2. So x is equal to natural log of 3 minus 1 over 2, or you can write it as 1 half natural log of 3 minus 1. Now let's spend some time reviewing logs and how to evaluate them. What is log base 2 of 8? What is that equal to?
So 2 to the what power is equal to 8? How many 2's do you have to multiply to get to 8? It turns out that you need 3 2's to multiply to get to 8, because 2 to the 3rd power is equal to 8. There's a property of logs. Let's say if you have log base A of B is equal to C. You can change it to its exponential form.
By using this process a raised to the C is equal to B likewise 2 raised to the third power is equal to 8. Now what about this one? Log base 3 of 9. What is that equal to? So how many 3's do you have to multiply to get to 9? You need 2. Because 3 squared is equal to 9. So try these.
Log base 3 of 27. log base 2 of 32 and log base 5 of 25 so how many threes do you have to multiply to get to 27 3 times 3 times 3 is 27 so you need to multiply three 3s because 3 to the third power is 27 now how many twos do you need to get to 32 you need five twos 2 times 2 is 4 times 2 is 8 times 2 is 16 times 2 is 32 so 2 to the fifth power is 32 and we know that 5 squared is 25 so log base 5 of 25 is 2 Now what about, let's say, log 10. If there's no number here, it's always assumed to be a 10. Log base 10 of 10 is 1. Whenever these two numbers are the same, they will cancel. going to be 1. 10 to the first power is 10. What about log of 100? Log of 100 is 2, because 10 squared is 100. Keep in mind, we still have this invisible 10 here.
Log of 1000 is is 3, you basically have three zeros. Log of a million, which has six zeros, is equal to 6, assuming it's base 10. Log of 0.01 is equal to negative 2, because 10 raised to negative 2, which is 1 over 100, that's also equal to 0.01. So log base 10 of 0.0001, that's equal to negative 4. Log base 10, 1 over 1,000, which has three 0's, that's negative 3. This is the same as log base 10.001, that's equal to negative 3. Now what about this one, log base 4 of 16?
4 times 4 is 16, so it takes two 4s, one multiplied, to get to 16. So this is 2. Now, based on that, what's log base 4, 1 over 16? And what's log base 16 of 4? And what's log base 16, 1 over 4? So, 4 squared is 16, but 4 to the negative 2 is 1 over 16. Because if you have a negative exponent, the 16 is going to go from the top to the bottom. So this is equal to negative 2. Whenever you have a fraction, it's going to be negative.
Log base 16 of 4, this is going to turn into a fraction. So it's going to be 1 half. As you can see, there's a 2 somewhere. It could be negative, positive, or in a fraction. Log base 16, 1 over 4, it's going to be negative 1 half.
So anytime you have a fraction here, this is going to be negative. And whenever this number is larger, you're going to have a fraction on the right side. So let's try some more examples.
What's log base 5 of 125 log base 5 of 1 over 125 and log 125 of 5 and log base 125 1 over 5 try those examples So how many 5's do we need to multiply to get to 125? 5 times 5 is 25, and 25 times 5 is 125. So we need 3 5's. So this is 3. Now because we have a fraction here, it's going to be negative 3. Now notice that the numbers are reversed.
Since this number is larger, we're going to have a fraction on the outside. So this is positive 1 over 3. It's positive because we don't have a fraction here. But here we do have a fraction, so it's going to be negative 1 over 3. So whenever you have a fraction, it's going to be negative.
And whenever this number is larger, if the base is larger, then you can have a fraction on the right side. Now let's say if we have this equation. Log base 3 of x plus 5 is equal to 2. How can we solve for x? Now... Keep in mind, log base 2 of 8 is 3 because we could change it into its x-minus-4 and say 2 to the third power is equal to 8. So, that's what we want to do in this particular problem.
3 raised to the second power is equal to what's inside. In this case, 3 squared is equal to x plus 5. Now, we can solve for x. 3 squared is 9, so 9 is equal to x plus 5. If we subtract 5 from both sides, x is equal to positive 4. Now, what about this one? Log base 4, 5x plus 6 is equal to 3. So use the same technique to solve for x in this particular problem. 4 raised to the third power is equal to what's inside.
So 4 cubed is equal to 5x plus 6. So 4 times 4 is 16 times another 4 is 64. And if we subtract both sides by 6... 5x is going to be equal to 64 minus 6 which is 58. So now all we need to do is divide both sides by 5. And so x is equal to 58 over 5. If you want to turn that into, let's say a decimal, you can write it as 55 over 5 plus 3 over 5. 55 over 5 is 11. 1 5th is basically 0.2, so 3 times 0.2, that's 0.6. So 58 over 5 is 11.6. Now you may find it advantageous to know a few fractions and their decimal value.
1 half is basically 0.5. 1 third is 0.333 repeated. 1 fourth is 0.25. 1 over 5 is 0.2.
1 over 6 is about 0.167, or 0.16 repeating. Only the 6th part is repeating. 1 over 7 is about 0.143. 1 over 8 is exactly 0.125.
5 1 over 9 is 0.1 repeating and 1 over 10 is 0.1 if you know these fractions then it's gonna be fairly easy to like convert other fractions into decimals so for example if I wish to convert 4 fifths into a decimal this is 4 times 1 over 5 so it's 4 times 0.2 which is 0.8 So, let's say if I want to convert 7 over 4 into, let's say, a decimal. I would write it as 4 over 4 plus 3 fourths. So, this is 1. Now, this is 1 times 3, or 1 plus 3 times 1 fourth. and 1 fourth is 0.25.
3 quarters is 75 cents, so this is about 1.75. Now granted, you can always use long division to convert a fraction into a decimal. So, the value on the bottom is going to go on the outside.
So, how many times does 4 go into 7? 4 goes into 7 one time. 4 times 1 is 4. And 7 minus 4 is 3. Now, 4 doesn't go into 3. So we need to add a 0. How many times does 4 go into 30? 4 goes into 30 7 times, because 4 times 7 is 28. So we have 2 remaining, and we're going to add a 0. How many times does 4 go into 20?
4 goes into 30 7 times. into 25 times and so if we put minus 20 the remainder is going to be 0 so this is the answer so 7 over 4 is 1.75 so you can always use long division to convert any fraction into a decimal now let's go back to logs So now let's say if we have log base x of 9 is equal to 2. What is the value of x? So using the same technique, x squared is equal to 9, we can solve for x. So if we square root both sides, x is going to be positive 3. Now typically, when you square root 9, you should get plus or minus 3, but I've never seen a negative base.
So I'm going to go with positive 3. Now there are some other properties of logs and natural logs that you need to be familiar with. For example, the natural log of A plus the natural log of B is equal to the natural log of A times B. The natural log of A minus the natural log of B is equal to the natural log of A divided by B. Now, in front of ln, we have a positive sign, so it goes on top. That's why a is on top.
In front of ln, we have a negative sign, so b goes on the bottom. And of course, you've seen this formula. ln is equal to 2 ln. You can take the exponent and move it to the front, if needed. Try this one.
Log base 2 of x plus log base 2 of x plus 2 is equal to 3. So how would you solve for x if you're given an equation? that looks like this. So what you want to do is combine the two log expressions into a single log expression. Since they're added, we need to multiply x times x plus 2. So if we distribute x to x plus 2, to it's going to be x squared plus 2x and so now we can convert the logarithmic equation into an exponential one. 2 raised to the 3 is equal to everything inside so therefore x squared plus 2x is equal to 2 to the third and 2 to the third 2 times 2 times 2 that's equal to 8 So now that we have this expression, what should we do to solve for x?
What you want to do is move the 8 to the left side, because if you do, you'll have a trinomial with a leading coefficient of 1, and so we can factor. What two numbers multiply to negative 8, but add to positive 2? This is 4 and negative 2. 4 times negative 2 is negative 8. 4 plus negative 2 is positive 2. So to factor it, it's going to be x plus 4 times x minus 2. Therefore, x is equal to negative 4 and... x is equal to positive 2. Now you need to check for extraneous solutions because you can't have a negative number inside a log.
So if we focus on the expression log x plus log x plus 2 If we plug in negative 4, we're going to have a negative value inside the log, which we don't want. So we're going to have to get rid of the solution. That's an extraneous solution. So therefore, the only correct answer is positive 2. So let's check our work.
So the original equation was log base 2 of x plus log base 2 of x plus 2 is equal to 3. So if we plug in 2, we're going to have... log base 2 of 2 plus log base 2 plus 2 is 4 and this should equal 3. So what's log 2 of 2? Well 2 to the first power is 2 so the 2's cancel.
cancel, we just get 1. Log base 2 of 4, how many 2's do we have to multiply to get to 4? It takes 2 2's, so it's just going to be 2, because 2 squared is 4. And as you can see, 1 plus 2 equals 3. And so, the answer is correct where x is equal to 2. Let's try another equation that involves logs. Go ahead and try to solve this one. Log base 3 of x plus 12 minus log base 3. x minus six is equal to one. So just like before, we're going to combine it into a single log expression.
But because we have a minus sign, instead of multiplying x plus 12 and x minus six, we need to divide the two. The one that's positive goes on top, and the one that's negative goes on the bottom. So x plus 12 is going to be on top, and x minus six is going to be in the bottom of the fraction.
So now let's convert... this logarithmic equation into an exponential equation. So 3 raised to the first power is equal to everything inside.
So x plus 12 divided by x minus 6. is equal to 3 to the first power which is the same as 3 or 3 over 1 so let's cross multiply 3 times x minus 6 if we distribute the 3 it's going to be 3x minus 18 and 1 times x plus 12 that's going to remain as x plus 12 So let's subtract both sides by x and let's add 18 to both sides. So 3x minus x is 2x and 12 plus 18 is 30. Dividing both sides by 2, we can see that x is equal to 15. So let's check. our answer. Just make sure that it's not an extraneous solution. So, here's the original expression, and let's go ahead and substitute.
Well, that's supposed to be a 6, not a 16. let's substitute 15 with X so 15 plus 12 that's 27 and 15 minus 6 is 9 So what's log base 3 of 27? How many 3's do you have to multiply to get to 27? 3 times 3 times 3 is 27 or 3 to the third power is 27 so log base 3 of 27 is 3 Now how many twos needs? multiply to get to 9 need to 3 3 squared is 9 so log base 3 of 9 is 2 and as we can see 3 minus 2 is indeed equal to 1 so the answer is 15 for X because the equation is true if you plug in 15 for X By the way, the video that you're currently watching is the two-hour trailer version of a longer five-hour video. Now, in that five-hour video, I'm also going to go over solving inequalities, including absolute value functions, system of equations.
Simplifying Rational Expressions and Radicals 2, Complex Imaginary Numbers, How to Write the Equation of the Line in Standard Form, Slope Intercept Form, and using the Point Slope Equation. and using parallel and perpendicular lines. In addition, I'm going to talk about how to graph linear equations, quadratic equations in vertex form, also cubic functions, radicals, how to find the domain and range of those functions, and a lot of other topics.
So you just got to see them. So I'm going to post a link to that video, and you can take a look at it. Try this one.
Log base 2, 7x minus 3 is equal to log base 2, 5x plus 13. so notice that we only have two log expressions we don't have a number by itself like for example in the other problems we had a log let's say 3x plus 4 equals the number that doesn't have a log. In a situation like this, you want to get just one log expression on one side and keep the number on the other side, and you can change it to its exponential form. But here we have a log on both sides of the equation, and we don't have a standalone number like 7 or 5 in any side.
So in a situation like this, because the bases are the same, what's inside of the log must be equal to each other. So we can set 7x minus 3 equal to 5x plus 3. 13 so let's add 3 to both sides and let's take away 5 from both sides 5x so 7x minus 5x is 2x 13 plus 3 is 16 16 divided by 2 is 8 so that's the answer for this particular problem Now what about this one? Log base 4, x plus 1 is equal to log base 4, 5x plus 1 minus 1. So here we have a constant, the negative. one so because we have a constant we're going to leave it on the right side we need to move this log expression to the left side so we need to subtract both sides by log base 4 5x plus 1 if you move it from the right to the left side it's going to change sign it's going to go from positive to negative so if we move it to the left side it's going to be log base 4x plus 1 minus log base 4 of 5x plus 1 and that's equal to negative 1 So, in a situation like this where you have two logs on one side and a constant on the other, you want to combine the two log expressions into a single log.
And since we have a subtraction sign, it's going to be division. So log base 4, this has a positive sign in front of it, so it's going to go on top. x plus 1 divided by 5x plus 1. And this is equal to negative 1. So now we could change it into its exponential form. So 4 raised to the negative 1 is equal to what's inside. So x plus 1 over 5x plus 1 is equal to 4 to the negative 1. So you need to keep in mind that 4 to the negative 1 is the same as 1 over 4. So therefore, 1 over 4 is equal to x plus 1 over 5x plus 1. So let's cross multiply.
1 times 5x plus 1 is 5x plus 1. 4 times x plus 1, if you distribute the 4, it's going to be 4x plus 4. So, let's subtract both sides by 4x, and let's subtract both sides by 1. So, 5x minus 4x, that's 1x, or simply x. 4 minus 1 is 3. So, that's the answer. x is equal to 3. So, let's say if you get a question that asks you to condense or write as a single log the following expression.
Log of x plus log of y squared minus log z. How would you do it? To write it as a single log expression, let's write a single log. Now, everything that has a positive sign in front is going to go on top. If it has a negative sign, it's going to go on the bottom.
So, x and y squared are going to go on top. Z is going to go on the bottom. And that's how you can condense it into a single log expression.
But now, let's say, we'll try this example. So, let's say if you have log x plus... log y cubed minus log z squared minus log r cubed. Try this before I give you a new example. So this is going to be log, and these two are positive.
So x and y cubed are going to go on top. z squared and r cubed are going to go on the bottom. And so this is the answer.
Now, what if there's coefficients? Let's say if we have 2 log x plus 3 log y minus 1 third log z. At this point, what can we do to combine it into a single log? If you have coefficients, the first thing you should do is move them to the exponent position. So, we can rewrite it as log x squared plus log y to the third.
minus log z to the 1 3rd, or cube root of z. So now we can write it as a single log expression. So it's x squared y cubed over z to the 1 3rd, or we can write it as log x squared y to the 3rd cube root of z. Both ways work.
So now sometimes... You need to work in the opposite direction. You need to expand a single log expression. So go ahead and expand log base 3, x to the 3rd, y to the 4th, z to the 5th, r squared.
So on top, we have x and y. So it's going to be log base 3, x cubed, plus... log base 4, y to the 4th. Now, z and r are in the denominator, so they're going to have a negative sign in front of it. So, it's going to be minus log z to the 5th, and minus log r squared.
Now, the next thing we need to do is take the exponent and move it back to the front. So we now have 3 log x. This shouldn't be a 4. It's really supposed to be base 3. Each one of these is base 3. So 3 log x plus 4 log y with a base 3 of course minus 2 log base 3 of r minus 5 log base 3 of z. So now we've expanded it. That is the answer in expanded form.
So let's try another question just like that. So let's say if we wish to expand log cube root xy squared divided by z. Now, if there's no number here, keep in mind it's base 10, but we don't need to write it.
So, the first thing I would do is convert the radical into an exponential fraction. So, what we now have is log xy squared over z raised to the 1 third power. So, let's move the exponent to the front first. So, it's 1 third.
log xy squared over z so now we can expand log xy squared over z so it's going to be the log x plus log y squared but minus log z since z is on the bottom now we can move the the exponent 2 to the front so the final answer is 1 3rd times log X plus 2 log Y minus log Z. So that's how you can expand this particular logarithmic expression. So let's move away from logs. Consider the equation 1 plus 3 over x is equal to 40 over x squared. If you receive an equation like this, what can you do to solve for x?
What's the first thing that we should do? Well, what I would do is multiply everything by x squared, just to get rid of all the fractions. So, let's distribute x squared into everything within both sides of the equation.
So, x squared times 1 is simply x squared. Now, what's... x squared times 3 of x so if we multiply x squared or x squared over 1 times 3 of x we can cancel an x so we have 1x left over times 3 and that's going to be 3x now 40 over x squared times x squared, the x squareds will cancel, and so all that you have left over is 40. So now let's move 40 from the right to the left side. So it's going to change from being positive to negative. Notice that we have a trinomial with a leading coefficient of 1. So to solve for x, we can factor it or we can use the quadratic equation.
So what two numbers multiply to negative 40 but add to positive 3? So this is none other than positive 8 and negative 5. So to factor it, it's going to be x plus 8 times x minus 5. So therefore, x is equal to negative 8 and positive 5. So these are the solutions to this particular equation. What about this one?
x to the fourth minus 3x squared minus 10 is equal to 0. How would you solve for x in this particular problem? So if you see a question like this, you want to factor by substitution. Let's replace a with the middle term x squared. So we can rewrite the expression as Well, if a is x squared, that means a squared is x to the fourth. So a squared minus 3a minus 10 is equal to 0. So what two numbers multiply to negative 10 but add to negative 3?
So this is going to be negative 5 and 2. Negative 5 times 2 is negative 10, but negative 5 plus 2 is negative 3. So to factor it, it's a minus 5 times a plus 2. So therefore, we can say that a is equal to positive 5 and a is equal to negative 2. And since a is equal to x squared, that means x squared is equal to 5, but x squared can never be negative 2. Whenever you square a number, it's always positive. For example, 3 times 3 is positive 9, and negative 3 times negative 3 is also positive 9. So when you square a number, it can never be negative 2. So x squared can never be negative 2. So we can eliminate this solution. So let's focus on this one, x squared equals 5. So solving for x, we can take the square root of both sides. So x is equal to plus or minus root 5. And these are the solutions to the original equation. How about this one?
e to the 2x Minus 6 e to the x plus 8 is equal to 0. So just like before, we're going to solve it by substitution. Let's set the middle term, or the middle variable, equal to a. So a is equal to e to the x. Therefore, a squared is equal to e to the 2x. So a squared minus 6a plus 8 is equal to 0. So what two numbers multiply to a positive 8 but add to negative 6?
This is negative 4 and negative 2. So, to factor it, it's going to be a minus 4 and a minus 2 is equal to 0. Therefore, a is equal to positive 4 and a is equal to 2. Since a equals e to the x, e to the x is equal to 4 and e to the x is equal to 2. Now, to solve for x, we need to take the natural log of both sides. So, let's start with the first one. Natural log e to the x is equal to 4, natural log of 4. So we can take the x and move it to the front.
So x ln e is equal to ln 4. And keep in mind, ln e is 1. So therefore, what we have is x times 1, or simply x, is equal to ln 4. So using the same process for the other equation, which looks very similar to this one, we can see that x will equal ln 2. So the answers are ln 2 and ln 4. So that's how you can solve that particular problem. Now let's say if we have the rational equation 6 over x plus 2 plus 4 over x plus 1 is equal to x minus 5 divided by x minus 2. So if you're given an equation that looks like this, how would you solve for x? Feel free to pause the video and try this particular example.
Let's modify that equation. So let's keep the left side the same, 6 over x plus 2 plus 4 over x plus 1. But let's just make the right side equal to 4. Okay, so how would you solve this particular equation? Now the first thing I would do is multiply by the common denominator, that is x plus 2 times x plus 1. x plus 2 times x plus 1 multiplied by 6 times x plus 2. Notice that the x plus 2's cancel.
So what we have left over is 6 times x plus 1. Now, if we multiply 4 over x plus 1 times x plus 2 x plus 1, the x plus 1's cancel. And so what remains is going to be 4. times x plus 2. And then these two times 4 is simply going to be 4 times x plus 2 times x plus 1. So on the right side, let's distribute the 6 and the 4. So it's going to be 6x plus 6, 4x plus 8. And on the right side, let's FOIL x plus 2 and x plus 1. So x times x is x squared, and x times 1 is simply x, 2 times x, that's 2x, and 2 times 1 is 2. So now, let's combine like terms. We can combine the 6 and the 8 on the left side, and the 6x and the 4x.
So this is going to be 10x plus 14 on the left. We can combine x and 2x and make it 3x. And let's distribute the 4. 4 times x squared is 4x squared.
4 times 3x, that's 12x. And 4 times 2 is 8. So now let's move everything from the left side to the right side. So let's subtract both sides by 14, and let's subtract both sides by 10x.
So on the right side, I mean on the left side, all we have is 0. On the right side, it's going to be 4x squared plus 2x minus 6. So now we can factor the expression to solve for x. But before we do, let's take out a 2. So it's going to be 2x squared plus x minus 3. So now we have a trinomial where the leading coefficient is a 2. so let's multiply the lean coefficient 2 by negative 3 so this will give us negative 6 so we need two numbers that multiply to negative 6 but add to the middle term 1 so this is going to be 3 and negative 2 So we're going to have 2x squared minus 2x plus 3x minus 3. So I replace the middle term 1x with negative 2x and 3x, and they add up to 1x. I put the 2's together and the 3's together, so it will be easier to find the GCF. So the GCF, or the greatest common factor between 2x squared and 2x, is simply 2x.
2x squared divided by 2x is x. negative 2x divided by 2x is negative 1. And if we take out a 3, we're going to have x minus 1 as well. So we can see that we have a common factor, x minus 1. And what remains on the outside is going to go in the next parentheses, so 2x plus 3. So we can set x minus 1 and 2x plus 3 equal to 0. So let's go ahead and solve for x.
So for this one x is equal to plus 1 and for the other one it's a negative 3 divided by 2. So x can be any one of these solutions.