in this video we're going to go over a basic introduction slash overview of college algebra so let's begin with some basics what is x squared times x to the fifth power what do you do to the exponents when you multiply common bases x squared times x to the fifth power is x to the seventh power when you multiply by a similar base you are allowed to add the exponents x squared can be thought of as x times x you're multiplying two x variables together x to the fifth power is basically five x variables together combined you have a total of seven x variables which represents that what about division what is x to the fifth power divided by x squared when you divide you need to subtract the top number by the bottom number 5 minus 2 is 3. again if you expand it it can make sense x to the fifth power is 5x is multiplied to each other x squared is just two x's so you can cancel two of them which leaves three on top so let's try another example what is uh x to the fourth divided by x to the seventh so taking the top number subtracting the bottom number four minus seven is negative three and whenever you have a negative exponent you can move the x variable from the top to the bottom and as you do that the sign will change the negative three will change into a positive three so this is the same as one over x cubed now let's expand it x to the fourth is basically four x variables multiplied to each other x to the seventh represents seven x variables we can cancel four on top four on the bottom and so we're left with three on the bottom which is x cubed now what is x cubed raised to the fourth power whenever you raise one exponent to another exponent you are allowed to multiply them three times four is twelve so the answer is x to the twelfth power one way you could think about it imagine x cubed raised to the fourth power is equivalent to four x cubes multiplied to each other that's what it really represents and each x cubed represents the multiplication of three x variables so when you combine them you have a total of 12 x variables multiplied to each other so that's why you get x to the 12th anytime you raise something to the zero power it's always equal to one that's something you just have to commit to memory now let's talk about simplifying expressions and combining like terms if you were to see something like this on a test 5x plus 3 plus 7x minus 4. how would you simplify this expression notice the 5x and the 7x are like terms they both carry the variable x 5 plus 7 is 12 so 5x plus 7x is 12x now these two don't have an x variable attached to it so they're similar to each other 3 plus negative 4 is the same as 3 minus 4 which is negative one go ahead and try this one three x squared plus six x plus eight plus nine x squared plus seven x minus five feel free to take a minute pause the video and work on this example so these two are like terms 3 plus 9 is 12 and we can add these two 6 plus 7 is 13 and then we can add these two 8 plus negative 5 or 8 minus 5 is positive 3. here's another one that we could try five x square minus three x plus seven minus four x squared minus eight x minus eleven now the first thing we should do is we should distribute this negative sign to each of these three terms before we combine like terms if you want to avoid making mistakes now we don't have a negative sign in front of the first set of terms inside the first parentheses so we could just open it so what we have now is 5x squared minus 3x plus 7 and if we distribute the negative sign to everything on the right all the sides will change the positive 4x squared will now become negative 4x squared the negative 8x will now change to positive 8x and the same is true for the 11. it's going to change from negative 11 to positive 11. so now let's combine like terms 5x squared minus four x squared is one x squared negative three plus eight which is the same as eight minus three is positive five and seven plus eleven is eighteen so you're gonna get this now what if we want to multiply two expressions instead of adding and subtracting polynomials by the way this is a monomial that's one term two terms represent a binomial three terms represent a trinomial and if you have like many terms you can simply call it a polynomial it's good to be familiar with those expressions so here we're multiplying two binomials we need to use something called foil in foil the first letter f is for the first part we multiply the first two terms 3x times 2x is 6x squared and then you multiply the term on the outside 3x times negative 6 which is negative 18x and then the ones on the inside negative 5 times 2x that's negative 10x and then the last ones negative 5 times negative 6 which is positive 30. so now let's combine the two terms in the middle since they're like terms negative 18x plus negative 10x is negative 28x so this is the answer now how can we expand an expression that looks like this what would you do if you saw the mss to expand it the square means that we have two two x minus five factors multiplied to each other so we can use the foil technique again two x times two x two times two is four x times x is x squared two 2x times negative 5 is negative 10x and negative 5 times 2x is also negative 10x negative 5 times negative 5 is positive 25. now we can add like terms negative 10 plus negative 10 is negative 20. and so this is the answer now what about solving linear equations how can we solve for x what number plus six is eleven it turns out that five plus six is eleven so x is equivalent to five but to find the answer notice that we see x plus six the opposite of addition is subtraction so we need to subtract both sides by six six minus six is zero so these two cancel you can bring down the x and eleven minus six is five so that's how you can solve it now what if x is multiplied to a number what can we do to solve for x the opposite of multiplication is division so you want to divide both sides by four four divided by four is one so you get one x which is simply equal to x eight divided by four is two and so that's how you can solve for x so now what if we have a combination of multiplication and addition in the equation what can we do so first we should subtract both sides by five don't divide by three not yet otherwise you'll have a fraction these two will cancel we could bring down the 3x 26 minus 5 is 21. next let's divide both sides by three 21 divided by 3 is 7 and so this is the answer try this one so what do you think we should do here in this problem there's many ways you can go about solving it you can distribute the 4 to the 2x to the negative 7 or you could subtract both sides by 8. i'm going to subtract both sides by eight so on the left we have four times two x minus seven and on the right we have twenty minus eight which is twelve now at this point you can distribute the 4 or you can divide both sides by 4. if you do that you'll be left with 2x minus 7 is equal to 12 divided by 4 which is 3. and now we can add 7 to both sides so we're going to have is 2x which is equal to 3 plus 7 that's 10 and if we divide both sides by 2 10 divided by 2 is 5. so x is 5. and you could check your work to make sure you have the right answer if we replace x with 5 the left side should equal 20. so let's go ahead and do that two times five is ten ten minus seven is three four times three is twelve and twelve plus eight is twenty so it works now let's move on to solving and graphing inequalities so let's say if we have the expression two x plus five is greater than 11. now when you solve it imagine as if this is an equal sign all the rules still apply so let's solve for x let's begin by subtracting both sides by 5. so 2x is greater than 11 minus 5 which is 6. next let's divide both sides by 2. so x is greater than 3. now how can we plot this on a number line so let's put zero in the center and somewhere to the right we have a three for the expression x is greater than three we need to shade to the right of three if it was less than three we need to shade towards the left now it's greater than three but not greater than or equal to three if it was greater than or equal to three we would have to draw a closed circle but since it's only greater than three we need to use an open circle and we're going to shade towards the right so that's how you can graph this particular inequality by the way if you have to write the answer in interval notation make sure you understand that negative infinity is all the way to the left and positive infinity is to the right so we're starting at 3 and we're stopping at infinity so you can write the answer from 3 to infinity in interval notation let's try another example go ahead and solve for x find its value so let's begin by subtracting both sides by seven so 13 minus seven is six next let's divide both sides by three six divided by three is two so two is to the right of zero now because it's equal to or greater than 2 we need to use a closed circle and we're going to shade it to the right so the answer in interval notation is from 2 to infinity but it includes 2. so we need to use a bracket instead of a parenthesis whenever you have a closed circle use a bracket for open circles use parentheses and infinity you should always use parentheses let's try another example let's say that negative three x is greater than or equal to nine solve for x so let's begin by dividing both sides by negative three negative nine i mean positive nine divided by negative three is negative three when you multiply or divide by a negative number the inequality changes sign so you have to flip it so that's the solution x is less than or equal to negative three so negative three is on the left side of zero and because we have the underline here we need to use a closed circle as opposed to an open circle and because it's less than we need to shade towards left so in interval notation the answer is from negative infinity to negative three but it includes negative three since we have a closed circle so whenever you write the answer in interval notation always start from the left and then write the answer on the right so always go from left to right try this one negative five x plus eight is greater than negative seven so let's begin by subtracting both sides by negative eight so negative five x is greater than negative seven minus eight which is negative fifteen now let's divide both sides by negative five so the inequality will change sign or direction so x is now less than positive three so let's use an open circle and we're going to shade to the left so the unit from left to right the value all the way to the left is negative infinity and the value all the way to the right is three so the answer in interval notation is negative infinity to three but it does not include three since we have an open circle so always write it from left to right now let's spend some time talking about absolute value expressions what is the absolute value of negative three and three the absolute value of positive three is positive three the absolute value of negative three is still positive three the absolute value will make any number and change it into a positive number however the absolute value of zero will just be zero but it will never give you a negative output so knowing that if the absolute value of x is equal to four what is the value of x x can be two answers it can be positive four or negative four the absolute value of positive four is four and the absolute value of negative four is positive four so whenever you're dealing with absolute value equations you need to write two equations in order to solve for x the next example will demonstrate that so let's say if the absolute value of 2x plus 3 is 11. to solve it write the original equation without the absolute value expression and for the second one make it equal to negative 11. so here let's subtract both sides by three 11 minus three is eight and then let's divide by two so the first answer is positive four now let's get the second answer negative eleven minus three is negative fourteen and if we divide this by 2 negative 14 divided by 2 is negative 7. so here's the second answer so x can equal 4 or negative 7. try this one let's say that the absolute value of three x minus one is greater than five so what can we do if we have an inequality and an absolute value expression so we're going to write two equations the original one is going to remain the same now the second one you need to do two things you need to change the direction of the absolute value expression and change the five into negative five and then solve for both equations so let's add one three x is greater than six and then let's divide by three so x is greater than two six divided by three is two here let's do the same thing let's add one to both sides negative five plus one is negative four and now let's divide by three so x is less than negative four thirds and negative four thirds is negative one point three three as a decimal so now we're going to do is plot the solution so here's two and negative one point three three is somewhere between negative one and two so x is greater than two so we have an open circle shaded to the right and it's less than a negative 1.33 this is really an or statement whenever you see it splits off into two directions like that so how can we write the answer in interval notation so we have two sections for the blue section the answer is negative infinity to negative four thirds for the red section it's two to infinity and you can connect these two with a union symbol so this is the solution in interval notation now what if you were to see a compound inequality that looks like this what can we do in order to solve for x so there's like three sides to the equation we're going to subtract all three sides by four negative two minus four is negative six these two will cancel and thirteen minus four is nine next let's divide all three sides by three so you can just do it simultaneously negative six divided by three is negative two and since we divided by a positive number the direction of the inequalities will not change nine divided by three is three and so we get this so now we can plot it on a number line so what this really means is that x is between negative 2 and 3. so this expression tells you that negative 2 is greater than or equal to x or if you read it backwards x is less than or equal to negative 2. so and then the other one is x is greater than or equal to 3. it turns out this is another or statement it's actually not between negative 2 or 3. so this is like the other one so the answers from negative infinity to negative two union three to infinity but it includes two and three so we need to use brackets let's try another example let's say two x plus five is less than or equal to let's say uh 13 but it's greater than or equal to uh one so this one should be like a an and inequality as opposed to an or inequality so let's subtract both sides by five so one minus five is negative four thirteen minus five is eight next let's divide both sides by two so x is greater than or equal to negative two but less than or equal to four so therefore x is between negative two and positive four so it's here the answer in interval notation is from negative two to four now let's talk about how to graph linear equations in slope intercept form slope intercept form looks like this y is equal to m x plus b m represents the slope b is the y intercept so let's say if we want to graph this equation y is equal to two x minus three so let's make a graph first so first plot the y-intercept so that's the point zero negative three or simply negative three on the y axis the slope is two which is the same as two over one the top number represents the rise the bottom number represents the run so what we're going to do is we're going to rise two units and only run one unit which will take us to the point one negative one so the next point is over here and then to find the second point go up two units over one which will take us to two comma one and then go up two over one that will take us to three comma three and then just connect these points with a line so that's how you can graph a linear equation in slope intercept form now what if we have a fraction let's say if we have the expression y is equal to negative three over four x plus five how can we graph this particular equation so we can see the slope is negative three over four so the rise is negative three the run is four the y-intercept is five so let's plot the y intercept at five to find the next point the rise is negative three which means we need to go down three and we need to travel four units to the right as long as you have two points it's enough to graph the equation acceptably so it looks like this now let's talk about how to graph linear equations if it's presented in standard form which is a x plus b y equals c now the best way to graph this equation is to use the intercept method so let's work on an example let's say that 2x plus 3y is equal to 6. so let's begin by finding the x-intercept to find the x-intercept replace y with zero so 2x plus three times zero is equal to six three times zero is just zero so two x is equal to six and x is string so the x intercept is three comma zero now let's find a y intercept to do that replace x with zero and solve for y now two times zero is zero so we just have three y is equal to six and if we divide by two i mean by three six divided by 3 is 2. so the y intercept is 0 comma 2. so what we're going to do is simply plot the x and y intercepts and then connect them with a line so the x intercept is 3 0 so that's x equals three on the x axis which is right here and on the y axis we need to plot two or zero comma two and then just connect these two points with a line so that's how you can graph an equation in standard form now it's your turn try this example let's say that 3x minus 4y is equal to 12. find the x and y intercepts and go ahead and plot it so let's start with the x intercept so let's replace y with zero so we can see that 3x is equal to 12 and if we divide both sides by three x is equal to four so the x intercept is four comma zero now let's find the y intercept and let's do that by replacing x with zero so three times zero is zero so we just have negative four y is equal to twelve and now let's divide by negative four so 12 divided by negative four is negative three so the y-intercept is 0 negative 3. so the x-intercept is 4 units to the right the y-intercept is 3 units down so 0 negative 3 is over here and then simply connect the two points with a line that's how you do it now let's talk about how to graph functions with transformations let's start with the absolute value of x function so to graph this function the origin is at 0 0. the general shape it looks like a v shape but you want to get the points right the number in front of x is a one so it has a slope of one on the right side and a slope of negative one on the left so as you go one to the right you need to go up one so you're gonna have the point one one and then one to the right up one so you have the point two two and three three and so forth the left side is a reflection of the right side relative to the vertex the vertex is at the origin zero zero so the graph looks like this now what about if we put a 2 in front of it so now the slope has a value of 2. so the origin is still the same but this time as we travel one unit to the right we need to go up two so the next point is going to be one comma two and then to the left is going to be the same and if we travel one more unit to the right we need to go up 2. so it's going to be 2 4 and negative 2 4. so it looks like this if you need to draw an accurate graph that's what you can do but from here on we're going to draw just a rough sketch so we know that the absolute value of x is just a v shape that opens this way with a slope of one but what if there's a negative sign in front of the absolute value of x what's going to happen if there's a negative sign it's going to reflect over the x-axis so instead of opening in the upward direction it's going to open downward now what if we have this expression absolute value x plus 2 on the inside what's going to happen to find out set the inside equal to zero if you solve for x you'll see that x is negative two so the graph is going to start at negative two which means that it shifts two units to the left and it's going to open upward so let's say if we have absolute value x minus three this is going to shift three units to the right if you set the inside equal to zero x will equal three so the new origin is at three comma zero and then it's going to open upward with a slope of one now what if the number is outside of the absolute value what happens in this case the 2 will cause it to shift 2 units up but it's still going to open in the upward direction so let's say if we have negative absolute value of x minus 1. so the graph is going to shift one unit down but because of the negative sign in front of the absolute value it's not going to open upward it's going to open downward so it looks like that so let's put it together what if we have the absolute value of x minus 3 plus two so this is going to shift two units to the right i mean three units to the right but up two units and it's going to open uh upward now let's change the parent function into a quadratic function or parabola the parent function for y equals x squared is a u shape instead of a v shape and it opens downward so based on your knowledge of transformations what's the shape of the graph negative x squared so putting a negative in front will cause it to reflect over the x-axis so it's going to open downward now try these two examples let's see if we have x minus two squared and x plus four squared so you know for the first one it's going to shift two units to the right and it's going to open upward for the second one it's going to shift 4 units to the left and it's going to open upward so try this one x squared minus one and negative x squared plus three so this one is going to shift one unit down and it's going to open upward the next one is going to shift three units up but because of the negative in front of the x squared it's going to open downward so now let's try this one x minus one squared plus two and three minus x plus two squared so for the first one it's going to shift one to the right up two so it starts here and because there's a positive in front it's going to open upward the next example is going to shift two to the left up three so it starts at negative two comma three and because the negative sign in front is going to open downward so that's how you can graph quadratic expressions using transformations now let's talk about how to solve quadratic equations by factoring let's begin with this quadratic expression x squared minus 25 is equal to zero how can we factor the expression and also solve for x so there's something called the difference of perfect squares x squared is a perfect square and 25 is a perfect square to factor it's going to be a minus b times a plus b the square root of x squared is x the square root of 25 is five one is going to be positive the other will be negative and that's how you factor it now let's set each factor equal to zero so let's set x plus five equal to zero and x minus five equal to zero so the first answer x is equal to negative five and for the second answer it's equal to positive five now let's work on another example x squared minus 36 is equal to zero the square root of x squared is x and what is the square root of 36 it turns out that 6 times 6 is 36 so that's the square root of 36 and 1 will be positive the other will be negative and then solve for x so x is equal to negative six and positive six try this one so what is the square root of 4 the square root of 4 is 2 and the square root of x squared is x so the square root of 4x squared is 2x the square root of 49 is 7 and this is going to be plus and minus now let's set each factor equal to zero so first let's subtract both sides by seven so we can see that 2x is equal to negative seven and then let's divide by two so for the first one x is equal to negative seven over two now here let's add seven to both sides so two x is equal to positive seven and then let's divide by two so x is seven over two so it's positive seven over two and negative seven over two what would you do if you see this expression how can you factor it and solve for x now the square root of 3 is not a nice number and the square root of 48 is not a perfect square however you could take out the gcf the greatest common factor which is three if we take out three three x squared divided by three is equal to x squared negative forty eight divided by three is negative 16. so now we have two perfect squares the square root of x squared is x the square root of 16 is 4. when it's positive the other is negative and based on the last examples we could easily tell that x will equal negative four and positive four you simply have to reverse the two signs now how can we factor a trinomial where the leading coefficient is one and let's solve for x at the same time to factor an expression that looks like this find two numbers that multiply to the constant term of 24 and then add to the middle coefficient of 10. so what are some numbers that multiply it to 24 we have 1 and 24 2 and 12 3 and 8 and 4 and 6. however only 4 plus 6 adds up to 10. so to factor it is simply going to be x plus 4 times x plus 6. and then to solve the um to solve for x simply change the sign so x is going to be negative 4 and negative 6 and that's it so let's work on a few more examples since this type of problem is very common feel free to pause the video and find two numbers that multiply to negative 24 but add to positive five so we have 1 and negative 24 2 and negative 12 3 negative 8 notice that 3 plus negative 8 is negative if we reverse it negative three plus eight is positive five but these two numbers still multiply to negative twenty four so it's going to be x minus three times x plus eight and therefore x is going to equal the reverse that's positive 3 and negative 8. here's another one x squared minus 2x minus 35 is equal to zero so what two numbers multiply to negative 35 and add to negative two seven times five is thirty five so we can do it as negative seven and five or seven and negative 5. 7 plus negative 5 adds up to positive 2 but negative 7 plus 5 adds up to negative 2. so this is the one we want to use so it's going to be x minus 7 times x plus 5 and so x will be equal to positive 7 and x will be equal to negative 5. let's try one more example where we have a trinomial with a leading coefficient of 1. this time we're not going to only factor it but we're going to get the answer using the quadratic equation so what two numbers multiply to 42 but add to negative 13. if we divide 42 by negative 1 we'll get negative 42. if we divide it by negative 2 we'll get negative 21. if we divide it by negative 3 we'll get negative 14. 4 doesn't go into 42 however we can divide 42 by negative six and that will give us negative seven and negative six plus negative seven adds to negative thirteen so we can factor it as x minus six times x minus seven so x is equal to positive 6 and positive 7. so let's get these two answers using the quadratic equation so this quadratic expression is in standard form that is it's ax squared plus bx plus c so a is the number in front of x squared so a is one b is negative 13 c is 42. to use the quadratic equation you need to know the formula it's negative b plus or minus b squared minus 4ac divided by 2a negative b is going to be negative times negative 13 plus or minus b squared negative 13 squared is 169. if you multiply negative 13 by negative 13 you will get positive 169. a is 1 and c is 42. so divided by 2a or 2 times 1 which is 2. so let's get rid of a few things negative times negative 13 is positive 13 and then it's plus or minus square root 169 and 4 multiplied by 42 is 168. and we have a two on the bottom so 169 minus 168 is one and the square root of one is equal to one so we have 13 plus or minus one divided by two so at this point we could set that expression equal to two separate expressions so we could say it's 13 plus one divided by two or we could say it's 13 minus 1 divided by 2. so as you can see we're going to get two answers 13 plus 1 is 14 and 13 minus one is twelve fourteen divided by two is uh seven and twelve divided by two is six so we get the same two answers that we had earlier now let's factor a trinomial where the leading coefficient is not one let's try this one two x squared plus five x minus 12 is equal to zero we're going to factor it solve for x and then we're going to confirm our answer using the quadratic equation so to factor an expression that looks like this first multiply the leading coefficient by the constant term 2 times negative 12 is negative 24 then find two numbers that multiply to negative 24 but that add to the middle coefficient of five so we know eight times three is 24 but one of these numbers has to be negative let's put the negative with the three eight times negative three is negative 24 but eight plus negative three is positive five now what we're going to do in this example is that we're gonna replace the 5x with 8x and negative 3x because 8x minus 3x adds up to 5x and then we're going to use a technique called factoring by grouping so let's get rid of this stuff when you have four terms you can factor by grouping if the first two terms has the same ratio as the last two terms for instance consider the two and eight eight divided by two is the same as negative twelve divided by negative three both answers equal four whenever you see that you can factor by grouping so in the first two terms let's take out the gcf the greatest common factor which is 2x 2x squared divided by 2x is x 8x divided by 2x is 4. now in the last two terms let's also take out the gcf which is negative 3. negative 3x divided by negative 3 is x and negative 12 divided by negative 3 is positive 4. so if you see these two factors if they're the same that means you're on the right track that particular factor write it once and then everything you see outside of that 2x minus 3 that's going to go in the next parenthesis so that's how you can factor it and you can foil these two binomials to check and make sure that you do indeed get this expression now let's set each factor equal to zero so here let's subtract both sides by four so we can see that x is negative four and then let's add three to both sides so two x is equal to three which means x is three over two so these are the two answers let's go ahead and try to get those two answers using the quadratic equation so we can see a is equal to 2 b is equal to 5 and c is equal to negative 12. so x is equal to negative b plus or minus square root b squared minus 4ac divided by 2a so b is 5 which means b squared or 5 times 5 is 25 minus 4 times a which is 2 times c which is negative 12 divided by 2a or 2 times 2. so this is negative five plus or minus twenty five now four times two is eight eight times twelve is ninety-six and because of the two negative signs it's going to be positive ninety-six and on the bottom two times two is four now 25 plus 96 is 121 and the square root of 121 is eleven so we have negative five plus or minus eleven divided by four now let's separate it into two expressions negative five plus eleven over four and negative five minus eleven over four so negative five plus eleven is the same as eleven minus five which is positive six negative five minus eleven is negative sixteen now six over four we can reduce it if we divide both numbers by two six divided by two is three four divided by two is two so the first answer is three over two which is an answer that we had before now negative sixteen divided by 4 is negative 4 and this is the other answer so as you can see we get the same answers using the quadratic equation or even by factoring the expression both methods work now let's try a harder example how can we factor this expression 12x squared minus 83x plus 66. now using the same technique we would have to multiply twelve by sixty six twelve times sixty six is a big number it's seven hundred and ninety two what two numbers multiply to seven ninety two but add to negative eighty three now this expression is factorable by the way so if you know an expression is factorable but you just don't know how to factor it what can you do is there a fallback method that you can apply it turns out that there is and that is the quadratic formula you can use the quadratic formula to work backwards to factor the expression so let's go ahead and do that in this particular example so let's make a list of the values that we have so a is 12 b is negative 83 and c is 66 so x will be equal to negative b that's negative negative 83 and then plus or minus square root b squared so 83 times 83 or 83 squared is 6889 minus 4 times a which is 12. times c which is 66 divided by 2a or 2 times 12. so negative times negative 83 is positive 83. 4 times 12 is 48 times 66 that's 3168 and 2 times 12 is 24. now let's subtract 6889 by 3168 so that's 3721 and the square root of 37 21 is 61. so we're going to have 83 plus or minus 61 over 24. so therefore we have two values 83 plus 61 over 24 and 83 minus 61 over 24. 83 plus 61 is 144. 83 minus 61 is 22. 144 divided by 24 is equal to a whole number 6 and 22 over 24 is not a whole number so we need to reduce the fraction let's divide it by 2 if we do that 22 divided by 2 is 11 half of 24 is 12 so the other answer is 11 over 12. so using the solutions we can now factor the expression so here's what we need to do write two equations x is equal to 6 and x is equal to 11 over 12. now what you want to do is you want to get 0 on one side of the equation so here let's subtract both sides by 6 and here we're going to subtract both sides by 11 over 12. so let's get rid of this so what we have is x minus six is equal to zero these two cancel and for this one x minus eleven over twelve is equal to zero so we don't need to change this factor however we do need to get rid of the fraction to do that let's multiply both sides by 12. so x times 12 is 12x 11 over 12 times 12 the 12's cancel giving you 11. so this is going to be minus 11 and 0 times 12 is 0. so these are the two factors so therefore to factor the expression it's going to be x minus 6 times 12x minus 11. and now let's make sure that this is indeed the right expression so let's foil it x times 12x is 12x squared x times negative 11 is negative 11x negative 6 times 12x is negative 72x and negative 6 times negative 11 is positive 66. now the two numbers that multiply to the product of 12 times 66 which was uh 12 times 66 was 792. remember we needed two numbers that multiplied to 792 but added to negative 83 it turns out that those two numbers are the two numbers in the middle negative 11 times negative 72 is positive 792. and they add to negative 83. so as you can see this technique works so if you can factor the expression but if you don't know how to factor it you can use the quadratic equation to get the solution and from the solution you can write the factors consider these two equations x squared is equal to 9 and x squared is equal to negative 9. let's solve for x in both equations to do that we can take the square root of both sides the square root of x squared is simply x the square root of 9 is plus or minus three now on the left side we're gonna have plus or minus the square root of negative nine what is the square root of negative nine it's important to understand that the square root of negative 1 is equal to an imaginary number in this case i so negative 9 can be written as 9 times negative 1. the square root of 9 is 3 and the square root of negative 1 is i so in this case instead of getting plus or minus 3 the answer is plus or minus 3i so it's positive 3i and negative 3i i is always equal to the square root of negative 1. i squared is equal to negative one now i to the third power which is the same as i times i squared that's equal to i and i squared is negative one so i to the third power is simply equal to negative i now what about i to the fourth power i to the fourth power is basically i squared times another i squared and each i squared is negative one so negative one times uh negative one is simply equal to one so make sure you know these four values i is equal to the square root of negative one i squared is negative one i cubed is the same as negative i and i to the fourth is one now let's talk about solving systems of equations so let's say if we have two linear equations 2x plus y is equal to 5 and 3x minus y is equal to zero and we have two variables to solve for x and y in order to solve for two variables you need two equations if you have three variables you need three equations to solve it whenever the x and y variables are aligned neatly as the way you see them it's best to use the elimination method to do that simply add the two equations y and negative y will cancel two x plus three x is five five plus zero is five so to solve for x uh let's divide both sides by five five divided by five is one so x is one so now what we're going to do is uh replace x with one in the first equation so two times one plus y is equal to five two times one is two and now we need to subtract both sides by two so five minus two is equal to three so the solution is one comma three let's try another example three x plus four y is equal to twenty five and uh five x minus three y is equal to 3. so let's focus on canceling the y variables notice that if we add the two equations now these two won't cancel the least common multiple of 3 and 4 is 12. if you're not sure what the least common multiple is you could just multiply 3 and 4 and that number will work so let's get to 12. let's multiply the 4 y by 3 and the second equation by uh 4. if we do that 4y times 3 is 12y negative 3y times 4 is negative 12y so they're going to be the same but they're going to have the opposite sign so they will cancel now we got to multiply everything by 3 in the first equation whatever you do to the left side you must also do to the right side 3x times 3 is 9x 4x i mean 4y times 3 is 12y and 25 times 3 is 75. now let's multiply the entire equation by 4. 5x times 4 is 20x negative 3y times 4 is negative 12y 3 times 4 is 12. so if we add the two equations we can see that the y variables will cancel 9 plus 20 is 29 75 plus 12 is 87 and if we divide 87 by 29 that's equal to 3 so x is 3. so now let's rewrite the first equation 3x plus 4y is equal to 25 and let's replace x with 3 and let's solve for y times three is nine twenty five minus nine is sixteen sixteen divided by four is four so y is equal to four so the answer is three comma four x comma y now what if you were to see two equations like this in this case it's best to use the substitution method we can replace y with 2x plus 5 since they're equal to each other and this will allow us to solve for x so let's subtract both sides by 2x 5x minus 2x is 3x and now let's add 4 to both sides five plus four is nine the last thing we need to do here is divide by three nine divided by three is three so x is three now once you have the value of x you can plug it into any one of these two equations and you should get the same y value so two times three is six plus five that's eleven and if we plug it into the other equation five times three is fifteen minus four is also eleven so the answer is three comma eleven now what if the two equations are presented differently than the last example let's say if it looks something like this which method do you think we should use in a situation like this we should still use the substitution method we can replace y with 7x minus 5 since they're equal to each other so let's go ahead and do that so we're going to have 2x plus 4 instead of 4y is going to be 4 times 7x minus 5 which is equal to 10 and now we can solve for x so let's distribute the four four times seven x is 28 x and four times negative five is negative 20. now we can combine like terms 2x and 28x is equal to 30x and now let's add 20 to both sides 10 plus 20 is 30. so 30x is equal to 30 and now we can divide both sides by 30. so we can clearly see that x is equal to one now once you have this value we could plug it into this equation to get you so y is equal to seven times one minus five seven minus five is two so y is two so the solution is one comma two the other method that you can use is you can graph the two equations when you graph them the solution is the point of intersection the point at which the two lines meet that is the answer to the question you just have to look up the x and y values and you could do it that way as well now let's say if we have the function f of x is equal to 5x plus 4. what is the value of f of 3 now this problem is pretty straightforward when x is 3 what is the value of the function all you got to do is replace x with 3 so it's 5 times 3 which is 15 plus 4 that's 19. but now what if you were to see an expression like this let's say if f of x is equal to 39 what is the value of x in order to find the value of x you need to know what the 39 represents the 39 is equal to the entire function it basically represents the y value y is equal to f of x so x is the number on the inside y is the number on the outside so in this particular case we need to replace f of x with 39 and solve for x 39 minus 4 is 35 and 35 divided by 5 is 7. so x is 7 which means f of 7 is equal to 39 now what is the value of f of 5 in this problem we're going to evaluate the function when x is equal to 5 two ways let's use it the standard way let's replace x with five five to the third power which is five times five times five that's one twenty five five squared is twenty five times four is a hundred six times five is thirty and one twenty minus 100 is 25 30 minus 14 is 16. 25 plus 16 is 41. so let's see if we can get this answer using another technique this other technique is called synthetic division so we know that f of 5 is 4. so we're going to put a 5 here and we need to place the coefficients 1 negative 4 6 negative 14 and decreasing order so let's bring down the one 5 times 1 is 5 and then add negative 4 plus 5 is 1. five times one is five six plus five is eleven five times eleven is fifty-five negative fourteen plus 55 or 55 minus 14 is 41. so the remainder is equal to the value of the function so that's another way in which you can evaluate difficult functions now let's review composite functions let's say f of x is 3x plus 5 and g of x is equal to x squared minus 4. so based on this information what is f of g of x notice that g is inside of f this expression can be written as f of g can look like this fog these two mean the same thing in order to find the value of f of g of x the expression we need to take g and insert it into f now f of x is three times x plus 5. but instead of writing the x we're going to replace x with x squared minus 4. so instead of writing 3x plus 5 is going to be 3 times x squared minus 4 plus 5. now let's distribute the 3. so it's 3x squared and 3 times negative 4 is negative 12 and then we can add negative 12 plus 5. so it's 3x squared minus 7. so f of g of x is equal to this expression now what about g of f of x which can be written as g of f so this time we need to take the function f and insert it into g so g is x squared minus 4. instead of writing x squared we're going to write 3x plus 5 squared so we need to foil uh the two expressions three x plus five squared is three x plus five times three x plus five three x times three x is nine x squared and 3x plus 5 or 3x times 5 is 15x and 5 times 3x is also 15x and 5 times 5 is 25. we can add like terms so 15x plus 15x is 30x 25 minus 4 is 21. so the composite function is equal to that now let's say if we wish to evaluate f of g of 2 what's the answer first we need to find g of 2. so let's replace x with 2 in this equation so it's two squared minus four two squared is four four minus four is zero so g of two is equal to zero so we can replace g of two of zero so now we're looking for f of zero which is three times zero plus five using this equation and that's equal to five so f of g of two is five let's try another example what is g of f of three so first let's evaluate f of three that's three times three plus five three times three is nine plus five is fourteen so f of 3 is 14. so now we got to find g of 14 which is 14 squared minus 4. now 14 times 14 or 14 squared is 196 minus 4 that's equal to 192. so if you see a number inside your final answer should equal a number if you see an x variable your final answer should be an expression in terms of x so just keep that in mind now let's briefly review inverse functions f of x is the original function and f to the negative one of x represents the inverse function so let's say that for the original function we have the points 2 3 4 negative 6 and negative 5 2. to find the points of an inverse function you need to switch the x and y values so it's going to be 3 2 negative 6 4 and 2 negative 5. now let's talk about the graphical relationship of an inverse function so let's say this function is f x the inverse function is going to look something like this and that's going to be f negative 1 of x now notice that these two functions are symmetrical about the line y equals x that's a property of inverse functions they will always be symmetrical about that line now let's say that f of x is equal to seven x minus five how can we find the inverse function in order to do that replace f of x with y and in your next step switch x and y and then solve for y so we need to add 5 to both sides let's move the negative 5 from the right side to the left side on the right side is negative on the left side it's going to be positive so it's x plus 5 is equal to 7y next let's divide both sides by 7. so the inverse function is x plus five over seven so at this point just replace y with the inverse function that's how you can find it now in order to prove if two functions are inverses of each other here's what you need to do let's call the inverse function g of x the composite function between f and g of x should equal x and also g of f of x should equal x as well if both of these conditions are satisfied then f and g are inverses of each other so let's go ahead and prove it so let's start with f of g of x so let's take g and insert it into f so instead of writing 7 x minus five it's going to be seven times x plus five over seven so notice that the sevens cancel leaving behind x plus five minus five and these cancel giving us x so the first expression is true f of g of x is x and so now let's check out g of f of x so this time we're going to take f insert into g so instead of writing x plus 5 divided by 7 it's going to be 7 x minus 5 which was in place of x and then plus 5 over seven so negative five plus five is zero seven x divided by seven is x so that's how you can prove if two functions are inverses of each other so that is it for this video if you like it feel free to comment subscribe share this video with your friends also if you want to find more algebra videos feel free to uh go to my channel and check out my playlist i have a lot of other topics i recognize that this video is only scratching the surface of all the algebra topics that you need to know so if you check out my playlist you can find other topics throughout the college algebra course that you might be taking i also have some other videos on like physics uh general chemistry pre-calculus trig calculus as well so you might find those helpful um if you check it out so thanks for watching and have a great day