Transcript for:
College Algebra Overview

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^2 * X 5th power what do you do to the exponents when you multiply common bases x^2 * X 5th power is X to the 7th power when you multiply by a similar base you allowed to to add the exponents x^2 can be thought of as x * X you're multiplying two x variables together x to the 5ifth power is basically five x variables together combine you have a total of seven X variables which represents St what about division what is X to the 5th power / x^2 when you divide you need to subtract the top number by the bottom number 5 - 2 is 3 again if you expand it it can make sense x to the 5th power is 5 X's multipli to each other X2 is just 2 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 4th / X to 7th so taking a top number subtracting the bottom number 4 - 7 is -3 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 negative3 will change into a positive three so this is the same as 1 /x Cube now let's expand it X to 4th it's basically four x variables multiply to each other x to the 7th 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 Cub now what is X Cub raised to the 4th power whenever you raise one exponent to another exponent you are allowed to multiply them 3 * 4 is 12 so the answer is X to 12 power one way you could think about imagine XB 4 power is equivalent to 4 XB MTI to that's what it really represents and each X Cube represents the multiplication of three x variables so when you combine them you have a total of 12 x variables multiply to each other so that's why you get x 12 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 + 3 + 7 x - 4 how would you simplify this expression notice the 5x and the 7x are like terms they both carry the variable X 5 + 7 is 12 so 5 x + 7 x is 12x now these two don't have an X variable attached to it so they're similar to each other 3 +4 is the same as 3 - 4 which is1 go ahead and try this one 3x^2 + 6 x + 8 plus 9 x^2 + 7 x - 5 feel free to take a minute pause the video and work on this example so these two are like terms 3 + 9 is 12 and we can add these two 6 + 7 is 13 and then we can add these two 8 +5 or 8 - 5 is POS 3 here's another one that we could try 5x2 - 3x + 7 minus 4x^2 - 8x - 11 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^2 - 3x + 7 and if we distribute the negative sign to everything on the right all the signs will change the POS 4x^2 will now become 4x2 the 8X will now change to positive 8X and the same is true for the 11 it's going to change from1 to POS 11 so now let's combine like terms 5x^2 - 4x2 is 1 x^2 -3 + 8 which is the same as 8 - 3 is positive 5 and 7 + 11 is 18 so you're going to get this now what if we want to multiply two expressions instead of adding and subtracting pols 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 polom is 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 you multiply the first two terms 3x * 2x is 6 x^2 and then you multiply the term on the outside 3 x * -6 which is -8x and then the ones on the inside -5 * 2x that's -10x and then the last ones5 * -6 which is POS 30 so now let's combine the two terms in the middle since they're like terms -8x + -10x is - 28x so this is the answer now how can we expand an expression that looks like this what would you do if you saw that on this s to expand it the square means that we have two 2x - 5 factors multiply to each other so we can use the foil technique again 2x * 2x 2 * 2 is 4 x * X is x^2 2x * -5 is -10x and5 * 2x is also -10x -5 * 5 is postive 255 now we can add like terms -10 + -10 is -20 and so this is the answer now what about solving linear equations how can we solve for x what number + 6 is 11 it turns out that 5 + 6 is 11 so X is equivalent to 5 but to find the answer notice that we see x + 6 the opposite of addition is subtraction so we need to subtract both sides by six 6 - 6 is zero so these two cancel you can bring down the X and 11 - 6 X is 5 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 4 / 4 is 1 so you get 1 x which is simply equal to x 8 / 4 is 2 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 - 5 is 21 next let's divide both sides by three 21 / 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 in A7 or you could subtract both sides by 8 I'm going to subtract both sides by eight so on the left we have 4 * 2x - 7 and on the right we have 20 - 8 which is 12 now at this point you can distribute the four or you can divide both sides by four if you do that you'll be left with 2x - 7 is equal to 12 / 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 + 7 that's 10 and if we divide both sides by two 10 / 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 2 * 5 is 10 10 - 7 is 3 4 * 3 is 12 and 12 + 8 is 20 so it works now let's move on to solving and graphing inequalities so let's say if we have the expression 2x + 5 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 five so 2x is greater than 11 - 5 which is 6 next let's divide both sides by two 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 uh 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 three and we're stopping at Infinity so you can write the answer from three 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 7 so 13 - 7 is 6 next let's divide both sides by three 6 / 3 is two so two is to the right of zero now because it's equal to or greater than two we need to use a closed Circle we're going to shade it to the right so the answer in interval notation is from 2 to Infinity But it includes two so we need to use a bracket instead of a a parenthesis whenever you have a closed Circle use a bracket for Open Circles you use parenthesis and infinity you should always use parenthesis let's try another example let's say that -3x is greater than or equal to 9 solve for x so let's begin by dividing both sides by3 9 I mean positive 9 / -3 is -3 when you multiply or divide by negative number the inequality changes sign so you have to flip it so that's the solution X is less than or equal to3 so-3 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 is L than we need to shade towards left so in interal notation the answer is from negative Infinity to3 But it includes ne3 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 -5x + 8 is greater than -7 so let's begin by subtracting both sides by8 so -5x is greater than -7 - 8 which is-15 now let's divide both sides by5 so the inequality will change sign or Direction so X is now less than POS 3 so let's use an open circle and we're going to shade to the left so VI it 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 3 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 -3 and 3 the absolute value of positive3 is pos3 the absolute value of3 is still pos3 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 4 what is the value of x x can be two answers it can be positive4 or4 the absolute value of pos4 is 4 and the absolute value of4 is pos4 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 + 3 is 11 to solve it write the original equation without the absolute value expression and for the second one make it equal to1 so here let's subtract both sides by three 11 - 3 is 8 and then let's divide by two so the first answer is pos4 now let's get the second answer1 - 3 is -14 and if we divide this by two -14 / 2 is -7 so here's the second answer so X can equal 4 OR7 try this one let's say that the absolute value of 3x - 1 is greater than 5 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 NE five and then solve for both equations so let's add one 3x is greater than 6 and then let's divide by 3 so X is greater than 2 6 / 3 is two here let's do the same thing let's add one to both sides -5 + 1 is -4 and now let's divide by 3 so X is less than 4/3 and 4/3 is 1.33 as a decimal so now what we're going to do is plot the solution so here's two and 1.33 is somewhere between 1 and 2 so X is greater than 2 so we have an open circle shaded to the right and it's less than 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 interal notation so we have two sections for the blue section the answer is infinity to 4/3 for the red section it's 2 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 -2 - 4 is -6 these two will cancel and 13 - 4 is 9 next let's divide all three sides by three so you can just do it simultaneously -6 / 3 is -2 and since we divide it by a positive number the direction of the inequalities will not change 9 / 3 is 3 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 -2 and 3 so this expression tells you that -2 is greater than or equal to X or if you read it backwards X is less than or equal to -2 so 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 -2 or three so this is like the other one so the answer is from negative Infinity to -2 Union 3 to Infinity But it includes two and three so we need to use brackets let's try another example let's say 2x + 5 is less than or equal to let's say uh 13 but it's greater than or equal to uh 1 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 1 - 5 is-4 13 - 5 is 8 next let's divide both sides by two so X is greater than equal to -2 but less than or equal to 4 so therefore X is between -2 and pos4 so it's here the answer in interval notation is from -2 to 4 now let's talk about how to to graph linear equations in slope intercept form slope intercept form looks like this Y is equal to mx + b m represents the slope B is the Y intercept so let's say if we want to graph this equation Y is = 2x - 3 so let's make a graph first so first plot the Y intercept so that's the 03 or simply -3 on the Y AIS the slope is 2 which is the same as 2 over 1 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 one1 so the next point is over over here and then to find a second Point go up two units over one which will take us to 2 comma 1 and then go up to 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 = -3 4x + 5 how can we graph this particular equation so we can see the slope is -3 over 4 so the rise is-3 the run is four the Y intercept is five so let's plot the Y intercept at five to find an X point the rise is3 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 ax plus b y equal 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 + 3 Y is equal to 6 so let's begin by finding the x intercept to find the x intercept replace y with zero so 2x + 3 * 0 is equal to 6 3 * 0 is just 0 so 2x is equal 6 and X is string so the x intercept is 3 comma 0 now let's find a y intercept to do that replace x with 0 and solve for y now 2 * 0 is zero so we just have 3 Y is equal to 6 and if we divide by two I mean by 3 6 ID 3 is 2 so the Y intercept is 0a 2 so what we're going to do is simply plot the X x and y intercepts and then connect them with a line so the x intercept is 3 0 so that's x = 3 on the x- axis which is right here and on the Y AIS we need to plot two or 0 comma 2 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 - 4 Y 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 3 x is equal to 4 so the x intercept is 4 comma 0 now let's find a y intercept and let's do that by replacing x with 0 so 3 * 0 is 0 so we just have -4 Y is equal to 12 and now let's divide by4 so 12 id4 is-3 so the Y intercept is 03 so the x intercept is 4 units to the right the Y intercept is three units down so 03 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 the general shape it looks like a vshape 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 going to have the point 1 one and then one to the right up one so you have the point 2 two and 3 three and so forth the left side is a reflection of the right side relative to the vertex the vertex is at the origin 0 0 so the graph looks like this now what about if we put a two in front of it so now the slope has a value of two 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 1 comma 2 and then to the left it's going to be the same and if we travel one more unit to the right we need to go up two so it's going to be 2 four and -24 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 vshape 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- axxis so instead of opening in the upward Direction it's going to open downward now what if we have this expression absolute value x + 2 on the inside what's going to happen to find out set the inside equal to Zer if you solve for x you'll see that X is -2 so the graph is going to start at -2 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 xus 3 this is going to shift three units to the right if you set the inside equal to zero X will equal 3 so the new origin is at 3 comma 0 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 two will cause it to shift two units up but it's still going to open in the upward Direction so let's say if we have a negative absolute value of x minus one 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 = x^2 is a u shape instead of a vshape and it opens downward So based on your knowledge of Transformations what's the shape of the graph negative x^2 so putting a negative in front will cause it to reflect over the xaxis so it's going to open downward now try these two examp examples let's say if we have x - 2^ 2 and x + 4 2 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 four units to the left and it's going to open upward so try this one x^2 - 1 and x^2 + 3 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 s it's going to open downward so now let's try this one X x - 1^ 2 + 2 and 3 - x + 2^ 2 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 -2 comma 3 and because of the negative sign in front is going to open downward so that's how you can graph uh quadratic expressions using Transformations now let's talk about how to solve quadratic equations by factoring let's begin with this quadratic expression x^2 - 25 is equal to 0 how can we Factor the expression and also solve for x so there's something called the difference of perfect squares X squ is a perfect square and 25 is a perfect square to factor it's going to be a minus B * a + b the square root of x^2 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 equals to zero so let's set x + 5 = 0 and x - 5 = 0 so the first answer X is = to5 and for the second answer it's equal to positive five now let's work on another example x^2 - 36 is equal 0 the square otk of x^2 is X and what is the sare < TK of 36 it turns out that 6 * 6 is 36 so that's the square root of 36 and one will be positive the other will be negative and then solve for x so X is equal to -6 and positive 6 try this one so what is the square root of four the square < TK of 4 is 2 and theare otk of x^2 is X so theare otk of 4x^2 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 -7 and then let's divide by two so for the first 1 x is equal to 7 /2 now here let's add seven to both sides so 2x is equal to postive 7 and then let's divide by two so X is 7 over2 so it's posi 7 over2 and Nega 7 over2 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 3x^2 / 3 is equal to x^2 -48 / 3 is -16 so now we have two perfect squares the square root of X2 is X the sare OT of 16 is 4 one is positive the other is negative and based on the last examples we could easily tell that X will equal ne4 and positive4 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 to 24 we have 1 and 24 2 and 12 3 and 8 and 4 and 6 however only 4 + 6 adds up to 10 so to factor it it's simply going to be x + 4 * x + 6 and then to solve the um to solve for x simply change the sign so X is going to be -4 and -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 -4 but add to positive 5 so we have 1 and -4 2 and -2 3 and8 notice that 3 +8 is5 if we reverse it -3 + 8 is POS 5 but these two numbers still multiply to -4 so it's going to be x - 3 * x + 8 and therefore X is going to equal the reverse that's POS 3 and8 here's another one x^2 - 2x - 35 is equal to 0 so what two numbers multiply to -35 and add to -2 7 * 5 is 35 so we can do it as -7 and 5 or 7 and neg5 7 +5 adds up to positive2 but -7 + 5 adds up to -2 so this is the one we want to use so it's going to be x - 7 * x + 5 and so X will be equal to POS 7 and X will be equal to5 let's try one more example where we have a trinomial with a leading coefficient of one 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 -3 if we divide 42 by- 1 we'll get -42 if we divide it by -2 we'll get -21 if we divide it by -3 we will get -4 4 doesn't go into 42 however we can divide 42 by -6 and that will give us -7 and -6 + -7 adds to -3 so we can Factor it as x - 6 * x - 7 so X is equal to POS 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^ 2 + BX + C so a is the number in front of X2 so a is 1 B is -3 C is 42 to use the quadratic equation you need to know the formula it's netive B plus or minus b^2 - 4 a c / 2 a b is going to be * -3 plus orus b^ 2 -3 2 is 169 if you multiply -3 by3 you will get positive 169 a is 1 and C is 42 so divid by 2 a or 2 * 1 which is two so let's get rid of a few things * -3 is pos3 and then it's plus or minus Square < TK 169 and 4 * 42 is 168 and we have a two on the bottom so 169 - 168 is 1 and theare < TK of 1 is equal to 1 so we have 13 + or- 1 divid two so at this point we could set that expression equal to two separate Expressions so we could say it's 13 + 1 / two or we could say it's 13us 1 / 2 so as you can see we're going to get two answers 13 + 1 is 14 and 13 - 1 is 12 14 / 2 is uh 7 and 12 / 2 is 6 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 2x^2 + 5x - 12 is equal to0 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 * -12 is -4 then find two numbers that multiply to -4 but that add to the middle coefficient of five so we know 8 * 3 is 24 but one of these numbers has to be negative let's put the negative with the 3 8 * -3 is -4 but 8 +3 is POS now what we're going to do in this example is that we're going to replace the 5x with 8X and -3x because 8x - 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 have has the same ratio as the last two terms for instance consider the two and 8 8 / 2 is the same as -12 /3 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^2 ID 2x is X 8X / 2x is 4 now in the last two terms let's also take out the GCF which is -3 -3x / -3 is X and -12 / -3 is pos4 so if you see these two factors if they're the same that means you're on the right track that particular Factor right at 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 is4 and then let's add three to both sides so 2x is equal to 3 which means X is 3 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 could see a is equal to 2 B is equal to 5 and uh C is equal to -12 so X = to B plus orus < TK b ^ 2 - 4 a c / 2 a so B is 5 which means b^ 2 or 5 * 5 is 25 - 4 * a which is 2 * C which is -12 / 2 a or 2 * 2 so this is -5 plus or minus 25 now 4 * 2 is 8 8 * 12 is 96 and because of the two negative signs it's going to be positive 96 and on the bottom 2 * 2 is 4 now 25 + 96 is 121 and the square root of 121 is 11 so we have ne5 plus or- 11 / 4 now let's separate it into two expressions -5 + 11 over 4 and5 - 11 over 4 so5 + 11 is the same as 11 - 5 which is POS 65 - 11 is -6 now 6 over 4 we can reduce it if we divide both numbers by two 6 ID 2 is 3 4 ID 2 is 2 so the first answer is 3 over two which is an answer that we had before now -16 / 4 is4 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 facture this expression 12x^2 - 83 3x + 66 Now using the same Technique we would have to multiply 12 by 66 12 * 66 is a big number it's 792 what two numbers multiply to 792 but add to 83 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 83 and C is 66 so X will be equal to B that's 83 and then plus or minus Square < TK b^2 so 83 * 83 or 83 2 is 6,889 - 4 * a which is 12 * C which is uh 66 / 2 a or 2 * 12 so * 83 is postive 83 4 * 12 is 48 * 66 that's 3,168 and 2 * 12 is 24 now let's subtract 6,889 by 3168 so that's 3,721 and the square root of 3721 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 - 61 over 24 83 + 61 is 144 83 - 61 is 22 144 / 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 two if we do that 22 / 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 zero on one side of the equation so here let's subtract both sides by six 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 - 6 is equal to 0 these two cancel and for this one x - 11 /2 is equal to uh 0 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 x * 12 is 12 x 11/ 12 * 12 the 12's cancel giving you 11 so this is going to be minus 11 and 0 * 12 is 0 so these are the two factors so therefore to factor the expression it's going to be x - 6 * 12x - 11 and now let's make sure that this isn't indeed the right expression so let's foil it x * 12x is 12 x^2 x * -11 is -1x -6 * 12x is - 72x and -6 * -1 is POS 66 now the two numbers that multiply to the product of 12 12 * 66 which was uh 12 * 66 was 792 remember we needed two numbers that multiplied to 792 but added to 83 it turns out that those two numbers are the two numbers in the middle -1 * -72 is positive 792 and they add to 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^2 isal to 9 and x^2 = to9 let's solve for x in both equations to do that we can take the square root of both sides the otk of x^2 is simply X the < TK of 9 is plus or minus 3 now on the left side we're going to have plus or minus the square < TK of9 what is the square root of9 it's important to understand that the square root of1 is equal to an imager number in this case I so9 can be written as 9 * -1 the < TK of 9 is 3 and theare < TK of1 is I so in this case instead of getting plus orus 3 the answer is plus orus 3 I so it's positive 3 i and-3 i i is always equal to the square root of1 I 2 is equal toga 1 now I to the 3 power which is the same as I * i^ 2 that's equal to I and i^2 is1 so I to the 3 power is simply equal to I now what about I to the 4th power I to the 4th power is basically i^ 2 * another i^ 2 and each I 2 is1 so1 * 1 is simply equal to 1 so make sure you know these four values I is equal to the < TK of1 i^ 2 is 1 I Cub is the same as negative I and I to 4 is one now let's talk about solving systems of equations so let's say if we have two linear equations 2x + y is equal to 5 and 3x - Y is equal to Z 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 2x + 3x is 5 5 + 0 is 5 so to solve for x uh let's divide both sides by five 5 ID 5 is 1 so X is 1 so now what we're going to do is uh replace x with one in the first equation so 2 * 1 + y is equal to 5 2 * 1 is 2 and now we need to subtract both sides by two so 5 - 2 is equal to 3 so the solution is 1 comma 3 let's try another example 3x + 4 Y is equal to 25 and 5x - 3 Y is equal 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 four is 12 if you're not sure what the least common multiple is you could just multiply 3 and four 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 four if we do that 4 y * 3 is 12 y -3 y * 4 is -12 y so they're going to be the same but they're going to have the opposite sign so they will cancel now now we got to multiply everything by three in the first equation whatever you do to the left side you must also do to the right side 3x * 3 is 9x 4X I mean 4 y * 3 is 12 Y and 25 * 3 is 75 now let's multiply the entire equation by 4 5x * 4 is 20x -3 y * 4 is -12 y 3 * 4 is 12 so if we add the two equations we can see that the variables will cancel 99 + 20 is 29 75 + 12 is 87 and if we divide 87 by 29 that's equal to 3 so X is three so now let's rewrite the first equation 3x + 4 Y is equal to 25 and let's replace x with 3 and let's solve for y 3 * * 3 is 9 25 - 9 is 16 16 / 4 is four so Y is equal to 4 so the answer is 3 comma 4 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 + 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 - 2x is 3x and now let's add four to both sides 5 + 4 is 9 the last thing we need to do here is divide by three 9 ID 3 is 3 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 2 * 3 is 6 + 5 that's 11 and if we plug it into the other equation 5 * 3 is 15 - 4 is also 11 so the answer is 3 comma 11 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 - 5 since they're equal to each other so let's go ahead and do that so we're going to have 2X + 4 instead of 4 y it's going to be 4 * 7 x - 5 which is equal to 10 and now we can solve for x so let's distribute the four 4 * 7 x is 28x and 4 * -5 is -20 now we can combine like terms 2X and 28x is equal to 30X and now let's add 20 to both sides 10 + 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 y so Y is equal to 7 * 1 - 5 7 - 5 is 2 so Y is 2 so the solution is 1 comma 2 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 ofx is equal to 5x + 4 what is the value of f of3 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 * 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 ofx 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 ofx 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 ofx with 39 and solve for x 39 - 4 is 35 and 35 ID 5 is 7 so X is 7 which means F of 7 is equal to 39 now what is the value of f of five 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 5 to the 3 power which is 5 * 5 * 5 that's 125 5^ 2 is 25 * 4 is 100 6 * 5 is 30 and 125 - 100 is 25 30 - 14 is 16 25 + 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 five is four so we're going to put a five here and we need to place the coefficients 1-4 614 in decrease in order so let's bring down the one 5 * 1 is 5 and then add -4 + 5 is 1 5 * 1 is 5 6 + 5 is 11 5 * 11 is 55 -14 + 55 or 55us 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 ofx is 3x + 5 and G of x isal to x^2 - 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 ofx is 3 * x + 5 but instead of writing the X we're going to replace x with x^2 - 4 so instead of writing 3x + 5 it's going to be 3 * x^2 - 4 + 5 now let's distribute the 3 so it's 3x^2 and 3 * -4 is -12 and then we can add -12 + 5 so it's 3x^2 - 7 so F of G of X is equal to this expression now what about G of f ofx 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^2 - 4 instead of writing x^2 we're going to write 3x + 5^ 2 so we need to foil uh the two expressions 3x + 5^ 2 is 3x + 5 * 3x + 5 3x * 3x is 9 x^2 and 3x + 5 or 3x * 5 is 15x and 5 * 3x is also 15x and 5 * 5 is 25 we can add like terms so 15x + 15x is 30X 25 - 4 is 21 so the composite function is equal to that now let's say if we wish to evaluate F of G of2 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 2^ 2 - 4 2^ 2 is 4 4 - 4 is 0 so G of 2 is equal to Z so we can replace G of 2 with 0 so now we're looking for f of 0 which is 3 * 0 + 5 using this equation and that's equal to five so F of G of 2 is five let's try another example what is g of f of3 so first let's evaluate F of three that's 3 * 3 + 5 3 * 3 is 9 + 5 is 14 so f of3 is 14 so now we got to find G of 14 which is 14^ 2 - 4 now 14 * 14 or 14 s 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 ofx is the original function and F to the1 ofx represents the inverse function so let's say that for the original function we have the points 2 three 4 -6 and 52 to find a points of an inverse function you need to switch the X and Y values so it's going to be 3 2 -64 and 25 now let's talk about the graphical relationship of an inverse function so let's say this function is f ofx the inverse function is going to look something like this and that's going to be F1 of X now notice that these two functions are symmetrical about the line yals X that's a property of inverse functions they will always be symmetrical about that line now let's say that f ofx is = to 7X - 5 how can we find the inverse function in order to do that replace f ofx with Y and in your next step switch X and Y and then solve for y so we need to add five to both sides let's move the Nega five 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 + 5 is equal to 7 y next let's divide both sides by seven so the inverse function is uh x + 5 over 7 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 ofx 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 seven X - 5 it's going to be 7 * x + 5 over 7 so notice that the Sev cancel leaving behind x + 5 - 5 and these cancel give it us X so the first expression is true F of G of X is X so now let's check out G of f ofx so this time we're going to take F insert it into G so instead of writing x + 5 / 7 it's going to be 7 x - 5 which was in place of X and then + 5 over 7 so 5 + 5 is 0 7 x / 7 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 uh share this video with your friends also if you want to find more uh 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 scratch the surface of all the algebra topics that you need to know so if you check out my playlist you can find other topics uh throughout the uh College alible 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 uh have a great day