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Contents

- 1 Doubling Booth Area
- 2 One Foot Longer and Wider
- 3 One Foot Longer and Wider
- 4 Increasing Area
- 5 Increasing Area
- 6 Nonlinear Behavior
- 7 Nonlinear Behavior
- 8 Equation for Area
- 9 Equation for Area
- 10 Switching Sides
- 11 -Keep your zs straight
- 12 -Keep your zs straight
- 13 -Sum of Squares
- 14 -Sum of Squares
- 15 -Linear Factors
- 16 -Linear Factors
- 17 -Find x Intercepts
- 18 -Solving Graphically
- 19 -Solving Graphically
- 20 -Solving Algebraically
- 21 Switching Sides
- 22 -Solving Algebraically
- 23 -Solving Harder Equation
- 24 -Solving Harder Equation
- 25 -Points of Intersection
- 26 -Points of Intersection
- 27 Equivalent Equations
- 28 Equivalent Equations
- 29 Distributing Multiplication
- 30 Distributing Multiplication
- 31 Matching Terms
- 32 Factoring
- 33 Factoring
- 34 Finding a and b
- 35 Finding a and b
- 36 More Factoring
- 37 More Factoring
- 38 a and b Added and Multiplied
- 39 a and b Added and Multiplied
- 40 Calculate a and b
- 41 Calculate a and b
- 42 Switching the Sign of 8
- 43 Switching the Sign of 8
- 44 Positive Product but Negative Sum
- 45 Positive Product but Negative Sum
- 46 More a and b
- 47 More a and b
- 48 Finally Factoring
- 49 Finally Factoring
- 50 More Practice
- 51 More Practice
- 52 Another Sign Change
- 53 Another Sign Change
- 54 Factoring for Grant
- 55 Factoring for Grant
- 56 Solving Grants Equation
- 57 Solving Grants Equation
- 58 Side Length Addition
- 59 Side Length Addition
- 60 Linear Factors
- 61 Linear Factors
- 62 Greatest Common Factor
- 63 Greatest Common Factor
- 64 Factor Again
- 65 Factor Again
- 66 More Common Factors
- 67 More Common Factors
- 68 Factor Fully
- 69 Factor Fully
- 70 Add to B
- 71 Add to B
- 72 Add and Multiply m and n
- 73 Add and Multiply m and n
- 74 Factor Pairs of Terms
- 75 Factor Pairs of Terms
- 76 Finish Factoring
- 77 Finish Factoring
- 78 Factoring Yet Again
- 79 Factoring Yet Again
- 80 Two Methods for Factoring
- 81 Pairs of Factors
- 82 Pairs of Factors
- 83 Which Coefficients
- 84 Which Coefficients
- 85 Both Methods Work
- 86 Full on Factoring
- 87 Full on Factoring
- 88 No More Common Terms
- 89 Common Terms First
- 90 Common Terms First
- 91 Which are fully factored
- 92 Which are fully factored
- 93 Factoring Cubics by Grouping
- 94 Factoring Cubics by Grouping
- 95 Pairing Terms
- 96 Pairing Terms
- 97 Fully Factored Form
- 98 Fully Factored Form
- 99 Factor All the Way
- 100 Factor All the Way
- 101 GCFs of 1
- 102 GCFs of 1
- 103 Grouping One More Time
- 104 Grouping One More Time
- 105 Difference of Cubes
- 106 Difference of Cubes
- 107 Multiply it out
- 108 Multiply it out
- 109 Sum and Difference of Cubes
- 110 Sum and Difference of Cubes
- 111 Factor the Cubic
- 112 Factor the Cubic
- 113 The best variable is m
- 114 The best variable is m

Right now, Grant's booth area at the Expo is 10 feet wide by 15 feet long. That gives him a total area of 150 feet squared. Since area is just length times width, which is 10 times 15. Now this whole slingshot game is taking up a ton of space. In fact, he's already way outside his boundaries in terms of where the ball is ending up. He's also worried that if people throw really poorly, they're going to hit people in other booths. That would definitely not be good marketing for Grant's Gleaming Glasses. The person in charge of the Expo told Grant that he can, in fact, expand the space that he is in under one condition. He has to increase his dimensions by the same amount In both directions, if he's going to make this any bigger. For now, I'm going to call that link that we're going to add on, x. Grant, ambitious as always, wants to double the area of his booth. That way he'll have 300 square feet of space to do whatever he wants with. The question is, how much does he need increase these dimensions by? What does x need to equal to make this happen?

Let's start out by figuring out what the area would be if Grant increased each side by 1 foot.

All that Grant needs to do is to change the length measurement by 1 foot higher and to also increase the width measurement by 1 foot. Instead of having feet.

So increasing 1 foot in each direction increased the total area from 150 to will happen then if we increase the dimensions by yet another foot? Do you think the area will increase by another 26 feet squared exactly? Do you think it'll increase by more than, than this? Or do you think it will increase by less than this?

Well, let's try it out to see what happens. Adding 2 to each side length gives us 12 times 17 or 204 square feet. That's 28 bigger than before, not 26. So I guess the area will increase by more than 26 feet squared. Interesting.

Let's take a moment to examine these values of x, the amount we're increasing each dimension by, and A, the total area of Grant's display space, or at least the ones that we've seen so far. The original area, before we added anything to either dimension, was 150 square feet. When we added 1 foot in each direction, the area changed 176 square feet. So in the transition from this point to this point, x went up by 1 and A went up by 26. Let's see what happens in the next transition. Once again, x went up by 1. But this time, A increased by 28 instead of 26. This is what we call nonlinear behavior. Remember that for a straight line, which we can write as y equals mx plus b, m, the slope, is going to be the same, regardless of which points we pick. The change in x is always going to be the same. The change in x value is always going to have the same proportion to change in y value. The change between two x values and the change between two y values is always going to obey the same ratio. That is not what's happening here. This equation no longer applies. A change of 1 foot in dimension doesn't always increase the area by the same amount. This is a pretty interesting problem to have. Remembering that area is equal to length times width, can you come up with an equation that we can use to figure out what value of x will give us an area of 300 square feet? So think about how x will be involved in both the length and the width of this new area.

Well we know that we want A over here to equal 300. Now for the length, which was originally 15 feet, you want to tack on an extra distance of x to it. So our length will be 15 plus x. We do a similar thing for our width. It's going to be you have these factors in because of the commutative property of multiplication.

I could draw the dimensions of Grant's new space like this. In orange we have the original layout, and these blue lines represent the addition that he wants to make, in order to get, to a total area of 300 square feet. What I'd like you to do now, is to rewrite this equation. So that you carry out the distribution or multiplication on the right hand side. So multiply these two polynomials together, and also combine like-terms on the right side, then write your new equation down here. Again, just simplify the right side. Leave the left side alone.

We know that we're leaving the left side alone. So I'll just go ahead and write 300 again. And then we know we need to distribute multiplication by each of the terms in the first set of parentheses to each of the terms in the second set of parentheses. So let's start with the 15. I need to multiply it by the 10 and by the x. Then I do the same thing with the x. That gets multiplied by both of these terms as well. From there, we just simplify within each term and combine like terms. Awesome. But we know that I love everything written in standard form for polynomials. So that means we need the term with the highest degree first. So we get 300 equals x squared plus 25x plus 150.

Now x, which is the amount of length and width we're adding to the booth, is the variable that we're going to want to solve for in this equation. Right now, we don't know how to do this. There's this x squared term, and it's kind of confusing to think about what you do with that. So let's learn. First of all, I don't like having constant terms on both sides of the equal sign. We can pretty easily make that not be the case. If I want to get rid of the 300 over here on the left, then what am I going to be left with on the right? Fill in this box with the expression that belongs in the right-hand side of the equation that will be left after I legally get rid of 300 from the left.

Let's do 1 last example using sum and difference of cubes. If we have the expression 125Z cubed minus 1, how can we factor this? Remember that the variables that we picked in the initial original equations are just place holders. It does not matter that we're using the variable Z here and that there's 1 present over here. All that matters in a given expression is the relationship of the 2 terms to one another. The fact that this variable and this variable happen to have the same name does not mean that they are necessarily connected.

This time we have a difference of 2 cubes, and the thing that's playing the role of the y cube down here is 125z cubed. This means, that anywhere I have y written up here in the general form, I need to write the cube root of 125z cubed, the cube root of 125 is 5 and the cube root of z cubed is just z so y will be replaced by 5z. Now, instead of z cubed, which was subtracted initially, we have 1, which means that instead of z, I'll put a 1. So again, like I said in the question video, the z cubed up here, in this expression we're interested in, is not at all related to the z cubed in the general expression. These are just placeholder names. Okay, let's plug this stuff in. The thing in the form of y minus z down here is 5z minus 1 and our trinomial is going to be 25z squared plus 5z plus 1. Awesome. That was really great.

Over the past few minutes, we've talked about how the sum of two cubes and the difference of two cubes can be factored. And we learned earlier about how the difference of two squares can be factored. Seems logically then like we should also be able to factor the sum of two squares, something in the form a squared plus b squared. What is the factored form of that though? Let's talk about this. Earlier we saw a difference of two squares that was x squared minus 1. So let's turn this into the sum of two squares instead, to see if that is something we can factor. So let's consider x squared plus 1 instead of x squared minus 1. So I'd like you to think through how we might factor x squared plus 1. And if you can find a way to do it, check yes and write in the factors. And if you can't find a way to do it, then check no.

The answer is no, there is a another way to factor x square plus 1. We can rewrite this using a middle term as x squared plus 0x plus 1. And if we try to factor this, we're looking for what two numbers a and b add equals 0, and multiply it to equal 1. However, if a plus b equals 0 and that means that a is equal to negative b, that means one of the most negative, and one of the most be positive. And any positive number multiply by negative number gives us the negative number, so their product can't be 1. Now of course, the product would be zero, then this equation would be true, but then their product would be 0 not

In the last lesson we learned that linear factors of polynomials tell us about x-intercepts of the graphs that go along with those polynomials. So if we have an equation like y equals x minus 3 times x plus 1 It's going to have x intercepts of three, zero, and negative one zero. So this is a big part of our motivation for talking about factoring so much. If we can factor a polynomial then we already know all of the places where it hits the x axis. We learned before the linear factors of a polynomial tell us about where the polynomial hits the x axis. On the last quiz, we learned that x squared plus one is not factorable, however, I'm going to expand our statement about x squared plus one and tell you that it is not only not factorable, but it actually has no real linear factors. I know that we're not totally clear on what this means yet, and you'll learn in a few lessons. But let's just explore this briefly for now, touch on it a bit. Considering this, how many places do you think on the graph of y equals x squared plus one touches the x axis?

The answer is 0. The graph of y equals x squared plus one doesn't touch the x axis at all. Let's take a peek at the graph. So here is a beautiful graph of y equals x squared plus one, you can see that the lowest point on the parabola is above the x axis. So, the fact that X squared plus 1 doesn't have any real linear factors means that its graph doesn't intersect the X axis. This is a great example of a lot of the things we've been talking about in this lesson.

So let's tie all of this work on factoring back to graphs and solving equations. We started off with the equation of a curve, y equals x squared plus times x plus 2. If we wanted to find the x-intercepts of this curve, we just need to set y equal to 0. And we know that the two values of x that could possibly make this true are if x equals negative 3 or if x equals negative 2. So thinking about this, factoring gives us the power not only to find x-intercepts but also to solve equations. Let's look into this a little bit more.

Let's say that we want to solve the equation x squared plus 8 x plus 15 equals plotting the intersection of two different curves. And our second option is to solve this purely algebraically. Let's start out with the first option. So, what two lines should be graph in order to solve this equation graphically? What two lines do we want to intersect?

We just need to set y equal to each side of the equation. So our two lines or curves are y equals x squared plus 8x plus 15 and y equals 3. Let's look at what the intersection of these lines actually looks like. So here's a lovely graph, and we can see two points of intersection. At these two points, we'll be able to find x values that satisfy this equation right here.

Our second option, however, is to solve this just algebraically. We've solved quadratic equations before, when we're trying to find x intercepts. The only difference between those situations, and this one, is that we have a 3 over here. Remember that a quadratic equation, is anything that can be written in the form Ax squared plus Bx plus C equals 0. The issue here, is that we don't have know you just subtract 3 from both sides, that gives us X squared plus 8x plus twelve equals 0. Since you know how to solve this using factoring, I'd like you to tell me the possible solutions for X. In the event that you find X can take on multiple values to make this equation true, please just write all those values in this box but separate them with commas.

All we need to do to get rid of the 300 on the left is subtract it from both sides. Then we have 0 equals x squared plus 25x minus 150. So, here's an equation. We want to solve for x. Let's take a couple minutes to figure out how we can actually do that.

Factoring this just gives us x plus 6 times x plus 2 equals 0. We know that we'll make this equation true if either x plus 6 is equal to 0 or if x plus 2 is equal to 0. So we can solve for x in both of those cases. So x can either be negative 6 or it can be negative 2.

Here's one that I'd like you to do on your own. We have the equation 3x squared plus 7x minus 10 equals 2x squared plus 5x plus 5. And I'd like you to solve for x. Now I know this probably looks really complicated right now because there are a lot of terms on both sides of the equation. But for your first few steps just start out by trying to get it in the general form for a quadratic equation. So with all the terms on one side and with like terms combined, and then just 0 on the other side of the equation.

Moving everything over to one side gives us the quadratic equation x squared plus 2 x minus 15 equals 0. From here, we can factor to find the x values that make this equation true. This factors to the quantity x plus 5 times the quantity x minus 3. And the product of those 2 equals 0. Still, that means to make this true, x can either equal negative 5 or x can equal positive 3. Now, let's think about what we're really solving for here. And if we were to do this graphcially what we'll working on. Well, here we have two polynomial set equal to another. Just like in the previous problem, we can solve this problem graphically by setting this side equal to y. And this side equal to y, and then finding the intersection of those two curves. Let's try it out. You can see here that we have two points of intersection on this graph between these two curves. We have one right here, which we see does in fact have an x coordinate of negative 5. And we have one right here, which, if we follow it down, does in fact have an x coordinate of positive 3.

What I'd like you to tell me now is what the coordinates of these two points of intersection are. Now I know that this probably seems a little bit tricky to do on the graph, but we can do it algebraically. So thinking about what these parabolas are actually graphs of, which equations they're graphs of, try to solve for the x and y coordinates of these points.

We know that in order to graph these two curves, we simply set y equal to each side of the equation that we originally started with. So the two lines that we have are this orange one, whose equation is y equals 3x squared plus 7x minus these points of intersection are points that lie on both of these curves. So that means the coordinates these points satisfy each of these equations. We already know the x coordinates of each point so all we need to do now to figure out the corresponding y coordinates is to pick either of these equations and plug in each x coordinate. So I'm kind of feeling this one right here. Doesn't matter which one you choose though. You'll get the same answer regardless of which equation you pick. So starting with this equation I'm going to plug in the negative 5 to see what y value corresponds to that x value on both curves. When I plug in x equals negative 5 I end up with y equals 30. So that means that the coordinates of this point must be negative five comma 30. Now let's do the same thing with x equal three, for this point. Solving for y when x equal three gives us 38, so we have the point three comma 38 as our other point of intersection.

You might remember that in the last lesson, we had two equations, y equals negative x squared over 8 plus 4x minus 14, and y equals negative 1/8 times the quantity x minus 4 times the quantity x minus 28. And these two equations were equivalent to one another. We also saw that if we wanted to look at the quadratic equations that corresponded to the equations for parabola, that there was some benefit in using this form of the quadratic equation, because from it we could read off the x-intercepts of the parabola. We were just looking at this equation that we modified to deal with the dimensions of Grant's expanding floor space. We had 0 equals x squared plus 25x plus 150. Now what is this? Is this a linear equation, is this an equation for a parabola, or is this a quadratic equation?

This is a quadratic equation. We know it's not linear because it has an x squared term in it. And we know it's not an equation for a parabola because there's only one variable, x. There's no y over here that we'd be able to relate to x to graph the parabola on a coordinate plane.

Now that we have formally recognized that this is a quadratic equation, it would be great if we could get this in the form x minus a times x minus b so that we would know what values of x satisfy this equation. It's way easier to solve if we have two factors multiplied together than if we have this ugly squared term. So let's take a second and just look at this product for a moment. Now actually, to make your life easier, I'm going to make these plus signs instead of minus signs for a second. So you have x plus a times x plus b. As a first step in exploring this, let's just multiply these two quantities out. Can you rate this in the standard form for a polynomial?

We get x squared plus ax plus bx plus ab.

So you're pretty proficient at this point at multiplying polynomials together. And then of course simplifying what you get. But if we can work from having two binomials. Binomials are just polynomials with two terms in them. We can work from having those multiplied together to having something that looks sort of quadratic. We should be able to start with something quadratic and work backwards, to having binomials multiplied together. And the problem with Grant, we're trying to go that direction. So let's say we have some simple quadratic expressions, like x squared, plus 5x, plus 6. Can we figure out how to factor this, based on patterns in the above equation? The x squareds correspond directly, and the 6, and the ab, are both constant terms, so they must go together. If this middle term, the 5x and ax plus bx that are going to be a little bit more tricky. So let's focus on this part of the term.

Can you figure out a common factor to pull out of ax plus bx? Write this in factored form.

There's an x in both terms, so you can pull that out, leaving us with x times the quantity a plus b. Of course, you could also write this differently, as a plus b times x. In fact, I prefer this way, so that's what I'm going to use. That way our x is behind our coefficients. That's pretty.

Now we can come into our earlier equation and replace ax plus bx with our new quantity. Now we can compare a bit more easily. In fact, let's compare these expressions directly. Let's say for the moment that they're equal to one another, so that we can directly compare the coefficients. If this is the case, then we must have a plus b equals 5, since these are the corresponding coefficients of the x to the first terms. We also know that ab must equal 6. So you have two numbers a and b that add to give us 5 and multiply to give us 6. What do you think those two numbers are?

2 plus 3 is 5 and 2 times 3 is 6, so our numbers must be 2 and 3.

Considering the answer to the last quiz, that 2 and 3 are the numbers that add to equal 5 and multiple to equal 6, and also reflecting on the factored form of this polynomial up here, the general one, what's a way we can write x squared plus 5x plus 6 in this form? Think about what a and b each equal, and then try to replicate that in these two parantheses.

It doesn't matter what order these factors go in. One of them is x plus 2, and one of them is x plus 3.

Let's say we have the polynomial x squared plus 8x plus 15, and we want to factor it. For the moment, I'm going to say that the factors are x plus a and x plus b. So their product is just this polynomial. Now thinking about how multiplication of binomials works, what number do a and b need to add to equal, and what number do they need to multiply to equal? Think about the quadratic expression on the left.

Remember that if we multiply these two binomials together, we get a general answer of x squared plus the quantity a plus b times x plus ab. So a plus b, we can see, is the coefficient of an x to the first term, which in the expression that we're trying to match is 8. So a plus b equals 8. Now a times b shows up in the constant term of this expression, which in the expression we're going for is

Now that we know that these two numbers a and b need to add to equal 8 and multiply to equal 15, can you fill in what they are down here? This is going to give us the factored form of x squared plus 8x plus 15. Doesn't matter which number goes in which box, as long as you get the right combination of two numbers.

So let's look at some numbers that multiplied, equal 15. Well,of course we have 15 and 1, but 15 plus 1 is not 8, so that can't be right. What else? 3 times 5. 3 times 5 equals 15. And, look at that. 3 plus 5. If we plug those two numbers in over here, equals 8. So, one of these must be 3 and one has to be 5. So, now we know that x squared plus 8x plus 15 equals x plus 3 times x plus 5, awesome. Now get ready for a little bit more practice, with factoring just like this.

We just looked at the polynomial x squared plus 8x plus15. But what if I change this very slightly. Instead, I'd like us to look at x squared minus 8x plus 15. Let's see if we can factor this. As before, I'm going to say that this is equal to the product of two binomials. One of which is x plus some number a, and the other of which is x plus some number b. In this case, what do a and b need to multiply to equal. And what do they need to add to equal?

We know that a times b is just equal to this constant factor back here of 15. And a plus b wound up being the coefficient of the middle term, the x to the first term. And remember that coefficient includes the sign, which makes it negative 8. Great.

So here we have two numbers that multiply together, to equal a positive number, but add together to equal a negative number. So, in the general case, not just this specific one we're looking at, if we have any two numbers that multiply to equal a positive number, but that add to equal a negative number, what can we say about those two numbers? Are they both positive, are they both negative, is one positive and the other negative? Is one zero, or are both of them zero? Pick which of them you think is the best choice.

Both of the numbers must be negative. For example, negative 1 times negative statement can be said about any two negative numbers.

So going back to our earlier problem, if a times b equals 15, and a plus b equals negative 8, then what are the values of a and b. Just to remind you in case it's not completely clear right away, it doesn't which number we call a and which number we call b. As long as we have the two right numbers that satisfy these two equations.

We saw earlier that 15 is equal to 3 times 5, but we know that this time a and b need to both be negative. Well that's easy. 15's just equal to negative 3 times negative 5. And sure enough, those also add to equal negative 8. Order again here does not matter. It doesn't matter which one we call a and which one we call b. Just matters that we got these two numbers. Great job.

So finally, how can we factor x squared minus 8x plus 15? Think about the values that you just got for a and b, and how they fit in the form of binomials that are multiplied together to give quadratic expressions. This is a little bit tricky. So just give it a couple tries if you need to.

Since we had a and b equaling negative 3 and negative 5, our answer is x minus 3 times x minus 5.

What if we have x squared minus 5x minus 6? First tell me what a times b and a plus b must be, and then use those two numbers to figure out what our factors must be. Think really carefully about what signs you need on these two numbers.

If we're looking for 2 numbers, a and b, to add to our xes, and our 2 binomials up here. They need to multiply to equal negative 6. And they need to add to equal negative 5. This is pretty tricky. When I think about 6 and 5, I think about 3 and 2 as numbers that might work. Let's see. If we have negative positive 2. But negative 3 plus positive 2 equals negative 1, which is not negative 5. Hm. We need something different. Well, how about 6 and 1? Negative 6 times positive 1. There we go. And negative 6 plus 1 equals negative 5, so that's perfect. So our factors will be x minus 6, and x plus 1.

What if instead I want us to look at x squared plus 5x minus 6? If one of these binomials is something like x plus a and the other is x plus b, then what is a times b and what is a plus b? And then also, what are these factors?

A times b is negative 6, and a plus b is positive 5. Now again, I like to start off with the multiplication bit. 3 and 2 don't work again here, because we'd have to have one of them negative, one of them positive, and then they wouldn't add to 5 anymore. But, what if this time we try negative 1 times positive 6, instead of negative 6 times positive 1. 6 minus 1 is 5, so that works. So a and b must be negative 1 and 6, either way around. So we'll have x minus 1 times x plus 6.

Finally, I think we're ready to help Grant. Remember that he left us with this equation, x squared plus 25 x minus 150 equals 0. Remember that x here stands for the number of feet that Grant wants to add to each side of the booth. He's going to add the same amount to each side. So, how can we use our new found factoring skills to rewrite the left side of this equation? Please fill in the two binomial factors that multiply to equal the left side.

We know that if we want to write this in the form x plus a times x plus b, that in this case, a times b will equal negative 150, and a plus b will equal other needs to be positive. So what are some things that multiply it to equal negative 150? Well, there's negative 1 times 150. There's negative 3 times 50. There's negative 5 times 30. Oh, it looks like that one will work since, as we know, 25 is equal to 30 minus 5. So the equation now should read x plus 30 times x minus 5 equals 0.

Now we want to solve for x. We learned last lesson that when two things were multiplied together to equal zero, one of them needs to equal zero. So either we need to have a value of x that will make this binomial equal to zero or we need to have a value of x that will make this binomial equal zero. Which two values of x then, satisfy this equation?

X can either equal negative 30 or positive 5 in order to make this equation true.

Awesome. We have a solution. Actually, we have two solutions. We have two different values of x, either of which would make this equation true. At any given time, we don't use both of them. As always when we're substituting in values for variables, we're consistent about what we plug in. If we're going with x equals negative 30, we put that in every spot where x is. Let's just double-check to make sure that this is what we want to have happen. The left-hand side of the equation, if we plug in x equals negative 30, will end up being 0 times negative 35. Even though this factor over here is not 0, anything times 0 is 0. Something very similar happens if we plug in x equals 5, but I'll leave that up to you. And the question is, considering what x stands for, remember, it's the length that we're adding to each side of Grant's booth, only one of these two answers really mean makes sense. Thinking about what x stands for in our original equation, remember it's the length that Grant wants to add in feet to each side of his display booth at the expo, only one of these two answers really makes sense. So, pick one of them, and tell me how much longer each side of the booth should be.

5 feet. We can't add a negative to something to make it bigger. We can of course check this answer, too. Remember that we started off with the equation 15 plus x times 10 plus x equals 300. Let's see if 5 satisfies this. For the left-hand side, we'll have 15 plus 5 times 10 plus 5, which is just 20 times 15, which is 300. And the right-hand side is just equal to 300. So, we're good. 5 feet is the right answer.

Here's another quadratic polynomial, x squared minus 9. And even though this poly looks a little bit different, from the other quadratic polynomials we've seen, we can still write it as the product of two linear factors. Remember that, being quadratic means that a polynomial is of second degree. And being linear means that an expression is of first degree. So what two linear factors multiply to give us x squared minus 9?

x squared minus 9 is equal to the quantity x plus 3 times the quantity x minus expression that we had, that the two numbers we insert here as being added to x, need to multiply to give us negative 9 and need to add to give us 0. And sure enough, 3 times negative 3 equals negative 9, and 3 plus negative 3 or 3 minus 3 equals 0.

Here is yet another expression, in particular, a quadratic polynomial. But this time the coefficient of the x squared term isn't just a 1. It's a 2. Because this coefficient is no longer a 1, we run the risk of having all three terms here have a common factor other than 1, a factor that we can pull out of the entire expression. I'd like you to rewrite this expression with the greatest common factor that all three terms share pulled out of the expression and the rest of the expression that it multiplies in the second box.

We can see that there's a common factor of 2 in all three terms here. Each of these is divisible by 2, so we can write 2 times a polynomial. The polynomial that we get when we divide each term by 2. That way when we redistribute multiplication by 2 to each of these terms, we end up with the original expressions. We have 2 times the quantity x squared plus 6 x plus 5.

The next step is to factor this polynomial that's inside the parentheses. What does our expression look like then?

Factored out, this expression is 2 times the quantity x plus 5 times the quantity x plus 1. We can check that these are equal by redistributing the multiplication on the right side.

With the polynomial 2 x squared plus 12 x plus 10, we got really lucky in how we got to factor. Because there was a common factor of 2 between all three of these terms. But you can imagine that for some polynomials, whose coefficients in front of the x squared term aren't one, this won't be the case. Let's say for example that we have an 11 here instead of a 12. Then we can no longer begin the problem by dividing all the terms by this 2. So we're not going to end up with a quadratic expression with a 1 in front that we know how to factor right away. We need to figure out another method for factoring. I think we should backtrack to the thing that we first started factoring with. Let's look at the general case of a quadratic polynomial that's been factored, and whose leading coefficient will be 1. We have x plus a and x plus b as our linear factors. Let's go through multiplying this out again. I know we've done it many times before. Okay, so multiplying this out we get x squared plus the quantity a plus b, times x plus ab. What if instead of a and b we want to play with real numbers? But maybe in playing with those real numbers, I want to keep the equation for the quadratic in this form. Let's say we start off with a polynomial x squared plus 5x plus 6. We know that we can rewrite this, recognizing that 2 and 3 are factors of 6 and also add equal 5. As x squared plus, the quantity 2 plus 3, times x, plus 6. Which pretty simply becomes x squared plus 2x plus 3x plus 6. Great. Now this probably all seems simple leading up until now. But now I'm going to ask you to do something a little bit different. The thing that I'd like you to do that probably will feel a little bit strange is look first at just the first two terms here. And find a factor between the two of them that's in common. Pull it out and then write the rest of the equation in here, and then do the same thing for the second pair of factors. Find a common factor and factor it out. And then the expression you have after that, right here.

x squared and 2x have a common factor of x. So for the first two terms we can rewrite them as x times the quantity x plus 2. The second two terms have a common factor of 3. And that will leave x plus 2 inside these parenthesis. Awesome. Maybe now you can see that there's a method to my madness.

Now at last, I'd like you to write this initial expression in its fully factored form. To do this, look at the two terms that we have, in the step right above this one, and see if they have a common factor. Then that will be a factor that you can pull out and put in one set of parentheses and then figure out what's left over.

Our two terms are x times the quantity x plus 2 and 3 times the quantity x plus 2. Both of these have a factor of x plus 2 in them. So we can pull that outside. And what's left over? Well, an x and added to that a 3. Awesome. That's our answer.

So far, we've been dealing with quadratic polynomials, like x squared plus 5 x plus 6, to figure out how to factor this. We've been looking for two numbers that add to equal the middle term coefficient, and multiply to equal the final coefficient. So, in the case of this polynomial, we end up with 2 and 3 as our two coefficients. What if I told you, though, that this quantity a,b, in other words, the product of the two numbers that add to equal the middle coefficient, is also equal to the product of two coefficients from the original expression. Let's just play around with this for a second, though. I'm going to name each of the coefficients of each term in this three term expression A, B, and C. So, A in this case is 1, B is 5, and C is 6. So before we found the two numbers that add to equal this term, they were 2 and 3. But what if I told you that these multiply to equal not just this final term, but actually the product of two of the coefficients in this expression over here. So the two numbers we found to equal this coefficient B, multiply to equal what? Do they multiply to equal A squared, B squared, C squared, AB, BC, or AC?

The answer is AC. Since A is equal to the invisible 1 that's written right here, and we know that 2 times 3 equals 6, we also know that 6 times 1 equals 6. And 6 and 1 are just C and A, so their product is A times C.

So the two numbers that we're going to need to help us with factoring add to equal this middle coefficient but multiply to equal the product of the two outer coefficients. Now I wonder if this works if we don't have an expression just like this, where the coefficient out here is 1, in front of the x squared term. Let's say we have something different. Maybe we have 8x squared plus 26x plus in front of the x squared term is different this time than what we've dealt with before. The question I have for you then is, what are the numbers n and m for this expression right here? What two numbers add to equal 26 and multiply to equal 8 times 15?

20 and 6 add to equal 26 and multiplied equal 120, which is the same as 8 times 15.

Now we can use those two numbers, 20 and 6, and insert them, as a sort of expansion, of this middle coefficient, for the x, for the first term. Now we're getting much, much closer to being able to factor. I'm going to take one step further in simpifying this for you, and then we'll have a little quiz. We know that this expression is equal to 8x squared plus 20x plus 6x plus 15. And now I'd like you to begin the Factoring. To do that, I'd like you to start off by looking at the first two terms, pulling a common factor out of them, and then doing the same thing for the second two terms, with their common factor. Write the new version of the expression in these two boxes right here.

First we're going to consider just 8x squared plus 20x. In the two coefficients, there's a common factor of 4. And in the variables, there's a common factor of x. So we have 4x times what's left over in each term, 2x plus parts of each term, since there's no variable portion of 15. And 6 and 15 are both divisible by 3. And then we're left with, surprise, 2x plus 5. Wonderful. So we have 4x times the quantity 2x plus 5 plus 3 times the quantity 2x plus 5.

We are so close to being done. I'd like you to take the one last step to fully factor this expression, 8x squared plus 26x plus 15. Write the two factors in these two boxes down here.

We now have two things added together. We have 4 x times the quantity 2 x plus common factor of 2 x plus 5, so we can start by pulling that outside. And then, we can multiply that, and then we can rate the things that it multiplies. 4 x plus 3. So the final factored form of 8 x squared plus 26 x plus 15 is just this expression, the quantity 2 x plus 5 times the quantity 4 x plus 3. Awesome. Let's check that just to make sure. We multiply 2 x by each of these terms, and then we multiply 5 by each of these terms. All we need to do now is add like terms, and voila, we have verified that this factored form is in fact equal to the expression we started out with. Awesome.

Let's do a quick overview of the method we used to get from this quadratic polynomial, which does not have 1 as its coefficient of the first term, to its fully factored form. So you started off by finding two numbers that add to equal this middle coefficient. And then multiply it equal the product of the two outer coefficients. Let's see if we can try that method out with another problem. What if we have the expression 5x squared minus 28x minus 12? How would you fully factor this quadratic polynomial? Remember the method that we used for the last one. And if you need to review it, just go back and peek at the past videos.

The first thing we're going to do is look for two numbers that add to equal negative 28. So we can write this middle coefficient in expanded form and factor from there. We know that those two numbers also need to multiply to equal the product of these outer two coefficients. So, 5 times negative 12 in this case, which is negative 60. So, what adds to equal negative 28 but multiplies to equal negative 60? Well, 60 is just equal to 30 times 2. So, we have negative 30 times positive 2, then negative 30 plus 2 equals negative 28. So those must be our two numbers. That means we can rewrite this expression as 5 x squared plus the quantity 2 minus 30 times x minus 12. I personally think that's the hardest part of the problem, is figuring out how to rewrite that middle coefficient. From here, it's pretty straightforward. Now, we just factor within the first two terms and then within the second two terms. We recognize a common factor of 5 x plus 2. And finally, we get to our official answer of 5 x plus 2 times x minus

We just used one method for factoring this polynomial right here, 5x squared minus 28x minus 12. And because I know that factoring polynomials that don't have leading coefficient of 1 is a little bit tricky, I want to take a second to go more in-depth into the logic of figuring out what numbers belong in our linear factors. Saying that a polynomial like this one, of degree 2, is factorable, it means that we can write it in this form. As two linear factors, or factors with an x to the first power in them, multiplied together. And the general form of a linear factor is just like this. You have some coefficient times x plus some constant term, what we've been talking about is how to figure out what numbers belong here. Now, of course, one way to do this is to factor by grouping like we have been that entailed expanding this middle term into two terms that added together to equal it and then grouping pairs of terms together, so considering this first pair as a chunk, and the second pair as a separate chunk, and pulling the greatest common factor that we can out of either those pairs. This is why we call this factoring by grouping, because we're grouping these terms and grouping these terms. We can use a slightly different kind of logic though, to get to the same answer. If we've already written out the form that we want the equation to end up in, we can look directly at what we might want from these different coefficients and constant terms based on what we start out with. We know that the way we end up with this x squared term here is by multiplying the term with an x to the first power from each binomial in it. So, that means that whatever the coefficient is here, multiplied by this coefficient, must equal 5. We can write out all the different factors of 5, we can either have 1 times 5 or we could have negative 1 times negative 5. Now, another thing we know is that the two constant terms multiplied together, equal the constant term in the original expression. So, we can think about the factors of negative 12 as well. There are a ton of different ones here. I guess I will write them all out though just for your pleasure, and voila, there we go. I hope I didn't miss any. Now we know that the way we get this negative 28, the coefficient of the middle term here, is by multiplying whatever this number is times this number. And then, adding that to this number times this number. That means, we are going to have either 5 times some number plus 1 times some number equals to negative 28 or for the other pair over here, negative 5 times something minus 1 times something equals negative 28. Now, since I rather have two positive coefficients for x having the two negative ones, let's just select to use 5 equals 1 time 5. This makes our job a lot easier because that only leaves 1 combination of factors over here that we can plug in. You just have to figure out which one it is. And, of course, as we already saw in the previous quiz, we know that this has to be negative 6 and this has to be positive 2. I can fill in the coefficients in front of the x's, and then since I know that the two outer terms multiply when we're trying to find this middle coefficient, and the negative 6 goes in this slot, and the 2 goes in this slot, since the inner terms are contributing to this as well.

Let's try this again. What if we have the polynomial 6 x squared minus 7 x minus and to figure out like we did before what belongs in these different slots, we need to first find out what the factors of 6 are that could go here, and here, and what the factors of negative 5 are. To know what could go here or what could go here. So let's do that. For both 6 and negative 5, I'd like you to write out all of the pairs of integers that multiply to equal each number. So put one pair per box. You're not going to fill in all the boxes, just so you know. And write the pairs of numbers with just a comma separating them.

The pairs of integers that multiply to give us 6 are 1 and 6, negative 1 and negative 6, 2 and 3, and negative 2 and negative 3. The integers that multiply to give us negative 5 are just 1 and negative 5, and negative 1 and 5.

Now, you can imagine that if I have a linear factor that has a negative coefficient in front of the X that I can actually pull that negative out and instead write the factor in an equivalent from with a positive coefficient for X. The negative can be reflected in the other factor that we're multiplying by. This basically means that I can choose which coefficients go in front of our X term. And if I want, I can decide that I want those to be positive. That's definitely what I would prefer, just for ease of looking at our factorization. So let's just go ahead and remove the negative number pairs that multiply to equal 6. This makes our lives much easier. Now we only have two pairs of coefficients to check to put in these slots instead of four. So let's do just that. Let's check them. I've written out for you two possible ways of factoring our original expression of 6x squared minus 7x minus 5. Actually only one of these 2 ways is possible because you know that only one of these combinations of coefficients for x actually works. So for the one that works you're going to figure that out by figuring out which number belongs here and here or here and here depending on which one is right and then on the equation that's not going to work at all based on these first 2 coefficients. You're going to simply bubble in wrong for the 2 constant term slots. So basically, try out each of the factors of negative 5 that you found earlier. And once you find out which one works, fill in the proper numbers and then rule out the other entire equation.

So it turns out that this first combination of coefficients for x 6 and 1 does not work, this is the wrong one we can only know that though by trying out these different numbers, now if we could have put a negative one here and the negative one here that our middle term would have become 6x trans negative 1 or negative -5. We could do that same exact process for each of the 4 numbers in this slot figuring out what they would need to be over here to come up with -7 to eventually rule out all of the possibilities up here. Going with the same process down here, for the factors that have 3x and 2x in them, shows us that we need a -5 here and a 1 here .So that to get our middle term, we have 3x times 1 minus 10x, or negative 7x. Negative 5 times 1 is negative 5. So this is what works.

Now I know that this last method that we used might have felt a little bit tedious, and it definitely is a fair amount of work, but for certain equations, it's not too bad to work with. I think that it also really underscores exactly how you get from the original expression you have to the factored form. It also really highlights the connection between the different terms in each factor in the factored form. And how they connect back to the original expression. So even if you always factor by grouping from here on out this still a good experience to have had. It's your choice and frankly both methods do use some similar reasoning. You're still thinking about what numbers add to equal the middle coefficient, and multiply it to equal the product of the two outer coefficients.

Let's look at another example. How would you factor 4x squared plus 4x plus

The first thing we need to do, as always, is think about what adds to equal expression as well. Then something interesting happens when we try to factor the first two terms and the second two terms. The first two are pretty normal, we pull out a 2x from the 4x squared plus 2x, but then when we get to 2x plus 1 Well there are no common factors that 2x and 1 have aside from 1. So that's exactly what we write, 1. And then the last two terms, 2x plus 1. Conveniently, this still works out perfectly and we get 2x plus 1 times 2x plus 1. Which we can also write as 2x plus 1 the quantity squared.

Looking back at the past four polynomials that we've factored, we can notice something super important about each of the factors that we got. And that is that there are no common terms left within any of these factors. So we don't see anything like 3x minus 9 in here, for example. Because if we saw that, we would know to rewrite it as 3 times the quantity x minus 3. So you always want to double check at the end of your factoring process that each of your factors is, in fact, fully factored itself. Keep that in mind with this upcoming problem.

Keeping in mind that we don't want any common factors to be within any of the final factors we end up with that multiply to create a polynomial, how would you factor this polynomial? I'll give you a little hint, and that hint is that if there are any common factors that all of the terms have, you should start the problem off by pulling those out first.

To start off, we need to figure out what two numbers add to equal 10, and multiply to equal negative 24. It took me a little while to come up with these, since there are so many different sets of numbers that multiplied equal negative we use those in the expression, you can go through just like we normally do. This gives us the quantity, x plus 2, times the quantity 6x minus 2. But remember what we said about factoring. We're not done fully factoring if there are any common terms left within any of our factors. Notice that 6 and negative order of our factors, but this looks a little bit more normal. And here's our final answer. 2 times the quantity x plus 2, times the quantity 3x minus 1. There is however, one other way we could have done this. We can notice, right off the bat, that all of our coefficients in the original expression have a common factor of, guess what, 2. Now this expression inside here is pretty easily factorable as well, and we won't run into the problem that we did earlier, of having common factors left within our factors at the end. Go ahead and try it on your own to see if doing the problem this way gives you the same answer as the other way did. Hint, it should end up exactly the same, since we're starting with the same expression.

Which of the expressions down here are fully factored? Please check off all of the ones that you think are. A couple of these might be tricky, so be careful.

Only three of these are actually fully factored, these three down here. 3x plus 2 times 5x minus 3, x plus 1 times x minus 1, and surprisingly perhaps, 3 times the quantity x squared plus 1. In the first choice, the second factor, 6x plus 9, is actually the same as 3 times the quantity 2x plus 3. So since we can factor this more, it's not fully factored. Similarly, 10x minus 15 can be factored to equal 5 times the quantity 2x minus 3. X squared minus 1, as we learned earlier, can be factored as x plus 1 times x minus 1. And the expression x squared minus 4x plus 4, which we see multiplied by 5 here, is the same as x minus 2 squared.

We've been working pretty much solely with polynomial expressions of degree 2 since we started talking about parabolas, like this one right here. However, we know from earlier on in the course that expressions can have terms with even higher powers. Let's take a second to look at one of those right now. How about something like this, x cubed minus x squared plus three x minus three. Now, my question for you is, can we factor this? Well, we have been talking about factoring by grouping a fair amount and we already have four terms here, so let's try it out. Maybe it will work. Let's consider the first two terms together as a pair. And then the last two terms is a pair. And for each of these pairs, please write in the box below it what the greatest common factor that we can pull put is. To make my life easier in the future, I'm going to call greatest common factor, the initials GCF.

We can divide both x cubed and negative x squared by x squared. And we can divide both 3x and negative 3 by 3.

So now we know that we can pull out an x squared from the first two terms. And when we do that, we need to multiply it by two more terms in here. So that when we redistribute the multiplication, we end up with what we started with. So what I'd like you to do now, is fill in what belongs in each set of parenthesis here. Now that we've pulled out our greatest common factors of x squared for the first pair, and 3 for the second pair of terms.

The first two terms, x cubed minus x squared, are equal to x squared times the quantity x minus one. The last two terms, three x minus three, are equal to three times the quantity x minus one.

Now we have two terms. And I'd like you to write this in fully factored form by pulling out the common factor that these share.

The factor that's shared by these two terms is this x minus 1. So you can write that in, and now it belongs in the second set of parentheses is what's left over and we divide each term by x minus 1. So you have an x squared here and then plus 3 from this term. So that means that we've modified this cubic expression this third degree polynomial. To equal one linear factor times one quadratic factor.

So factoring by grouping definitely works for certain higher degree polynomials, just like it works for certain quadratic polynomials. This time, I'd like you to try factoring by grouping for this cubic expression, all on your own. Just move through this slowly and carefully, you can definitely do it, and write you final answer in this box.

Since we're factoring by grouping, we want to look at these first two terms first, 2z cubed minus 3z squared. The only common factor that they share is a z squared. So we can rewrite them as z squared times 2z minus 3. Now, onto the last two terms, negative 6z plus 9 has a common factor of negative 3. Now, of course, we could have pulled out a positive three instead and left this as a negative 2z plus 3, but I'd rather have a positive coefficient in front of my variable, that's just a personnel preference thing. Now we just pull out the common factor between the two terms we have left, that common factor is 2z minus factored.

I'd like us to just go through a couple more examples using factoring by grouping with higher order polynomials. I know this isn't really that different from factoring quadratic polynomials, but I think it's worth it to get some practice. How would you factor 4x cubed minus 12x squared plus 2x minus 6? Make sure that at the end you check the factors you've multiplied together to see if any of them have common factors within them.

The first thing that I notice when I look at this, and really the first thing that you should check for whenever you look at any polynomial expression, is that there's a common factor of 2, in every term in the expression. So let's start by pulling that out. Now we only need to concern ourselves with what's inside these parenthesis, just have to make sure that we don't lose track of this 2. Now it's time to factor by grouping. You can pull out 2x squared, out of these first 2 terms, and the last 2 terms are a little bit tricky, because they only share a factor of, well, 1, so that is what we will write. Luckily, we still see that we have a common factor, between the two terms we have left. The fully factored form of this is then, 2 times the quantity x minus 3 times the quantity of 2x squared plus 1. Now in the event that you forgot to pull out the common factor of 2 at the very beginning, all is not lost, we can still solve the problem that way too, lets check it out. If we start off by grouping, instead of pulling the common factor of 2, then we can just proceed by factoring by grouping as normal. We end up with x minus 3, times 4x squared plus 2. At this point, however, when we think that we're fully factored. We have to take that last step to double check, and make sure that neither of these factors has a greatest common factor greater than 1. We see, however, that this term, 4x squared plus 2, has a common factor of 2 between its terms. Then we pull that out. And that is what gives us our final factored form, which of course is the same answer that we got doing the problem the other way. Awesome job.

Let's do one final factoring by grouping example together. How would you factor x cubed plus 4 x squared minus x minus 4? Now, this one is definitely a little bit slippery. Think really critically as you're going through the problem and double-check at the end that there are no fractable expressions left in your final answer. Good luck.

As before, we pull out a common factor from each pair of terms, and that gets us to a point where we have x plus 4 times x squared minus 1. Now, it's really easy to think that this is the final factored form, but we are not done right now. We have to recognize that the term x squared minus 1 is just a difference of squares. We know that if we have a squared minus b squared, that can be factored as a plus b times a minus b. We can do the same thing for this expression, x squared minus 1 is equal to x plus 1 times x minus 1. This orange expression is our final factored form. This rule is definitely a really tricky problem, so great job.

Like we were just talking about, when you look at the expression, X squared minus 1, it's sometimes hard to remember that it's factorable because it looks so simple. It's kind of not that intuitive, that expression that has 2 terms. Can be factored to equal the product of 2 expressions that each have 2 terms. I know that I was pretty blown away when I first learned about the differences of squares though, we had the difference of two cubes. Something like Y cubed minus Z cubed. Since the difference of two squares is factorable, maybe this is too. I've given you 3 choices here, 2 of which are possible factorizations of Y cubed minus Z cubed, and the third choice, which is that this expression is not factorable. Please pick what you think is the best choice.

The second choice is the right one. We can check this by multiplying out every term in the first set of parentheses by every term in the second set of parentheses. If you do this, you can see that it will reduce to this expression.

So we know that this second choice is equal to y cubed minus z cubed. So now I'm really curious what this first choice simplifies to? Let's find out, please fill in, in this box what this expression simplifies to.

Our first shot at multiplying each of the terms in the first expression by every term in the second set of parenthesis, gives us this long disgusting expression. But wait, maybe it's not really that gross, let's take a closer look. When we notice which terms we have that are like terms, we see a minus y squared z and a plus y squared z. So these directly cancel each other out. We do the same thing with the other 2 middle terms positive yz squared and negative yz squared. All that we're left with in the end is a really simple expression, then. This long factorized form is actually just equal to the sum of 2 cubes. Y cubed plus z cubed. This is really incredible. Math is just so gorgeous.

So, here's a summary of what we learned over the last two quizzes. We found a general expression for the sum of two cubes and a general expression for the difference of two cubes. We can see here that each of these factorized forms has a binomial, a polynomial with 2 terms, multiplied by a trinomial, a polynomial with 3 terms. The binomial that's involved in the factorization of each expression, has the same sign between its two terms, as the original expression did between its two terms. So, y cubed plus z cubed has a factor of y plus z and y cubed minus z cubed as a factor of y minus z. The middle term of each trinomial, each quadratic expression, has a sign opposite that original sign between the terms. See we have a minus yc in the adding up the cubes and plus yc in the subtracting. Other than that, these factorizations are the same. So, now that we have this wonderful general framework for dealing with the sum and differences of cubes, let's put it into action. Let's look at the expression x cubed minus 8. Now, compared with the general equations that we were looking at before, what in this case is playing the role of y, and what is playing the role of z?

This is a difference of two cubes, and the first term, the positive term, is in the slot where y cubed initially was. So that means that x is used here in place of y. Instead of z cubed, we have 8. So that means that 8 is equal to z cubed. If we take the cube root of both sides of this equation, we get 2 on the left side and z on the right side. So z must equal 2.

Now that we've figured out, what we're using instead of y and z, we just need to substitute our new values, into the equation up here for the difference of 2 cubes. That's going to give us the factored form of x cubed minus 8. So, please write that out.

We can see here that the binomial we have is y minus z, which down here is going to mean x minus 2. So that's what belongs in this first set of parantheses. Then the trinomial is y squared which is x squared why don't you substitute in x, plus y c which is really x times 2, but I'm going to write that as 2x. And then plus z squared. Which is just plus 4. Awesome! I know it might feel a little bit silly to find a factorization that's so much more complicated than the initial expression. But once we talk more about higher order polynomials, you'll understand what the importance of writing out factorizations like this is. I'll give you a hint that we've already talked about. Each linear factor is showing us an x-intercept of a graph. There's a ton of exciting stuff ahead. So just get ready.

Here's another expression for you to factor. M cubed plus 27. M's a pretty great letter, so we might as well use it as a variable. Just as we did in the last example, be super careful about what we're going to substitute in for y and z in our original equation. Or rather, in the general form of the equation.

I still like to write out what y and z are equal to? If we initially had y cubed plus z cubed, then y is equal to m here and z is equal to the cubed root of 27, which is just 3. Three cubed gives us 27, great, that checks off. Now our binomial is just going to be y plus z or m plus 3, and our trinomial is going to be m squared minus 3m plus 9, beautiful.