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Contents

- 1 Athenas Building
- 2 Athenas Building
- 3 Area of Roof
- 4 Area of Roof
- 5 The Necessity of Polynomial Division
- 6 Divide by Factors
- 7 Focus on Leading Terms
- 8 Focus on Leading Terms
- 9 Multiply Entire Divisor
- 10 Multiply Entire Divisor
- 11 Subtract
- 12 Subtract
- 13 Squared Term
- 14 Squared Term
- 15 Bring Down the x
- 16 Bring Down the x
- 17 Subtract Both Terms
- 18 Subtract Both Terms
- 19 First Degree Terms
- 20 First Degree Terms
- 21 Find the Product
- 22 Find the Product
- 23 Final Subtraction
- 24 Final Subraction
- 25 Write as a Product
- 26 Write as a Product
- 27 Factor the Quadratic
- 28 Factor the Quadratic
- 29 Solve with Long Division
- 30 Solve with Long Division
- 31 Missing Terms
- 32 Missing Terms
- 33 Set up the Problem
- 34 Set up the Problem
- 35 Do that Division
- 36 Do that Division
- 37 Fill in the Boxes
- 38 Fill in the Boxes
- 39 Find the Remainder
- 40 Find the Remainder
- 41 The Remainder Theorem
- 42 Find the GCF
- 43 Find the GCF
- 44 Factor the Right
- 45 Factor the Right
- 46 GCF of 1
- 47 What are the zeros
- 48 What are the zeros
- 49 Form of the Fractions
- 50 Form of the Fractions
- 51 The Rational Zero Theorem
- 52 Possible Rational Zeros
- 53 Possible Rational Zeros
- 54 Are they really zeros
- 55 Are they really zeros
- 56 Is the third time the charm
- 57 Is the third time the charm
- 58 Find the Factor
- 59 Find the Factor
- 60 Divide out the Factor
- 61 Divide out the Factor
- 62 Product of Polynomials
- 63 Product of Polynomials
- 64 Does it work
- 65 Does it work
- 66 Check the Factor
- 67 Check the Factor
- 68 Finally a Factor
- 69 Finally a Factor
- 70 The Facts of Factors

Athena has a new problem that she's really stumped by. Her boss at the architecture firm, left her with an important project. Working on the plans for this awesome new building that's going to go on Grant's business property. Now, this building is special because it's a triangular prism. That means that if we look at the building from above, the roof is shaped like a triangle. In this case, it's actually a right triangle. All the faces down here though, are rectangles. Athena worked really hard to come up with an equation for the volume of this building, depending on each of the side lengths. This one, this one and this one. But unfortunately, she had a disaster with a cup of green tea and the sheet of paper she was working on. Now, all that she has left on her paper is the equation for volume in the way that she expressed the length of one side, the height here, which she labeled 2x plus 8. Now, this expression for volume is found by finding the area of the right triangle, that we see here on the roof. And multiplying that by the height. Let me write that out for you. There we go. A nice word equation for you. So, let's see if we can combine all this information together. Given that Athena already figured out this expression for the total volume of the building and she knows that the height is 2x plus 8. And also, if we decide to call the area of this top triangle A, how could you write a new equation, replacing this one in black, that only involves the variables X and A? Just use these new expressions for volume, height and area and put them in the proper places, as shown in this block equation.

First things first, we replace the volume here with this cubic expression. Then the area is just capital A, and then the height is just 2 x plus 8. And that is our brand new equation. Wonderful.

What Athena really wants to do now is to find the area of this triangle that we saw at the top of the building. This quantity a in our new equation for volume. But how? Well, her life would be made much easier if she didn't have to worry about this side length that's tacked onto the right side of the equation, 2x plus 8. Now, if we want to get rid of this factor of 2x plus 8 on the right hand side. What should we do to that equation? Should we subtract 2x plus 8 from the right-hand side? Should we subtract it from both sides? Should we divide the right-hand side by 2x plus 8, or should be divide both sides by this factor?

We need to divide both sides by 2x plus 8. Right now, this factor is multiplied in on the right hand side. So, we need to divide by it to get rid of it. However, we know that we need to do the same thing to both sides of our equation to keep a statement of equivalence. Once we do that, A will be all alone on the right hand side. Now we are going to leave this problem for this moment and you will get to solve it yourself in the practice problems

We've been doing a lot of multiplication in this course, especially multiplying polynomials together. We've seen that we can take linear factors, multiply them together and produce more complicated polynomials. With Athena's situation though, we're starting off with a complicated polynomial. In order to make it simple and find all of its factors, we need to figure out how to divide a polynomials.

In the last lesson, we left off with dealing with this function, f of x equals x cubed minus 4 x squared plus x plus 6. We had found that 2 is a zero of this function, since when we substitute it in, the entire function comes to equal 0. From this we learned, from the factor theorem, that x minus 2 must be a factor of the polynomial. So we're one step on our way to writing f of x in fully factored form. We know that eventually we'll be able to write it as x minus 2 times a bunch of other stuff. But what is that other stuff? Well, to find it, we're going to need to divide this expression by x minus 2. In particular we are going to use long division. That's what we'll look at next.

We'll set up our long division problem like this. We write the thing that we want to divide, underneath here, and the thing want to divide it by out here. Now, let's start by thinking how many times x minus 2 can go into the first term over here, the one with the highest degree. So, how many times can x minus x here, since the degree of x is higher than the degree of 2. So what do we need to multiple x by to get x cubed? Please put the answer right here.

x goes into x cubed x squared times. Or in other words, x times x squared equals x cubed.

Now what do we get if we multiply this entire quantity x minus 2 times what we wrote up here, x squared? Please write your answer in this box, and make sure that you write it in standard form for polynomials. So in other words write the highest degree term here and then the next highest here and so on and so forth.

Multiplying x squared times x minus two gives us x cubed minus two x squared.

Now we need to subtract this quantity we just wrote down here, from what is written above. So what is s cubed minus 4 x squared minus the quantity x cubed minus 2 x squared. Please write your answer down here. Note that setting up the problem this way, makes subtraction very easy, since like terms are lined up. Just be careful about our minus sign out here and the signs inside

All we get down here is negative 2 x squared. Our x cubeds gets cancelled out, and then negative 4 x squared minus a negative 2 x squared is the same as adding positive 2 x squared.

The next step is to bring down this x. So down here, for the expression, we're dividing by x minus 2 next, we're going to have negative 2 x squared plus x. We know, though, we only need to look at the first term here and the first term here. So, what do we need to multiply x by to get negative 2 x squared? Please write your answer in this box.

We need to write a negative 2x here, since negative 2x times x gives us negative 2x squared.

Now, to figure out what we should write down here, we just need to multiply the quantity x minus 2 times negative 2x. What is that product?

Negative 2 x times the quantity x minus 2, gives us the negative 2 x squared plus 4 x.

Now we need to complete the subtraction. What do we get for negative 2x squared plus x minus the quantity negative 2x squared plus 4x?

The difference is negative three x.

The next step is to bring down the 6. So that we have negative 3x plus 6, right here. Now, how many times does x go into negative 3x? Please write your answer's in this box.

x goes into negative three x negative three times.

Now, what is negative three times x minus 2? Type your answer here.

We get negative three x plus six.

Now last but not least, what do we get when we subtract these two quantities? What is negative 3x plus 6 minus the entire quantity negative 3x plus 6?

We get 0. Those are the two quantities inside the parentheses over here are the same. Anything minus itself is equal to 0.

Based on this long division we just did then, we can see that we divide x cubed minus 4x squared plus x plus 6, by x minus 2, we get what we wrote up here. x squared minus 2x minus 3. I wonder then, if there's a new way that we can write this original expression, the one we started out with. So, think about what we could do to both sides of this equation right here, to get our original expression, the cubic polynomial, by itself on the left hand side. What that would be on the right hand side of the equation? This will be a new way of expressing what we have on the left.

Since we're dividing by x minus 2 over here, and that's the only thing that's keeping the cubic polynomial from being by itself. We just need to multiply it by both sides of the equation by the quantity x minus 2. On the left-hand side, we'll be left with what we're trying to look for. And on the right-hand side, we'll have x minus 2 times x squared minus 2x minus 3.

Now that we know that f of x is equal to x minus 2 times x squared minus 2x minus 3, let's see if can go from here to writing the function in its fully factored form. Lucky for us, we know how to factor x squared minus 2x minus 3. So, please factor this and write the two factors in these two boxes. That will give us the fully factored form of f of x.

One of these factors needs to be x minus three, and the other needs to be x plus one.

Now let's do some long division practice. What is 2 x cubed minus x squared minus 19 x plus 12 divided by x minus 3? Please write your answer in here. And remember to write it in the standard form for polynomials. In order of descending degrees, starting with the highest.

When we carry out this long division, we end up with 2 x squared plus 5 x minus

I'd like you to keep getting more practice using long division, so let's do this problem as well. This time we're starting out with x cubed minus 1, and I'd like us to divide that by x minus 1. How do you write this using long division though? Well, you might think it would be this way, like before all I've done is taken the thing we want to divide, written our lovely long division surrounding it. And then written the thing we want to divide it by outside. This isn't going to work quite right in this situation of this polynomial though. Here's why. We only have two terms here, but this is a cubic polynomial. With other cubic's we've worked with, we've had a term for every single power starting with the highest power and descending. This time we don't have an x squared term and we don't have an x term. In order to do long division properly though, we still need to leave slots for those terms In the expression that we start out with. Instead we're going to need to write the polynomial like this. Note that by writing 0x squared and 0x for the x squared and x terms, we're not changing the value of the original expression we had underneath the division sign. But we are still leaving room for the terms of these powers. Now we're in great shape to do our division. So let's do it! Please write your final answer in this box.

When we go through this long division we end up with x squared plus x plus 1. You'll remember that you actually have seen this problem before when you learned how to factor, we learned a rule for factoring difference of cubes. We learned that x cubed minus 1 is equal to x minus 1 times x squared plus x plus

Thinking about what we did in the last quiz, if instead we wanted to solve x cubed plus 1 divided by x plus 1, and use long division to do it, how should we set up the problem? What should we write here under the long division sign and what should we write outside?

Underneath the division sign, we need to leave room for the x squared and x terms, even though their coefficients are zero. So, we'll write x cubed plus zero x squared, plus zero x plus 1. The x plus 1 doesn't need any changing at all. We can just write it as is outside the division sign.

Now that we've set up our problem, please solve it using long division.

When we go through the steps of long division, we end up with x squared minus x plus 1 as our answer. Again, this matches a rule that we learned earlier when we were factoring. We learned that x cube plus 1 is equal to x plus 1 times x squared minus x plus 1. And that's exactly what you found here. All we're doing this time is dividing both sides of this equation by x plus 1. That's pretty cool.

In all the long division problems we've done so far, we've gotten really nice clean answers. Because we have gotten answers that don't have a remainder. We've gotten a zero when we complete our final subtraction step down here. Now, the reason this happened for every division we've done so far is that, each of the things we've divided our original polynomials by, have been factors of those first polynomials. When we divide by a factor, there's nothing left over from the original expression that this doesn't go into. Well, what happens if we divide by something that isn't a factor? Well, then we get some number down here instead of zero. Whatever number we get down here is, as I said before, called the remainder. But if the remainder isn't equal to zero, we have some more stuff to think about. Let's see what happens. Earlier on, we were working with the function f of x equals x cubed minus 4 x squared plus x plus 6. Now, when we were looking for roots, or zeroes, we tried to see what happens when we let x equal 1. We found out that plugging that in gave us 4. Since when we plug this in we don't get 0, we know that 1 is not a 0 of this polynomial. And that means, in turn, that x minus 1 is not a factor. I wonder what would happen then, if we divide this polynomial by x minus 1? Let's find out. Here I've written out all the different boxes you're going to need to fill in to complete this long division. So please, write what belongs in each one, and then write your final answer up here. Please note that I've put subtraction signs and then parentheses around boxes. So make sure that you write the expression that belongs inside here, given that the subtraction sign is outside. Also, don't forget to bring down your x and your 6 from the original expression.

When we carry out this long division, what we end up with up here, is x squared minus 3 x minus 2. And note that down here, in our remainder spot, we have a 4 this time instead of a 0. This is proof that x minus 1 is not a factor of this polynomial. Since we have a 4 left over, this doesn't go into our original expression evenly.

So one thing that's really interesting to note here, is that, the remainder that we got, when we divided x cubed minus 4 x squared plus x plus 6 by x minus function. That's pretty interesting. This would be a pretty incredible coincidence, if it weren't actually caused by something. You'll also remember that, when we plugged in 2 to our function, we ended up with 0, and we know that x minus 2 is a factor, so if we divided by x minus 2, we would get 0 as a remainder. Seems like remainders in the numbers we get, when we plug in different values for a function, are very closely connected. Let's keep going with this idea. So, what will the remainder be if we divide the function we've been working with, instead, by x plus 3? Think for a second about what to do. And now, I'll give you a hint. Why don't you see what happens if you just find f of negative 3. You can still think about the connection we were seeing before, between remainders and plugging numbers into the function. And then if you want to, double check by doing the long division.

If we plug negative 3 into the function for x, we end up with negative 60 as a value of the function. So this should be, based on what we've seen before, the remainder that we get and we divide this by x plus 3. Let's see if it actually works. And, sure enough, completing our long division, we get a remainder of negative 60.

Now, the general form of what we've been seeing is called the Remainder Theorem. Over all, the remainder theorem tells us that the remainder we'll get when we divide a function f by x minus a is going to be equal to f of a. Exactly what we get when we plug in a for the value of x into the function. Now, their remainder actually has really interesting implications for graphs. We'll see that when we let a function equal the division of two polynomials and their remainder is not equal to zero we get some pretty cool graphs. We'll end up seeing things like this. The remainder's instantly involved in the way that the curve breaks into these different pieces. We'll come back to that later, though. We don't need to worry about this for now.

When the coefficient of the highest degree term in a polynomial isn't one we need to take some extra steps to be sure that we're factoring correctly. First things first you should always look for a common factor between all of the terms in the expression. So here for example we have f of x equals 10x squared Minus 30x minus 40. This might look kind of hard to factor but if we notice that there's a common factor that we can take out first, the rest of the function will be much more easy to factor. So what is the greatest common factor that is shared by all three terms on the right hand side of this equation. Note that I'm abbreviating greatest common factor as GCF. Please write that number in this box.

The common factor shared by all these trends is ten.

Let's finish factoring the right-hand side of the equation. I've given you a few boxes to fill in to help you out. So, just write in what belongs in each box. The last step is going to be the fully factored form.

The first step is to pull a factor of 10 out from each term. So, that gives us inside the parentheses here. When we do that, we get 10 times x minus 4 times x plus 1. Great, we are fully factored.

Now, what if we have something like this function? f of x equals 6x squared minus 7x minus 5. Once again, our leading coefficient, the coefficient of the highest degree term, is not equal to 1. It's 6. Unfortunately, this time, our situation's not as easily solvable as last time. We can't just factor out a 6 from each of our terms, since it's not a common factor. Well, what do we do now? I'll tell you already what the factors are. There 2x plus 1 and 3x minus

We just saw that the expression 6x squared minus 7x minus 5 factors to 2x plus different terms. And notice how the terms that are the same colors in the top expression and the one below it are related. The negative 5 here, comes straight from the positive 1 and the negative 5. And the 6 here, comes from the you tell me based on our factorization, what are the zeros of this function? As always, please separate your answers in this box by commas.

There are two 0sfor this function, one at x equals negative 1 half, and one at x equals 5 over 3.

So we use the factored form of our polynomial function to find its zeros. And interestingly enough both of these zeros were ratios. But where did these ratios come from? They're in a very specific form. I'll give you two options. Are each of these zeros a ratio of A factor of negative 5 over a factor of 6 or a factor of 6 over a factor of negative 5. Remember that 6 and negative 5 here are coming straight from our original function.

Both 0s are in the form over a factor of negative 5 over a factor of 6.

Now in fact, every time we have a polynomial function that has rational zeros, we can be sure that whatever those zeros are, they'll be in this form. They'll be the ratio of a factor of a constant term at the end of the expression, divided by a factor of the leading coefficient. The coefficient of the highest degree term. This looks like a super useful fact, and it's so useful that it has a special name. It's called the rational zero theorem. Now, we're going to have to add one little addendum to this rule in a second. But, first, let's just look at it as it is. This is an incredibly powerful tool. It's going to help us narrow down the zeros that we think a polynomial function could have, from an infinite number of possible numbers. All the rational numbers, to only numbers that fit this form. It's going to make our lives a ton easier.

Why don't we try an example using this rational zero theorem? Let's say we have this function. f of x equals 4 x squared minus 4 x plus 1. Now, I know that, based on the rational zero theorem, any rational zero of this polynomial must be a factor of 1 divided by a factor of 4. Since 1 is the constant term in the expression, and 4 is the leading coefficient. Well, 1 has factors of just 1 and negative 1. And four has factors of 1 and 4, and also their negatives. And also rational zero theorem says that this function can have. I'll give you one more hint. There are six different fractions that we should get. As always, please separate these answers by commas.

When we think about the possible ratios we could obtain, we get the following. We could take 1 over 1, and that would give us 1. We could also take negative 1 over 1. And that would give us negative 1. 1 over 4 gives us, well, 1 4th. 1 over negative 4 gives us negative 1 4th. And you'll notice that if we do negative 1 over 4, we get negative 1 4th again. And negative 1 over negative 4 gives us positive 1 4th again. So because we have both a positive and negative versions of each number in each set of factors, we're going to have a lot of overlapping factors. This is convenient, because it means that we'll get less in the end than it looks like we would get. The last two possible rational zeros are one half and negative one half.

Now, we know exactly how to check whether each of these numbers down here actually is a zero of this function. We just need to substitute each one in for x. And see which of these values give us f of x equals zero. Let's check the easy ones first. The first two on the list. 1 and negative 1. So, please tell me whether negative 1 and 1 are each zeros of this function. Just answer yes or no for each one.

And the answer to each of these questions is no. If we plug in negative 1 into our function, we get 9. And if we plug in 1, we get 1. Neither one of these numbers is 0, so that means neither one of these values of x is a root or a 0.

So you just checked 1 and negative 1 off our list of possible rational zeros for this function. We know that neither one of these works. So let's move o to one of our next four possibilities. Why don't we try out one half? Is one half a root or a zero of this function?

Well, to figure this out, we just need to plug 1 half into our function in place of x. And yes, in fact, we do get that f of 1 half equals 0. So this is a

Now that we found that 1/2 is a zero, we can say that x minus 1/2 is a factor of this polynomial. But really, we want our factors to have integer coefficients. And negative 1/2 is not an integer. I wonder if there's a way we could rewrite this, so that this factor gives us the same zero. But it does have integer coefficients. So what factor then gives us the same route as X minus 1/2 does but as integer coefficients and is still a factor of F of X equals 4X squared minus 4X plus 1. Please fill in the integers in these boxes. One more little tidbit for you to think about; When we had a route of X equals

The root that we need is 2x minus 1. We can get this by multiplying each term here by 2. The denominator of our fraction over here. If we set this equal to zero, we can double check that it gives us the same root. And, sure enough, it still gives us a zero of x equals 1/2. Awesome.

So now, we have one factor of f of x equals 4x squared minus 4x plus 1. We know that it's 2x minus 1. How can we find the other? Well, we can use long division. So, carry out this division using long division. Write your answer in this box.

When we do this long division problem, we get 2x minus 1. And as expected, we get a remainder of 0. Since we knew that 1 half was a 0 of this polynomial, and that 2x minus 1 was a factor. Pretty cool.

So reflecting on the long division problem we just did in the last quiz, how can the results of this problem give us a new way to rewrite the polynomial we started out with? 4x squared minus 4x plus 1. Can you write this instead as the product of two polynomials? Now just so you know I would like you not to write

when we divided it by 2x minus 1, we got 2x minus 1. We could check those by multiplying everything out. And sure enough, it works, wonderful.

This time, if we have the function 6x cubed minus 17x squared plus 11x minus 2, then do you think that x equals 2 over 6 is a 0? Please answer yes or no.

When we plug in 2 over 6 to our function, we do in fact get 0. So yes, 2 over 6 is a 0.

Now that we found that 2 over 6 is a zero of this function, does that mean that one of it's factors is 6x minus 2? Please tell me yes if you agree that this follows the pattern of previous relationships between zeros and factors, or tell me no if you think that the fact that 6x minus 2 is divisible by 2 But 6x cubed minus 17x squared plus 11x minus 2, is not as a problem.

And the answer is no. This is a really great reason, and a really great thing to think about. If 6x minus 2 were a factor, then we could rewrite this expression as follows. We could write it as 6x minus 2 times a bunch of other things. But then, that would mean that we could rewrite it as 2 times 3x minus or in other words, since we can't pull factor of 2 out of every single term here, that means 6x minus 2 can't be a factor of f of x.

We know that x equals 2 over 6 is a zero of this function, but we need to notice that 2 over 6 is not a fraction in simplest terms. It's a fraction that can be reduced. Both 2 and 6 have a factor of 2 in them. So this fraction is actually just equal to 1 3rd. This means that 1 3rd is actually a 0 of f of x. Now that we know that x equals 1 3rd is a 0, what factor of this polynomial do we know about? Please fill in the proper integers in these boxes.

Three x minus one is the factor we are looking for.

Let's summarize what we've learned in the past few quizzes. We can finally fill in the last detail of our rational zero theorem. If some polynomial, f of x has, rational zeros, then they will be of the form x equals plus or minus a positive factor of the constant term of f of x, over a positive factor of the leading coefficient of f of x. Now if we call this rational zero p over q, then p and q only have a common factor of 1. In other words, we have to write these zeros as fractions in lowest terms, or in simplest form. So, as we saw in the previous quiz, instead of having 2 over 6, we need to write 1 over 3 instead, or else we'll mess up our factors. Provided that these requirements are met, especially that p over q is in simplest terms, then q x minus p will be a factor of f of x. You'll have plenty of opportunities to practice using the rational zero theorem in our question section, after this lesson. Great job.