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### Course: Algebra 2 > Unit 4

Lesson 3: Dividing polynomials by linear factors- Dividing polynomials by linear expressions
- Dividing polynomials by linear expressions: missing term
- Divide polynomials by linear expressions
- Factoring using polynomial division
- Factoring using polynomial division: missing term
- Factor using polynomial division

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# Factoring using polynomial division: missing term

If we know one linear factor of a higher degree polynomial, we can use polynomial division to find other factors of the polynomial. For example, we can use the fact that (x+6) is a factor of (x³+9x²-108) in order to completely factor the polynomial. We just need to be careful because the polynomial has no x-term.

## Want to join the conversation?

- Would this polynomial have been completely factored if Sal had written the expression as (x+6)^2 * (x-3)?(15 votes)
- It's the same, but one of them is shorter(6 votes)

- at3:53, couldn't you write (x+6)(x+6)(x-3) as (x+6)^2(x-3) to simplify it, or would that turn the polynomial into a quadratic?(16 votes)
- I give credit to one guy named 94ohanamin, "Linear factors need to be first-degree equations, so writing it as (x+6)^2 * (x-3) would not work".(1 vote)

- what wpould happen in this case if there is a remainder?(5 votes)
- If there was a nonzero remainder, it would mean the polynomial is not divisible by x+6. So you would have to find a different factor.(6 votes)

- could it also be = to (X-3)*(X+6)^2(3 votes)
- Yes that is the same thing as what Sal wrote(2 votes)

- Can you write is as (x+6)^2(x-3)?(3 votes)
- The expressions are equal, and I don't think anybody would say that that isn't a product of linear factors, so you can.(1 vote)

- At the end of the video(3:55), if we wrote (x+6)^2 * (x-3) instead of (x+6)(x+6)(x-3), would it still be linear factors?(2 votes)
- Linear factors need to be first-degree equations, so writing it as (x+6)^2 * (x-3) would not work.(1 vote)

- What does P(X) or F(X) mean?(2 votes)
- Hi. Both represent a function, in your case, "P" and "F" respectively. "P(x)" can be read as "P in function of x" that is, x is the input and P(x) the output. Hope this helps.(1 vote)

- how do I factor the ones that don't seem to add and multiply consistently?(1 vote)
- If factoring is not working, it is likely because the quadratic can not be factored. This occurs when the roots are not integers. When this is the case, use the completing the square or quadratic formula methods to solve the problem.(2 votes)

- what If the question is like, "x to the six power, 15x to the third power, 45x and 200"? ps this is a equation that I made up and may not have a solution.(1 vote)
- You should say if they are being added or subtracted. x^6+15x^3+45x+200 would be quite different a graph than x^6+15x^3-45x+200.

Also, all polynomials have solutions, they just may be non real. Although, if the degree of the polynomial is odd that guarantees at least one real answer.

For yours they are indeed all nonreal(1 vote)

- How would you multiply this out to check your answer?(1 vote)
- We use the distributive law a bunch of times, and collect like terms when we can.

(𝑥 + 6)(𝑥 + 6)(𝑥 − 3)

= (𝑥(𝑥 + 6) + 6(𝑥 + 6))(𝑥 − 3)

= (𝑥² + 6𝑥 + 6𝑥 + 36)(𝑥 − 3)

= (𝑥² + 12𝑥 + 36)(𝑥 − 3)

= 𝑥²(𝑥 − 3) + 12𝑥(𝑥 − 3) + 36(𝑥 − 3)

= 𝑥³ − 3𝑥² + 12𝑥² − 36𝑥 + 36𝑥 − 108

= 𝑥³ + 9𝑥² − 108(1 vote)

## Video transcript

- [Instructor] We're told
the polynomial p of x, which is equal to this, has
a known factor of x plus six. Rewrite p of x as a
product of linear factors. Pause this video, and see if
you can have a go at that. All right, now let's
work on this together. Because they give us one of the factors, what we can do is say hey, what happens if I divide
x plus six into p of x? What do I have left over? It looks like I'm still
going to have a quadratic, and then I'll probably
have to factor that somehow to get a product of linear factors. So let's get going. So if I were to try to figure out what x plus six divided into, x to the third plus nine x squared, and now we're gonna have to be careful. You might be tempted to
just write minus 108 there, but then this gets tricky because you have your third degree column, your second degree column. You need your first degree column, but you just put your zero degree, your constant column here. So to make sure we have good hygiene, we could write plus zero x, and I encourage you to
actually always do this if you're writing out a polynomial so that you don't skip
that place, so to speak, minus 108. And so then you say, all right, let's look at the highest degree terms. X goes into x to the
third x squared times. X squared times six is six x squared. X squared times x is x to the third. We want to subtract. We've done this multiple times, so I'm going a little
bit faster than normal. Those cancel out. Nine x squared minus six x
squared is three x squared. Bring down that zero x. And then how many times does
x go into three x squared? Well, it goes three x times, and we would write it in this column. And notice, if we didn't keep this column for our first degree terms,
we'd be kind of confused where to write that
three x right about now. And so three x plus, times
six, I should say, is 18 x. Three x times x is three x squared. We want to subtract what we have in that, I guess that color is mauve,
light purple, not sure. And so we get three x squareds cancel out, and then zero x minus 18 x is negative 18 x. Bring down that negative 108. And so then we have x goes into negative 18 x negative 18 times. Negative 18 times six is negative 108. That's working out nicely. Negative 18 times x is negative 18 x. And then we want to subtract what we have in this
not-so-pleasant brown color, and so I will multiply
them both by a negative. And so I am left with zero. Everything just cancels out. And so I can rewrite p of x. I can rewrite p of x as being equal to x plus six times x squared plus three x minus 18. But I'm not done yet because
this is not a linear factor. This is still quadratic. So let's see, can I think of two numbers that add up to three and that when I multiply
I get negative 18? So I'll need different signs, and then the obvious one is positive six and negative three. And if that, what I just did,
seems like voodoo to you, I encourage you to review
factoring polynomials. But this I can rewrite
'cause negative six plus, or actually I should say positive six plus negative three is equal to three, and then positive six times negative three is equal to negative 18. So I can rewrite this as x plus six times x plus six times x minus three. And so there we have it. We have a product of linear factors. And we are done.