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### Course: Algebra 2 > Unit 3

Lesson 4: Factoring higher degree polynomials# Factoring higher-degree polynomials: Common factor

Discover the art of factoring higher degree polynomials. We start by pulling out common factors, then spot perfect squares. The key is seeing patterns and using them to simplify complex expressions. Sal demonstrates by factoring 16x^3+24x^2+9x as (x)(4x+3)^2.

## Want to join the conversation?

- Would x(-4x-3)(-4x-3)=x(-4x-3)^2 also be a solution to the problem it seems to work(20 votes)
- Yes, because when you multiply 2 negative numbers together, it's going to be positive. squaring something is basically the absolute value times the absolute value of the expression.(5 votes)

- I don't get FOIL .. what is it?(10 votes)
- It's a way of multiplying two binomials. It stands for first, outside, inner, and last.

So for a problem (a + b) (c + d), using FOIL we get ac + ad + bc + bd(20 votes)

- Would (4x^2+3x)(4x+3) be a solution to this?

I factored by grouping instead of using a common factor.(4 votes)- You answer isn't incorrect, but it is incomplete. When you are factoring, you need to ensure that your result can not be factored any further. Your first binomial is still factorable because it contains a common factor of "x" that needs to be factored out. If you do that, then your result would match the video: x(4x+3)(4x+3)

Hope this helps.(11 votes)

- how do I factor xy(x-2y)+3y(2y-x)^2?(1 vote)
- There are 2 ways:

1) Completely simplify the polynomial, then factor

Simplifying:`xy(x-2y)+3y(2y-x)^2`

=`x^2y-2xy^2 + 3y(4y^2-4xy+x^2)`

=`x^2y-2xy^2+12y^3-12xy^2+3x^2y`

=`4x^2y-14xy^2+12y^3`

Factoring...

-- Factor out GCF=2y:`2y [2x^2-7xy+6y^2]`

-- Factor trinomial using grouping. AC=12

Find factors of 12 that add to -7. Use -4 and -3

Use these to expand middle term`2y [2x^2-4xy-3xy+6y^2]`

=`2y [2x(x-2y)-3y(x-2y)]`

=`2y (x-2y) (2x-3y)`

2) Use the structure of the polynomial to factor.

This requires that you realize that`(x-2y) = -1(2y-x)`

or`(2y-x) = -1(x-2y)`

. I'm going to use the 2nd version.

If`(2y-x) = -1(x-2y)`

, then`(2y-x)^2 = (-1)^2(x-2y)^2`

=`1(x-2y)^2`

=`(x-2y)^2`

Using this, we can change the polynomial into:`xy(x-2y)+3y(x-2y)^2`

This has 2 terms with a common factor of (x-2y). Use the distributive property to factor it out.`xy(x-2y)+3y(x-2y)^2`

=`(x-2y)[xy+3y(x-2y)]`

Simplify the 2nd factor:`(x-2y)[xy+3y(x-2y)]`

=`(x-2y)[xy+3xy-6y^2]`

=`(x-2y)[4xy-6y^2]`

Then, factor out a GCF=2y from 2nd factor:`(x-2y)[4xy-6y^2]`

=`2y (x-2y)(2x-3y)`

Hope this helps.(12 votes)

- so at3:24he is explaining the answer where does the 2 go? and by 2 I mean the one from 2 times 4 times 3x.(6 votes)
- Ok, I need some help on this problem. 486 + 108x + 6x^2.

(I have already figured out that a is 9 and b is x. But, what is the number that is outside of the parentheses?)(4 votes)- Since all are even, they are divisible by 2, since all digits of each add to a number divisible by 3 (18, 9, and 6) they are divisible by 6, so 6 can come out to yield 6( x^2 + 18x + 81), since 9*9 = 81 and 9+9 = 18, you do in fact get 6(x+9)^2.(5 votes)

- what a FOIL method ?(1 vote)
- -
**F**irst ("first" terms of each binomial are multiplied together)

-**O**uter ("outside" terms are multiplied—that is, the first term of the first binomial and the second term of the second)

-**I**nner ("inside" terms are multiplied—second term of the first binomial and first term of the second)

-**L**ast ("last" terms of each binomial are multiplied)

Let me give an example.

(a + b)(c - d)

First: a * c

Outer: a * (-d)

Inner: b * c

Last: b * (-d)

And we add them together to get the result.

(a + b)(c - d) = a * c + a * (-d) + b * c + b * (-d)(8 votes)

- the way he does it is different that my teacher taught me is there more than one way to do this correctly?(4 votes)
- There are many ways to solve a problem, not only just one way. As long as it leads you to the correct answer, all's fine !(1 vote)

- At1:30, how did he get x((4x)2(4 votes)
- You mean, how he got to x( (4x)^2 + 2*4*3 + 3^2) ?(0 votes)

- If I get the same factors but have them in a different order, it gets counted as wrong. Does it really matter what order they are in?

For example

Isn't (x-2)(x-1) the same thing as (x-1)(x-2)?(2 votes)- The order does not matter. Did you maybe lose a common factor? What was the problem that you started with?(3 votes)

## Video transcript

- [Voiceover] So let's say that we've got the polynomial 16x to the third plus 24x squared plus nine x. Now what I'd like you to do is pause the video and see if you could factor this polynomial completely. Now let's work through it together. So the first thing that you might notice is that all of the
terms are divisible by x so we can actually factor out an x. So let's do that, and
actually, if we look at these coefficients, it
looks like, let's see, it looks like they don't have any common factors other than one. So it looks like the
largest monomial that we can factor out is just going to be an x. So let's do that. Let's factor out an x. So then this is gonna be x times. When you factor out an
x from 16x to the third you're gonna be left with 16x squared, and then plus 24x and then plus nine. Now this is starting to look interesting so let me just rewrite it. It's gonna be x times. This part over here looks interesting because when I see the 16x squared, this looks like a perfect square. Let me write it out. 16x squared. That's the same thing as four x squared, and then we have a nine over there, which is clearly a perfect square. That is three squared. Three squared. And when we look at this 24x, we see that it is four times three times two, and so we can write it as,
let me write it this way. So this is going to be plus two times four times three x. So let me make it, so two times four times three times three x. Now why did I take the trouble, why did I take the trouble of
writing everything like this? Because we see that it fits the pattern for a perfect square. What do I mean by that? Well in previous videos, we saw that if you have something of the form Ax plus B, and you were to square it, you're going to get Ax squared plus two ABx plus B squared, and we have
that form right over here. This is the Ax squared.
Let me do the same color. The Ax squared. Ax squared. We have the B squared.
You have the B squared. And then you have the two ABx. Two ABx right over there. So this section, this entire
section, we can rewrite as being we know it, what A and B are. A is four and B is three
so this is going to be Ax, so four x plus four x plus b, which we know to be three. That whole thing, that whole
thing squared, and now we can't forget this x out front
so we have that x out front. And we're done. We have
just factored this. We have just factored this completely. We could write it as x
times four x plus three and then write out and then say times four x plus three or we could just write x times the quantity four
x plus three squared. And so we've just factored it completely, and the key realizations here is well, one: what could I factor
out from all of these terms? I could factor out an x
from all of those terms, and then to realize what we had left over was a perfect square,
really using this pattern that we were able to
see in previous videos.