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## Algebra 2

### Course: Algebra 2>Unit 3

Lesson 4: Factoring higher degree polynomials

# Factoring higher-degree polynomials: Common factor

Sal factors 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
• 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.
• I don't get FOIL .. what is it?
• 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
• so at he is explaining the answer where does the 2 go? and by 2 I mean the one from 2 times 4 times 3x.
• 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?)
• 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.
• how do I factor xy(x-2y)+3y(2y-x)^2?
• 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.
• Would (4x^2+3x)(4x+3) be a solution to this?

I factored by grouping instead of using a common factor.
• 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.
• the way he does it is different that my teacher taught me is there more than one way to do this correctly?
• 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)
• 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)?
• The order does not matter. Did you maybe lose a common factor? What was the problem that you started with?