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# 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.
• 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?
• At , how did he get x((4x)2
• You mean, how he got to x( (4x)^2 + 2*4*3 + 3^2) ?
• Would this equal 4x3+6x+14x2+21
2x2(2x+7)+3(2x+7)?