## Factoring polynomials using the distributive property

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
onmyown
Posts: 18
Joined: Fri Dec 17, 2010 10:16 pm
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### Factoring polynomials using the distributive property

HI, I am having trouble getting the correct answer for this problem:
4x^2+8y-4xy-8x

I have made several attempts. Here is one of them:
4x^2+8y-4xy-8x
1) Grouped like terms together:
= (4x^2 + (-8x))+(8y + (-4xy))
2) Factored each term:
(2*2*x*x + (-1)*2*2*2*x) + (2*2*2*y + (-1)*2*2*x*y)
3) Found greatest common factor for each pair of like terms, and factored each pair by its GCF:
(4x*x+4x*(-2)) + (4y*2+4y*-x)
4) Used distributive property:
4x(x-2) + 4y(2-x)
5) Therefore, the solution to the problem is the binomial:
(4x+4y)(x-2)

However, this is wrong. The actual answer is (4x-4y)(x-2). So, I'm correct, except for the fact that 4y is positive when it should be negative.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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4) Used distributive property:
4x(x-2) + 4y(2-x)
5) Therefore, the solution to the problem is the binomial:
(4x+4y)(x-2)
How did the "2 - x" turn into "x - 2"? Is 5 - 3 the same value as 3 - 5?

onmyown
Posts: 18
Joined: Fri Dec 17, 2010 10:16 pm
Contact:

### Re:

4) Used distributive property:
4x(x-2) + 4y(2-x)
5) Therefore, the solution to the problem is the binomial:
(4x+4y)(x-2)
How did the "2 - x" turn into "x - 2"? Is 5 - 3 the same value as 3 - 5?
I got the right answer now. Thanks