# Algebra - Factoring

• May 4th 2008, 06:37 AM
lax600
Algebra - Factoring
I've tried to factor this for a long time but I can't get it right.

(x-2y)^3 + 4(2y-x)
• May 4th 2008, 06:38 AM
Moo
Quote:

Originally Posted by lax600
I've tried to factor this for a long time but I can't get it right.

(x-2y)^3 + 4(2y-x)

Hello,

(2y-x)=-(x-2y)

:D

What have you done ? It's better pointing out mistakes than writing the whole solution ^^
• May 4th 2008, 07:20 AM
lax600
well the directions were for me to factor it completely. Since the problem was (x-2y)^3 + 4(2y - x). I figured I had to use one of those formulas. So at first I multiplied a -1 into (2y - x) to make it the same as the first part. Then it became (x - 2y)^3 + 4(2y - x). So I checked the answer, and it said the answer was:

(x - 2y)(x - 2y + 2)(x - 2y - 2)

So then I became confused... Like I don't see there's three -2y's and how the 2 and -2 from 4 went into the parenthesis.... yeah.. hehe
• May 4th 2008, 07:33 AM
Moo
No, when multiplying by -1, it may yield this :

$(x-2y)^3 + 4(2y - x)=(x-2y)^3-4(x-2y)$

$=(x-2y)\left((x-2y)^2-\underbrace{4}_{2^2}\right)$

Now, do you know the formula : $a^2-b^2=(a-b)(a+b)$ ?
• May 4th 2008, 07:38 AM
lax600
yeah i know that formula... heh.. forgot that I had to multiply the 4 by -1 too.