Grouping Common Factors

**!!!** · March 14th 2008, 11:59 PM

How do I get from:

to:

**Moo** · March 15th 2008, 12:19 AM

Hello,

$(y^2+2y)^5 4(y-1)^3=[2(y^2+2y)^4 (y-1)^3] [2(y^2+2y)]$

$(y-1)^4 5(y^2+2y)^4(2y+2)=2(y-1)^4 5(y^2+2y)^4(y+1)$ $= [2(y^2+2y)^4 (y-1)^3][5(y-1)(y+1)]$

And you factorize

**!!!** · March 15th 2008, 12:49 AM

Do you mind showing it step by step, because I still can't grasp it.

**Moo** · March 15th 2008, 12:58 AM

$(y^2+2y)^5 4(y-1)^3 = 2 \times 2 (y^2+2y)^4 (y^2+2y) (y-1)^3 = [2(y^2+2y)^4 \times (y-1)^3] \times [2 (y^2+2y)]$

$(y-1)^4 5(y^2+2y)^4 (2y+2) = (y-1)^3 (y-1) 5 (y^2+2y)^4 2 (y+1) = 2 (y-1)^3 (y^2+2y)^4 (y+1)$ $= [2 (y^2+2y)^4 (y-1)^3] \times [5 (y-1) (y+1)]$

So the numerator will be :

$\left\{[2(y^2+2y)^4 \times (y-1)^3] \times [2 (y^2+2y)]\right\} - \left\{[2 (y^2+2y)^4 (y-1)^3] \times [5 (y-1) (y+1)]\right\}$ $= [2 (y^2+2y)^4 \times (y-1)^3] \times [2 (y^2+2y) - 5 (y-1) (y+1)]$

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