can i factor these completely?

$\displaystyle r^2(1+h)(1-h)+4(rh-1)$

$\displaystyle 2n^2+4ng+2g^2+n+g-3$

thanks for your time.

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- Mar 30th 2009, 05:19 AMprincess_21Factor completely
can i factor these completely?

$\displaystyle r^2(1+h)(1-h)+4(rh-1)$

$\displaystyle 2n^2+4ng+2g^2+n+g-3$

thanks for your time. - Mar 30th 2009, 05:28 AMNonCommAlg
- Mar 30th 2009, 06:32 AMprincess_21
- Mar 30th 2009, 06:44 AMNonCommAlg
- Mar 30th 2009, 07:15 AMprincess_21
oh, yes i got it. thank you very much!(Bow)(Bow)

but how did you get the factor of $\displaystyle 2n^2+4ng+2g^2+n+g-3$?? - Mar 30th 2009, 10:30 AMmasters
Hi princess_21,

Here's how this one breaks out:

$\displaystyle 2n^2+4ng+2g^2+n+g-3$

Factor out 2 in the 1st 3 terms:

$\displaystyle 2(n^2+2ng+g^2)+n+g-3$

The grouped trinomial above can be factored:

$\displaystyle 2(n+2)^2+(n+g)-3$

Now, to make things super simple, let u = (n-2).

$\displaystyle 2u^2+u-3$

This factors into:

$\displaystyle (2u+3)(u-1)$

Now, back substitute:

$\displaystyle (2(n+g)+3)(n+g-1)$

$\displaystyle (2n+2g+3)(n+g-1)$