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