1. ## Factor completely

can i factor these completely?
$r^2(1+h)(1-h)+4(rh-1)$

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

2. Originally Posted by princess_21

can i factor these completely?

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

$2n^2+4ng+2g^2+n+g-3$
$(n+g-1)(2n+2g+3)$

it didn't take me that long! now try to discover what i did!

3. Originally Posted by NonCommAlg
$(r-rh+2)(r+rh-2)$

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

it didn't take me that long! now try to discover what i did!
i'm really stucked here,
1. $r^2(1-h^2)+4(rh-1)$
then i got $r^2-r^2h^2+4rh-4$
i don't know how to factor this

4. Originally Posted by princess_21
i'm really stucked here,
1. $r^2(1-h^2)+4(rh-1)$
then i got $r^2-r^2h^2+4rh-4$
i don't know how to factor this
$r^2(1-h^2)+4(rh-1)=r^2-(r^2h^2 - 4rh + 4)=r^2 - (rh -2)^2.$ now use the identity: $a^2-b^2=(a-b)(a+b).$

5. oh, yes i got it. thank you very much!

but how did you get the factor of $2n^2+4ng+2g^2+n+g-3$??

6. Originally Posted by princess_21
oh, yes i got it. thank you very much!

but how did you get the factor of $2n^2+4ng+2g^2+n+g-3$??
Hi princess_21,

Here's how this one breaks out:

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

Factor out 2 in the 1st 3 terms:

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

The grouped trinomial above can be factored:

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

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

$2u^2+u-3$

This factors into:

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

Now, back substitute:

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

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