# Factor completely

• Mar 30th 2009, 05:19 AM
princess_21
Factor 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\$

• Mar 30th 2009, 05:28 AM
NonCommAlg
Quote:

Originally Posted by princess_21

can i factor these completely?

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

\$\displaystyle (r-rh+2)(r+rh-2)\$

Quote:

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

Quote:

it didn't take me that long! now try to discover what i did! (Wink)
• Mar 30th 2009, 06:32 AM
princess_21
Quote:

Originally Posted by NonCommAlg
\$\displaystyle (r-rh+2)(r+rh-2)\$

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

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

i'm really stucked here,
1. \$\displaystyle r^2(1-h^2)+4(rh-1)\$
then i got \$\displaystyle r^2-r^2h^2+4rh-4\$
i don't know how to factor this(Worried)
• Mar 30th 2009, 06:44 AM
NonCommAlg
Quote:

Originally Posted by princess_21
i'm really stucked here,
1. \$\displaystyle r^2(1-h^2)+4(rh-1)\$
then i got \$\displaystyle r^2-r^2h^2+4rh-4\$
i don't know how to factor this(Worried)

\$\displaystyle r^2(1-h^2)+4(rh-1)=r^2-(r^2h^2 - 4rh + 4)=r^2 - (rh -2)^2.\$ now use the identity: \$\displaystyle a^2-b^2=(a-b)(a+b).\$
• Mar 30th 2009, 07:15 AM
princess_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 AM
masters
Quote:

Originally Posted by princess_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\$??

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)\$