• Feb 12th 2006, 05:17 PM
soca615
I really need some help factoring some equations please. These don't neccessarily have to be in the real numbers either so :eek:

1.) 4x^4 + 3x^2 - 1
2.) 9c^4 + 5c^2 + 1
3.) x^3 + x^2 + 4
4.) (a+b)^2 + 7(a+b) + 12
• Feb 12th 2006, 08:39 PM
earboth
Quote:

Originally Posted by soca615
I really need some help factoring some equations please. These don't neccessarily have to be in the real numbers either so :eek:

1.) 4x^4 + 3x^2 - 1
2.) 9c^4 + 5c^2 + 1
3.) x^3 + x^2 + 4
4.) (a+b)^2 + 7(a+b) + 12

Hello,

to factorize those expressions use the the rule of Vieta:

to 1) $4x^4+3x^2-1=(2x-1)(2x+1)(x^2+1)$
to 2) $9c^4 + 5c^2 + 1=\left(c^2+\frac{1}{9} \right) \left(c^2+1 \right)$
to 3) $x^3 + x^2 + 4=(x+2)(x^2-x+2)$
to 4) $(a+b)^2 + 7(a+b) + 12=((a+b)+3)((a+b)+4)$

Bye
• Feb 12th 2006, 09:15 PM
topsquark
Quote:

Originally Posted by earboth
to 2) $9c^4 + 5c^2 + 1=\left(c^2+\frac{1}{9} \right) \left(c^2+1 \right)$

2) Isn't correct, sorry.
$9c^4 + 5c^2 + 1$ doesn't factor over the reals. The way I would approach this, then would be to set $9c^4 + 5c^2 + 1 = 0$. The solution is $c^2= -5/18 \pm \sqrt{-11}/18$, so $9c^4 + 5c^2 + 1$
$= 9(c^2+5/18 + \sqrt{-11}/18)(c^2+5/18 - \sqrt{-11}/18)$, or you can take the square root of $c^2$ (i.e. solve for c, not $c^2$) and string the four factors together. Don't forget the extra 9!

-Dan