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

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- Feb 12th 2006, 05:17 PMsoca615Help Me Factor PleaseI 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 PMearbothQuote:

Originally Posted by**soca615**

to factorize those expressions use the the rule of Vieta:

to 1) $\displaystyle 4x^4+3x^2-1=(2x-1)(2x+1)(x^2+1)$

to 2) $\displaystyle 9c^4 + 5c^2 + 1=\left(c^2+\frac{1}{9} \right) \left(c^2+1 \right)$

to 3) $\displaystyle x^3 + x^2 + 4=(x+2)(x^2-x+2)$

to 4) $\displaystyle (a+b)^2 + 7(a+b) + 12=((a+b)+3)((a+b)+4)$

Bye - Feb 12th 2006, 09:15 PMtopsquarkQuote:

Originally Posted by**earboth**

$\displaystyle 9c^4 + 5c^2 + 1$ doesn't factor over the reals. The way I would approach this, then would be to set $\displaystyle 9c^4 + 5c^2 + 1 = 0$. The solution is $\displaystyle c^2= -5/18 \pm \sqrt{-11}/18$, so $\displaystyle 9c^4 + 5c^2 + 1$

$\displaystyle = 9(c^2+5/18 + \sqrt{-11}/18)(c^2+5/18 - \sqrt{-11}/18)$, or you can take the square root of $\displaystyle c^2$ (i.e. solve for c, not $\displaystyle c^2$) and string the four factors together. Don't forget the extra 9!

-Dan