# Algebra II

• November 6th 2012, 09:22 AM
skweres1
Algebra II
Find the number a such that x-1 is a factor of x^3+ax^2-x+8
• November 6th 2012, 10:04 AM
emakarov
Re: Algebra II
• November 6th 2012, 10:14 AM
Re: Algebra II
Also, you could use Vieta's theorem.

When you have: $P(x)=x^3+px+qx+r$, you know that:

$x_1+x_2+x_3=-p$

$x_1x_2+x_1x_3+x_2x_3=q$

$x_1x_2x_3=-r$

... where $x_1,x_2,x_3$ are roots of $P(x)$.

So when you have $P(x)=x^3+ax^2-x+8$ and you know that 1 is root ( $x_1=1$), you can write three equations:

$1+x_2+x_3=-a$

$x_2+x_3+x_2x_3=-1$

$x_2x_3=-8$

... so you have 3 equations with three unknowns, and you can solve for $a$.
• November 6th 2012, 11:44 AM
HallsofIvy
Re: Algebra II
Well, if you really want to do it the hard way! Personally I would prefer to evaluate the polynomial at x= 1.
• November 6th 2012, 12:17 PM
skweres1
Re: Algebra II
Thanks guys that was a huge help!!