# Algebra II

• Nov 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
• Nov 6th 2012, 10:04 AM
emakarov
Re: Algebra II
• Nov 6th 2012, 10:14 AM
Re: Algebra II
Also, you could use Vieta's theorem.

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

\$\displaystyle x_1+x_2+x_3=-p\$

\$\displaystyle x_1x_2+x_1x_3+x_2x_3=q\$

\$\displaystyle x_1x_2x_3=-r\$

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

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

\$\displaystyle 1+x_2+x_3=-a\$

\$\displaystyle x_2+x_3+x_2x_3=-1\$

\$\displaystyle x_2x_3=-8\$

... so you have 3 equations with three unknowns, and you can solve for \$\displaystyle a\$.
• Nov 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.
• Nov 6th 2012, 12:17 PM
skweres1
Re: Algebra II
Thanks guys that was a huge help!!