Find the number a such that x-1 is a factor of x^3+ax^2-x+8

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- Nov 6th 2012, 09:22 AMskweres1Algebra II
Find the number a such that x-1 is a factor of x^3+ax^2-x+8

- Nov 6th 2012, 10:04 AMemakarovRe: Algebra II
Use the polynomial remainder theorem.

- Nov 6th 2012, 10:14 AMPaulAdrienMauriceDiracRe: 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 AMHallsofIvyRe: 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 PMskweres1Re: Algebra II
Thanks guys that was a huge help!!