Find the number a such that x-1 is a factor of x^3+ax^2-x+8
Use the polynomial remainder theorem.
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 $\displaystyle a$.