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**Prove It** If $\displaystyle x - 3$ is a factor, then the remainder when $\displaystyle f(x)$ is divided by $\displaystyle x - 3$ is $\displaystyle 0$. Therefore $\displaystyle f(3) = 0$.

If $\displaystyle f(x)$ is divided by $\displaystyle x - 2$, the remainder is $\displaystyle -20$. So $\displaystyle f(2) = -20$.

Using these pieces of information...

$\displaystyle f(3) = p(3)^3 - 3^2 + q(3) - 6 = 0$

$\displaystyle 27p - 9 + 3q - 6 = 0$

$\displaystyle 27p + 3q = 15$

and

$\displaystyle f(2) = p(2)^3 - 2^2 + q(2) - 6 = -20$

$\displaystyle 8p - 4 + 2q - 6 = -20$

$\displaystyle 8p + 2q = -10$.

So now you have 2 equations in 2 unknowns which you can solve simultaneously.

$\displaystyle 27p + 3q = 15$

$\displaystyle 8p + 2q = -10$

$\displaystyle 54p + 6q = 30$

$\displaystyle 24p + 6q = -30$

So $\displaystyle 30p = 60$

$\displaystyle p = 2$.

If $\displaystyle 8p + 2q = -10$

$\displaystyle 8(2) + 2q = -10$

$\displaystyle 16 + 2q = -10$

$\displaystyle 2q = -26$

$\displaystyle q = -13$.