Factorising cubic polynomial given 1 factor

Question

Given that $x^3-x^2-17x-15 = (x+3)(x^2+bx+c)$ where $b$ and $c$ are constants, find the values of $b$ and $c$.

I don't know how to easily solve this question. I could use polynomial long division but these seems overly complicated especially since at this point in the textbook I am using, that hasn't even been covered yet.

Answer 1

Multiply out and you'll get that

$$x^3 - x^2 - 17x - 15 = x^3 + (3+b)x^2 + (3b + c) + 3c $$

Thus we must have that $3c=-15, 3b+c=-17, 3+b = -1$. It's not hard to get that $c=-5$ and $b=-4$

Answer 2

\begin{align} x^3 - x^2 - 17 x - 15 &= x^3 + 3 x^2 - 4 x^2 - 12 x - 5 x - 15\\ &= x^2 (x + 3) - 4 x(x + 3) - 5(x + 3)\\ &= ( x + 3)(x^2 - 4 x - 5) \end{align}

Answer 3

Let $x_2,x_3$ be the other two roots, then by Vieta's formulas $-3+x_2+x_3=1$ and $-3x_2x_3=15$. Therefore $x_2+x_3=4$ and $x_2x_3=-5$, so $x_2,x_3$ are the roots of the quadratic $x^2-4x-5$ which must then match the other factor $x^2+bx+c$.