# Given that P(x) = x^4 + ax^3 - x^2 + bx - 12 has factors x - 2 and and x + 1, how do you solve the equation P(x) = 0?

This equation has $4$ solutions: ${x}_{1} = - 3$, ${x}_{2} = - 2$, ${x}_{3} = - 1$ and ${x}_{4} = 2$
According to Viete's Theorem if $P \left(x\right)$ has a factor of $\left(x - a\right)$ then $a$ is a root of this polynomial, so in this case this polynomial has 2 roots: ${x}_{3} = - 1$ and ${x}_{4} = 2$.
To find the other roots you have to divide $P \left(x\right)$ by $\left(x - 2\right) \left(x + 1\right)$. The result will be: ${x}^{2} + 5 x + 6$.