Question

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?

Answer 1

This equation has `4` solutions: `x_1=-3`, `x_2=-2`, `x_3=-1` and `x_4=2`

Explanation:

According to Viete's Theorem if `P(x)` has a factor of `(x-a)` 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(x)` by `(x-2)(x+1)`. The result will be: `x^2+5x+6`.

Then you can calculate the remaining roots.