How do you solve x^3+3x^2-10x-150 using the fundamental theorem of algebra?

Apr 18, 2016

Three roots of ${x}^{3} + 3 {x}^{2} - 10 x - 150$ are $\left\{5 , - 4 - i \sqrt{14} , - 4 + i \sqrt{14}\right)$

Explanation:

According to fundamental theorem of algebra, any polynomial of degree $n$ with complex coefficients has exactly $n$ roots, counted with multiplicity.

As $f \left(x\right) = {x}^{3} + 3 {x}^{2} - 10 x - 150$ is a polynomial of degree $3$, it has $n$ roots.

It is apparent that $5$ is one of the root as

$f \left(5\right) = {5}^{3} + 3 \times {5}^{2} - 10 \times 5 - 150 = 125 + 75 - 50 - 150 = 0$ and hence $\left(x - 5\right)$ divides ${x}^{3} + 3 {x}^{2} - 10 x - 150$.

Dividing by $\left(x - 5\right)$, we get ${x}^{2} + 8 x + 30$ or

${x}^{3} + 3 {x}^{2} - 10 x - 150 = \left(x - 5\right) \left({x}^{2} + 8 x + 30\right)$

using quadratic formula complex roots of ${x}^{2} + 8 x + 30$ are $\frac{- 8 \pm \sqrt{{8}^{2} - 4 \times 1 \times 30}}{2 \times 1}$ or $\frac{- 8 \pm \sqrt{64 - 120}}{2} = - 4 \pm i \frac{\sqrt{56}}{2}$

or $- 4 \pm i \sqrt{14}$

Hence, three roots of $f \left(x\right) = {x}^{3} + 3 {x}^{2} - 10 x - 150$ are $\left\{5 , - 4 - i \sqrt{14} , - 4 + i \sqrt{14}\right)$