# How do you find all the real and complex roots of x^3 - 2x^2 + 10x + 136 =0?

Jan 22, 2016

$x = - 4 , 3 \pm 5 i$

#### Explanation:

We should first try to find a real root of the polynomial. Any real root, since the coefficient of the high highest degree term is $1$, will be a factor of the final term, $136$.

Thus, our potential factors are $\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 17 , \pm 34 , \pm 68 , \pm 136$.

Attempt synthetic division or polynomial long division with the factors until you find one that doesn't give a remainder.

The only factor that will work is $x + 4$, which means the polynomial has a root of $\textcolor{red}{- 4}$.

We can determine the other two roots by dividing the original function and $x + 4$ to see the remaining quadratic.

$\frac{{x}^{3} - 2 {x}^{2} + 10 x + 136}{x + 4} = {x}^{2} - 6 x + 34$

To determine the remaining two roots, we can apply the quadratic formula to ${x}^{2} - 6 x + 34$.

x=(-b+-sqrt(b^2-4ac))/(2a)=(6+-sqrt(-100))/2=color(red)(3+-5i

The graph should have one real root at $- 4$ and two imaginary roots (it will only cross the $x$ axis once):

graph{x^3-2x^2+10x+136 [-15, 15, -2000, 2000]}