# How do you find all the real and complex roots of x^3 + 9x^2 + 19x - 29 = 0?

Jan 22, 2016

$x = 1 , - 5 \pm 2 i$

#### Explanation:

The potential real roots of a polynomial are the factors of $\frac{p}{q}$, where $p$ is the constant and $q$ is the coefficient of the term with the highest degree.

Since $p = 29$ and $q = 1$, the possible real root(s) must be factors of $\frac{29}{1} = 29$. This makes the problem significantly easier since the $29$ is prime, so the only possibilities for the roots are $\pm 1 , \pm 29$.

Use synthetic division or polynomial long division to find that $\left(x - 1\right)$ is a factor, which means that $\textcolor{red}{1}$ is a root:

$\frac{{x}^{3} + 9 {x}^{2} + 19 x - 29}{x - 1} = {x}^{2} + 10 x + 29$

The other two roots can be found through applying the quadratic formula on ${x}^{2} + 10 x + 29$:

x=(-b+-sqrt(b^2-4ac))/(2a)=(-10+-sqrt(-16))/2=color(red)(-5+-2i