# How do you use the Rational Root Theorem to find all the roots of x^3 + 9x^2 + 19x – 4 = 0?

Aug 23, 2016

The roots are: $- 4$ and $\frac{- 5 \pm \sqrt{29}}{2}$

#### Explanation:

$f \left(x\right) = {x}^{3} + 9 {x}^{2} + 19 x - 4$

By the rational roots theorem any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a factor of the constant term $- 4$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4$

Trying each in turn, we find:

$f \left(- 4\right) = - 64 + 9 \left(16\right) + 19 \left(- 4\right) - 4 = - 64 + 144 - 76 - 4 = 0$

So $x = - 4$ is a zero and $\left(x + 4\right)$ a factor:

${x}^{3} + 9 {x}^{2} + 19 x - 4 = \left(x + 4\right) \left({x}^{2} + 5 x - 1\right)$

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(1\right) \left(- 1\right)}}{2 \cdot 1}$
$= \frac{- 5 \pm \sqrt{29}}{2}$