# How do you find all the real and complex roots of x^3 - x^2 - 7x + 15=0?

Jan 29, 2016

Use the rational root theorem to help find one root, then factor that out to find the remaining two Complex roots:

$x = - 3$, $x = 2 \pm i$

#### Explanation:

$f \left(x\right) = {x}^{3} - {x}^{2} - 7 x + 15$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be expressible in the form $\frac{p}{q}$ for some integers $p$ and $q$ where $p$ is a factor of the constant term $15$ and $q$ is a factor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1$, $\pm 3$, $\pm 5$, $\pm 15$

Let's try the first few...

$f \left(1\right) = 1 - 1 - 7 + 15 = 8$

$f \left(- 1\right) = - 1 - 1 + 7 + 15 = 20$

$f \left(3\right) = 27 - 9 - 21 + 15 = 12$

$f \left(- 3\right) = - 27 - 9 + 21 + 15 = 0$

So $x = - 3$ is a root of $f \left(x\right) = 0$ and $\left(x + 3\right)$ is a factor of $f \left(x\right)$

${x}^{3} - {x}^{2} - 7 x + 15 = \left(x + 3\right) \left({x}^{2} - 4 x + 5\right)$

The remaining quadratic factor has negative discriminant, so no Real roots, but we can find its Complex roots using the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$= \frac{4 \pm \sqrt{- 4}}{2} = 2 \pm i$