# How do you find the zeros of x^3-x^2-7x+15=0?

May 19, 2016

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

#### Explanation:

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

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $15$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

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

Trying each of these in turn, we find:

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

So $x = - 3$ is a zero and $\left(x + 3\right)$ a factor:

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

We can factor ${x}^{2} - 4 x + 5$ by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - 2\right)$ and $b = i$ as follows:

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

Hence zeros:

$x = 2 \pm i$