# How do you find the zeros of f(x) = x^3 − 5x^2 − 3x + 15?

Aug 12, 2016

$f \left(x\right)$ has zeros $\pm \sqrt{3}$ and $5$

#### Explanation:

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

Notice that the ratio of the first and second terms is the same as the ratio between the third and fourth terms.

So this cubic will factor by grouping:

${x}^{3} - 5 {x}^{2} - 3 x + 15$

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

$= {x}^{2} \left(x - 5\right) - 3 \left(x - 5\right)$

$= \left({x}^{2} - 3\right) \left(x - 5\right)$

$= \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left(x - 5\right)$

Hence zeros:

$\pm \sqrt{3}$ and $5$