# How do you find the zeros of f(x) = 2x^3 − 5x^2 − 28x + 15?

Jul 7, 2016

Zeros: $x = \frac{1}{2}$, $x = 5$, $x = - 3$

#### Explanation:

$f \left(x\right) = 2 {x}^{3} - 5 {x}^{2} - 28 x + 15$

By the rational root theorem, since $f \left(x\right)$ has integer coefficients, any rational zeros 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 $2$ of the leading term.

That means that the only possible rational zeros of $f \left(x\right)$ are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm \frac{5}{2} , \pm 3 , \pm 5 , \pm \frac{15}{2} , \pm 15$

Trying the first one we find:

$f \left(\frac{1}{2}\right) = \frac{2}{8} - \frac{5}{4} - \frac{28}{2} + 15 = \frac{1 - 5 - 56 + 60}{4} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$2 {x}^{3} - 5 {x}^{2} - 28 x + 15$

$= \left(2 x - 1\right) \left({x}^{2} - 2 x - 15\right)$

To factor the remaining quadratic, find a pair of factors of $15$ which differ by $2$. The pair $5 , 3$ works, so:

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

So the two remaining zeros are $x = 5$ and $x = - 3$