# What are all the possible rational zeros for f(x)=x^3-10x^2+16x+15 and how do you find all zeros?

Sep 19, 2016

$f \left(x\right)$ has zeros $x = 3$ and $x = \frac{7}{2} \pm \frac{\sqrt{69}}{2}$

#### Explanation:

$f \left(x\right) = {x}^{3} - 10 {x}^{2} + 16 x + 15$

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 divisor of the constant term $15$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

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

We find:

$f \left(3\right) = 27 - 90 + 48 + 15 = 0$

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

${x}^{3} - 10 {x}^{2} + 16 x + 15 = \left(x - 3\right) \left({x}^{2} - 7 x - 5\right)$

We can find the zeros the remaining quadratic ${x}^{2} - 7 x - 5$ which is in standard form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 7$ and $c = - 5$ using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{7 \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(1\right) \left(- 5\right)}}{2 \cdot 1}$

$\textcolor{w h i t e}{x} = \frac{7 \pm \sqrt{49 + 20}}{2}$

$\textcolor{w h i t e}{x} = \frac{7}{2} \pm \frac{\sqrt{69}}{2}$