# What are all the possible rational zeros for f(x)=x^3+5x^2-48x-10 and how do you find all zeros?

Oct 20, 2016

The only rational zero is $5$.

#### Explanation:

As putting $x = 5$ in $f \left(x\right) = {x}^{3} + 5 {x}^{2} - 48 x - 10$

gives $f \left(x\right) = {5}^{3} + 5 \times {5}^{2} - 48 \times 5 - 10 = 125 + 125 - 240 - 10 = 0$

Hence, $5$ is a zero of $f \left(x\right)$ and $\left(x - 5\right)$ is a factor of $f \left(x\right)$

Hence $f \left(x\right) = {x}^{3} + 5 {x}^{2} - 48 x - 10$

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

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

As the determinant in ${x}^{2} + 10 x + 2$ is ${10}^{2} - 4 \times 1 \times 2 = 100 - 8 = 92$, which though positive, is not a square of a rational number, there are no other rational zeros.

Irrational zeros can be had using quadratic formula as $\frac{- 10 \pm \sqrt{100 - 4 \times 1 \times 2}}{2} = \frac{- 10 \pm \sqrt{92}}{2} = \frac{- 10 \pm 2 \sqrt{23}}{2}$

i.e. $- 5 - \sqrt{23}$ and $- 5 + \sqrt{23}$