# What are all the possible rational zeros for f(x)=x^2+8x+10?

Mar 28, 2018

The roots are $- 4 + \setminus \sqrt{6}$ and $- 4 - \setminus \sqrt{6}$.

#### Explanation:

You start with $f \left(x\right) = {x}^{2} + 8 x + 10$. First you have to establish its degree, which is $2$ because the highest exponent is $2$. That being said, you can apply the quadratic formula, namely:

$a {x}^{2} + b x + c = 0 \setminus \implies x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this particular case, $a$ is $1$, $b$ is $8$ and $c$ is $10$, and therefore:

$x = \setminus \frac{- 8 \setminus \pm \setminus \sqrt{{8}^{2} - 4 \setminus \cdot 1 \setminus \cdot 10}}{2 \setminus \cdot 1}$

$x = \setminus \frac{- 8 \setminus \pm \setminus \sqrt{64 - 40}}{2}$

$x = \setminus \frac{- 8 \setminus \pm \setminus \sqrt{24}}{2} = \setminus \frac{- 8 \setminus \pm \setminus \sqrt{2 \setminus \cdot 2 \setminus \cdot 6}}{2}$

$x = \setminus \frac{- 8 \setminus \pm 2 \setminus \sqrt{6}}{2} = - 4 \setminus \pm \setminus \sqrt{6}$

Hence, the two roots are $- 4 + \setminus \sqrt{6}$ and $- 4 - \setminus \sqrt{6}$.