# How do you find the zeros of a function f(x) = (x^3 + x^2 - 6x) /(4x^2 - 8x - 12)?

May 5, 2015

The zeros are $\left\{- 3 , 0 , 2\right\}$

First you have to calculate the domain of the function remembering that 0 cannot be the value of the denominator of a fraction.
So you have to find the solutions of $4 {x}^{2} - 8 x - 12 = 0$ and exclude them.

$4 {x}^{2} - 8 x - 12 = 0$
After dividing both sides by 4 you get
${x}^{2} - 2 x - 3 = 0$
$\Delta = {\left(- 2\right)}^{2} - 4 \cdot 1 \cdot \left(- 3\right) = 4 + 12 = 16$
${x}_{1} = \frac{2 - 4}{2} = - 1$
${x}_{2} = \frac{2 + 4}{2} = 3$

So the domain is D=RR-{-1;3}

Now you can calculate zeros by solving ${x}^{3} + {x}^{2} - 6 x = 0$

$x \left({x}^{2} + x - 6\right) = 0$
$x = 0 \vee {x}^{2} + x - 6 = 0$
$\Delta = {1}^{2} - 4 \cdot 1 \cdot \left(- 6\right) = 25$
$\sqrt{\Delta} = 5$
${x}_{1} = \frac{- 1 - 5}{2} = - 3$
${x}_{2} = \frac{- 1 + 5}{2} = 2$
So we have just calculated the zeros to be {-3;0;2}

Now we have to check if the values are in the domain of the function. The calculated zeros are neither -1 or 3 so they all are in the domain and we can finally write the answer:

The zeros of the function are {-3;0;2}