Question

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

Answer 1

The zeros are `{-3,0,2}`

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 `4x^2-8x-12=0` and exclude them.

`4x^2-8x-12=0`
After dividing both sides by 4 you get
`x^2-2x-3=0`
`Delta=(-2)^2-4*1*(-3)=4+12=16`
`x_1=(2-4)/(2)=-1`
`x_2=(2+4)/(2)=3`

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

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

`x(x^2+x-6)=0`
`x=0 vv x^2+x-6=0`
`Delta = 1^2-4*1*(-6)=25`
`sqrt(Delta)=5`
`x_1=(-1-5)/2=-3`
`x_2=(-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}`