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

1 Answer
May 5, 2015

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}#