finding real zero's

Question

Find the real zeros of g(x)=(12x² +16x-3)/(x+3)

 

Answer 1

The real zeros of a rational function will be found when the numerator is equal to 0.  

Factor the numerator.  12x² + 16x - 3.  Then set each factor equal to 0 and solve to find the real zeros.  

 

12x² + 16x - 3 = 0

(6x -1)(2x + 3) =0

6x - 1 = 0 2x+3 = 0

6x = 1      2x = -3

x = 1/6  x= -3/2

 

Also check to make sure neither of your answers would make the denominator zero.  If it does, then it is not a real zero.  Since the denominator is x + 3, x would have to equal -3 to make is zero.  Thus our two answers are the real zeros of the function g(x).  

Answer 2

For real zeros of g(x), 12x²+16x-3=0, (6x-10)((2x+3)=0 

The solutions of x=-1.5, and x=1/6, for both of these values denominator x+3 is non-zero

So real zeros are -1.5, 1/6