Question

What are the excluded values of x for x+4/ -3^2+12x+36

Answer 1

Answer:

The excluded values of x for $\dfrac{x+4}{-3x^2+12x+36}$ are:

                  $-2\ and\ 6$

Step-by-step explanation:

We are given a rational expression as follows:

              $\dfrac{x+4}{-3x^2+12x+36}$

We know that the excluded value of a rational expression are the possible values of x which makes the denominator of the rational expression equal to zero i.e. these are the zeros of the denominator.

The denominator could also be factorized as follows:

$-3x^2+12x+36=-3(x^2-4x-12)\\\\\\-3x^2+12x+36=-3(x^2-6x+2x-12)\\\\\\-3x^2+12x+36=-3(x(x-6)+2(x-6))\\\\\\-3x^2+12x+36=-3(x+2)(x-6)$

i.e. the zeros of the expression are:

$x=-2\ and\ x=6$

             Hence, the excluded values are:

                    -2 and 6

Answer 2

X+4 / -3x^2 + 12x + 36

= x + 4 / -3 ( x^2 - 4x - 12) 

= x - 4 / -3(x - 6)(x + 2)

the excluded values make the denominator = 0  

so they are 6  and -2  Answer