Question

What value is a discontinuity of x squared plus 8 x plus 4, all over x squared minus x minus 6?

Answer 1

Answer:

x=3

x=-2

Step-by-step explxanation:

Answer 2

Answer: $x=3\\x=-2$

Step-by-step explanation:

1. You have the following expression:

$\frac{x^{2}+8x+4}{x^{2}-x-6}$

3. The denominatoir cannot be zero, so, you must find its roots:

$x^{2}-x-6=0$

Factorizing it, you obtain:

$(x-3)(x+2)=0\\x=3\\x=-2$

 Then:

 $\frac{x^{2}+8x+4}{(x-3)(x+2)}$

4. Therefore, the values that are discontinuity of  $\frac{x^{2}+8x+4}{x^{2}-x-6}$ are:

$x=3\\x=-2$