what are the non-permissible value for the rational expression (x^2 + x -6) / (x^2 + x -12)?

a) x = 0 and x = -2
b) x = 0 and x = -4
c) x = 2 and x = -3
d) x = 3 and x = -4

The non-permissable values results in

a denominator of 0.Let X^2+X-12=0,
Solve for X by factoring:(X+4)*(X-3)=0
X+4=0, X=-4; X-3=0, X=3. Ans.= d

thanks henry

Please paki sagot nmn Po

To find the non-permissible values for a rational expression, we need to identify the values of x that would make the denominator equal to zero. Since dividing by zero is undefined, these values are considered non-permissible.

In this case, the expression is (x^2 + x - 6) / (x^2 + x - 12).

To find the non-permissible values, we set the denominator equal to zero:

x^2 + x - 12 = 0.

Next, we need to factorize the quadratic equation to solve for x:

(x + 4)(x - 3) = 0.

Now, we can set each factor equal to zero and solve for x:

x + 4 = 0, x - 3 = 0.

Solving for x in each equation:

x = -4, x = 3.

Therefore, the non-permissible values for the given rational expression are x = -4 and x = 3.

So, the correct answer is d) x = 3 and x = -4.