To determine for which values of [tex]\(x\)[/tex] the rational expression [tex]\(\frac{(x-5)(x+2)}{x+1}\)[/tex] equals zero, we need to focus on the numerator of the expression. This is because a fraction equals zero when its numerator is zero, provided the denominator is not zero.
The numerator of the given expression is [tex]\((x-5)(x+2)\)[/tex].
Set the numerator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
(x-5)(x+2) = 0
\][/tex]
For this product to be zero, at least one of the factors must be zero. Thus, we solve the equations:
[tex]\[
x - 5 = 0 \quad \text{or} \quad x + 2 = 0
\][/tex]
Solving these equations individually:
1. [tex]\(x - 5 = 0\)[/tex]
[tex]\[
x = 5
\][/tex]
2. [tex]\(x + 2 = 0\)[/tex]
[tex]\[
x = -2
\][/tex]
These are the potential solutions. However, we must also ensure that these values do not make the denominator zero, as this would be undefined. The denominator of the expression is [tex]\(x + 1\)[/tex].
Check the solutions in the denominator:
1. For [tex]\(x = 5\)[/tex]:
[tex]\[
x + 1 = 5 + 1 = 6 \quad (\text{not zero})
\][/tex]
2. For [tex]\(x = -2\)[/tex]:
[tex]\[
x + 1 = -2 + 1 = -1 \quad (\text{not zero})
\][/tex]
Both [tex]\(x = 5\)[/tex] and [tex]\(x = -2\)[/tex] do not make the denominator zero.
Therefore, the rational expression [tex]\(\frac{(x-5)(x+2)}{x+1}\)[/tex] equals zero for:
[tex]\[
\boxed{5 \text{ and } -2}
\][/tex]
So, the correct choices are:
A. -2
E. 5
