For which value(s) of [tex]x[/tex] will the rational expression below equal zero? Check all that apply.

[tex] \frac{(x-5)(x+2)}{x+1} [/tex]

A. -2
B. -5
C. -1
D. 1
E. 5
F. 2



Answer :

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