1Cademy - Evaluating $$\frac{x^2+8x+7}{x^2-4}$$ for $$x = 0$$, $$x = 2$$, and $$x = -1$$

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Evaluating $\frac{x^2+8x+7}{x^2-4}$ for $x = 0$ , $x = 2$ , and $x = -1$

Evaluate the rational expression $\frac{x^2+8x+7}{x^2-4}$ for three values of $x$ , illustrating three distinct outcomes: a defined fraction, an undefined expression, and a result of zero.

ⓐ When $x = 0$ : Substitute $0$ for $x$ to obtain $\frac{(0)^2+8(0)+7}{(0)^2-4}$ . Simplify the numerator: $0+0+7 = 7$ . Simplify the denominator: $0-4 = -4$ . The result is $\frac{7}{-4} = -\frac{7}{4}$ .

ⓑ When $x = 2$ : Substitute $2$ for $x$ to obtain $\frac{(2)^2+8(2)+7}{(2)^2-4}$ . Simplify the numerator: $4+16+7 = 27$ . Simplify the denominator: $4-4 = 0$ . Because the denominator equals zero, the expression is undefined for $x = 2$ .

ⓒ When $x = -1$ : Substitute $-1$ for $x$ to obtain $\frac{(-1)^2+8(-1)+7}{(-1)^2-4}$ . Simplify the numerator: $1-8+7 = 0$ . Simplify the denominator: $1-4 = -3$ . The result is $\frac{0}{-3} = 0$ .

This example highlights three important outcomes when evaluating a rational expression with quadratic polynomials in both numerator and denominator: a substitution can produce a normal fraction, a zero denominator renders the expression undefined, or a zero numerator (with a nonzero denominator) yields a result of zero.

Updated 2026-04-21

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OpenStax Elementary Algebra

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