For all real numbers x such that x \neq -3, x \neq 0, and x \neq 3, which of the following… — GRE General Test

For all real numbers

x

such that

x \neq -3

x \neq 0

, and

x \neq 3

, which of the following expressions is equivalent to

\frac{x^4 - 81}{x^3 + 3x^2} \div \frac{x^2 + 9}{x^2}

x - 3

Factoring the numerator

x^4 - 81

gives

(x - 3)(x + 3)(x^2 + 9)

and factoring the denominator

x^3 + 3x^2

gives

x^2(x + 3)

. Multiplying by the reciprocal

\frac{x^2}{x^2 + 9}

allows the factors

(x + 3)

(x^2 + 9)

, and

x^2

to cancel completely, leaving

x - 3

Rewrite division of rational expressions as multiplication by the reciprocal

\frac{x^4 - 81}{x^3 + 3x^2} \cdot \frac{x^2}{x^2 + 9}

Dividing by a fraction is equivalent to multiplying by its reciprocal.

Factor the numerator

x^4 - 81

using the difference of squares formula twice

x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)

Recognize

x^4 - 81

(x^2)^2 - 9^2

and then factor

x^2 - 9

further into

(x - 3)(x + 3)

Factor out the greatest common factor from the denominator

x^3 + 3x^2

x^3 + 3x^2 = x^2(x + 3)

Extract

x^2

from both terms of the binomial.

Substitute the factored expressions into the product and cancel common factors

\frac{(x - 3)(x + 3)(x^2 + 9)}{x^2(x + 3)} \cdot \frac{x^2}{x^2 + 9} = x - 3

The common terms

(x + 3)

(x^2 + 9)

, and

x^2

cancel out completely from the numerator and denominator for all

x \neq -3, 0, 3

Factoring polynomial expressions using difference of squares and common terms to simplify rational expressions.

Estimated Time:1m 30s

Question