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

For all real numbers $a$ and $b$ such that $b \neq 0$ and $a \neq -b$ , which of the following expressions is equivalent to $\frac{a^3b - ab^3}{a^2b + 2ab^2 + b^3}$ ?

$\frac{a(a - b)}{a + b}$ Cevap
B
$a - b$
C
$\frac{a(a + b)}{a - b}$
D
$\frac{a^2 + b^2}{a + b}$
E
$\frac{a - b}{a + b}$

Cevap

\frac{a(a - b)}{a + b}

Factoring the numerator yields

ab(a^2 - b^2) = ab(a - b)(a + b)

, and factoring the denominator yields

b(a^2 + 2ab + b^2) = b(a + b)^2

. Dividing both the numerator and the denominator by the common factor

b(a + b)

produces

\frac{a(a - b)}{a + b}

Adım Adım Çözüm

Factor out the greatest common monomial factor from the numerator and denominator.

The numerator

a^3b - ab^3

becomes

ab(a^2 - b^2)

, and the denominator

a^2b + 2ab^2 + b^3

becomes

b(a^2 + 2ab + b^2)

Extracting common terms simplifies polynomials prior to applying polynomial identities.

Apply polynomial algebraic identities to factor the remaining terms.

a^2 - b^2 = (a - b)(a + b)

and

a^2 + 2ab + b^2 = (a + b)^2

. Thus, the expression becomes

\frac{ab(a - b)(a + b)}{b(a + b)^2}

The difference of squares and perfect square trinomial identities allow complete factorization.

Cancel non-zero common factors from the numerator and denominator.

Canceling

b

and one factor of

(a + b)

leaves

\frac{a(a - b)}{a + b}

Dividing both numerator and denominator by

b(a + b)

yields the simplified expression.

Anahtar Kavram

Simplifying rational expressions using common monomial factoring, difference of squares, and perfect square trinomial identities.

Tahmini Süre:1m 30s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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