Question

Write as equivalent rational expression with LCD$(b+3)(b+3)(b-5)$of$\frac{4}{b^2+6b+9},\frac{2b}{b^2-2b-15}$.

Short answer

The result is$\frac{4b-20}{(b+3)(b+3)(b-5)},\frac{2b^2+6b}{(b+3)(b+3)(b-5)}$.

Step-by-step solution

Given Information

The given expression is$\frac{4}{b^2+6b+9},\frac{2b}{b^2-2b-15}$.

The LCD is$(b+3)(b+3)(b-5)$.

Factor the denominator of both expressions.

The factored form of first denominator.

$b^2+6b+9=(b+3)(b+3)$

The factored form of second denominator.

$b^2-2b-15=(b-5)(b+3)$

The fractions in factored form is$\frac{4}{(b+3)(b+3)},\frac{2b}{(b-5)(b+3)}$.

Multiply each denominator by the 'missing' factor and multiplyeach numerator by the same factor.

$\frac{4(b-5)}{(b+3)(b+3)(b-5)},\frac{2b(b+3)}{(b-5)(b+3)(b+3)}$

Simplify the numerators.

Multiply the numerator using distributive property.

$\frac{4b-20}{(b+3)(b+3)(b-5)},\frac{2b^2+6b}{(b+3)(b+3)(b-5)}$.