Question

Write as equivalent rational expression with LCD$(z-2)(z+4)(z+2)$of$\frac{9}{z^2+2z-8},\frac{4z}{z^2-4}$.

Short answer

The result is$\frac{9z+18}{(z+4)(z-2)(z+2)},\frac{4z^2+16z}{(z+2)(z-2)(z+4)}$

Step-by-step solution

Given Information

The given expression is$\frac{9}{z^2+2z-8},\frac{4z}{z^2-4}$.

The LCD is$(z-2)(z+4)(z+2)$.

Factor the denominator of both expressions.

The factored form of first denominator.

$z^2+2z-8=(z+4)(z-2)$

The factored form of second denominator.

$z^2-4=(z+2)(z-2)$

The fractions in factored form is$\frac{9}{(z+4)(z-2)},\frac{4z}{(z+2)(z-2)}$.

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

$\frac{9(z+2)}{(z+4)(z-2)(z+2)},\frac{4z(z+4)}{(z+2)(z-2)(z+4)}$

Simplify the numerators.

Multiply the numerator using distributive property.

$\frac{9z+18}{(z+4)(z-2)(z+2)},\frac{4z^2+16z}{(z+2)(z-2)(z+4)}$.