Answer
Factor: $\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{(2x+3)(2x-3)}{3x(2x+{3})}$
$\frac{2x-3}{3x}$
Work Step by Step
1. Factor out the numerator and denominator
${4x^2-9}$= $(2x)^{2}-3^{2}$
= $(2x+3)(2x-3)$
${6x^{2}+{9x}}$=${3x}{(2x+3)}$
$\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{(2x^2+3)(2x-3)}{3x(2x+{3})}$
2. Cancel out the common term: ${2x+3}$
$\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{(2x+3)(2x-3)}{3x(2x+{3})}$
$\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{2x-3}{3x}$
3. Final answer is
$\frac{2x-3}{3x}$
