Question

simplify.
\\(\dfrac{9u^2 - 4}{3u^2 + 19u - 14}\\)

Explanation:

Step1: Factor numerator (difference of squares)

$9u^2 - 4 = (3u-2)(3u+2)$

Step2: Factor denominator (quadratic trinomial)

$3u^2 + 19u -14 = (3u-2)(u+7)$

Step3: Cancel common factor

$\frac{(3u-2)(3u+2)}{(3u-2)(u+7)} = \frac{3u+2}{u+7}$, where $u
eq \frac{2}{3}, -7$

Answer:

$\frac{3u+2}{u+7}$ (with restrictions $u
eq \frac{2}{3}, -7$)