Question

study the polynomial expression.
\\(\frac{y^2 - 16}{3y^2 - 6y - 24}\\)
which expression is equivalent to this expression?
\\(\frac{y - 4}{3y - 6}\\)
\\(\frac{y + 4}{3y + 6}\\)
\\(\frac{y - 4}{2y + 24}\\)
\\(\frac{y + 4}{2y + 24}\\)

Explanation:

Step1: Factor numerator (difference of squares)

$y^2 - 16 = (y-4)(y+4)$

Step2: Factor denominator (first factor out 3)

$3y^2 - 6y -24 = 3(y^2 - 2y - 8)$

Step3: Factor quadratic in denominator

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

Step4: Rewrite and cancel common terms

$\frac{(y-4)(y+4)}{3(y-4)(y+2)} = \frac{y+4}{3(y+2)}$

Step5: Simplify denominator

$3(y+2) = 3y + 6$

Answer:

$\boldsymbol{\frac{y+4}{3y+6}}$ (the second option)