Question

For all $h\ne \pm 4$, simplify $\frac{3h^2-9h-12}{16-h^2}$.

• A. $\frac{-3(h+1)}{(h-4)}$
• B. $\frac{3(h-1)}{(h-4)}$
• C. $\frac{-3(h+1)}{(h+4)}$
• D. $\frac{(h+1)}{3(h+4)}$
• E. $\frac{3(h-1)}{(h+4)}$

Answer 1

The correct option is C $\frac{-3(h+1)}{(h+4)}$

The value of $\frac{3h^2-9h-12}{16-h^2}$

$=$ $\frac{3h^2+3h-12h-12}{(4-h)(4+h)}$

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

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

$=$ $\frac{-3(h+1)(4-h)}{(4-h)(4+h)}$

$=$ $\frac{-3(h+1)}{(4+h)}$

Hence, option C is correct.