Question

1. solve \\(\frac{x^2 + 7x + 12}{x^2 - 9} \div \frac{x^2 + 10x + 24}{x^2 - 3x - 54}\\). completely simplify your answer and state any restrictions on the variable.\

\\(\bigcirc \frac{x + 9}{x - 3}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)\
\\(\bigcirc \frac{x - 3}{x - 9}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)\
\\(\bigcirc \frac{x - 9}{x + 3}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)\
\\(\bigcirc \frac{x - 9}{x - 3}, x \
eq -4, x \
eq -3, x \
eq 3, x \
eq -6\\)

Explanation:

Step1: Rewrite division as multiplication

$\frac{x^2+7x+12}{x^2-9} \times \frac{x^2-3x-54}{x^2+10x+24}$

Step2: Factor all polynomials

$\frac{(x+3)(x+4)}{(x-3)(x+3)} \times \frac{(x-9)(x+6)}{(x+4)(x+6)}$

Step3: Cancel common factors

$\frac{\cancel{(x+3)}\cancel{(x+4)}}{(x-3)\cancel{(x+3)}} \times \frac{(x-9)\cancel{(x+6)}}{\cancel{(x+4)}\cancel{(x+6)}} = \frac{x-9}{x-3}$

Step4: Find variable restrictions

Denominators cannot be zero:
$x^2-9=0 \implies x
eq3, x
eq-3$
$x^2+10x+24=0 \implies x
eq-4, x
eq-6$
Original divisor numerator cannot be zero:
$x^2-3x-54=0 \implies x
eq9, x
eq-6$ (already included)
Original dividend numerator cannot be zero for simplification validity:
$x^2+7x+12=0 \implies x
eq-4, x
eq-3$ (already included)
Final restrictions: $x
eq -4, x
eq -3, x
eq 3, x
eq -6, x
eq9$ (note: the options list the core shared restrictions, and the simplified form matches the last option)

Answer:

$\boldsymbol{\frac{x-9}{x-3}, x
eq -4, x
eq -3, x
eq 3, x
eq -6}$ (the last option)