Question

1. multiply \\(\frac{x^2 + 7x + 12}{x^2 - 9} \cdot \frac{x^2 - 4x + 3}{x^2 + 8x + 16}\\). state any restrictions on the variables.\

\\(\bigcirc\\) \\(\frac{x + 1}{x + 4}\\); the variable restrictions are \\(x \
eq -4\\), \\(x \
eq -3\\), \\(x \
eq 3\\)\
\\(\bigcirc\\) \\(\frac{x + 1}{x - 4}\\); the variable restrictions are \\(x \
eq -4\\), \\(x \
eq -3\\), \\(x \
eq 3\\)\
\\(\bigcirc\\) \\(\frac{x - 1}{x + 4}\\); the variable restrictions are \\(x \
eq -4\\), \\(x \
eq -3\\), \\(x \
eq 3\\)\
\\(\bigcirc\\) \\(\frac{x - 1}{x - 4}\\); the variable restrictions are \\(x \
eq -4\\), \\(x \
eq -3\\), \\(x \
eq 3\\)

Explanation:

Step1: Factor all polynomials

1. $x^2+7x+12=(x+3)(x+4)$
2. $x^2-9=(x+3)(x-3)$
3. $x^2-4x+3=(x-1)(x-3)$
4. $x^2+8x+16=(x+4)^2$

Substitute into the expression:
$$\frac{(x+3)(x+4)}{(x+3)(x-3)} \cdot \frac{(x-1)(x-3)}{(x+4)^2}$$

Step2: Identify restrictions

Denominators cannot equal 0:

1. $x^2-9=0 \implies x

eq -3, 3$

1. $x^2+8x+16=0 \implies x

eq -4$

Step3: Cancel common factors

Cancel $(x+3)$, $(x-3)$, and $(x+4)$ from numerator/denominator:
$$\frac{\cancel{(x+3)}\cancel{(x+4)}}{\cancel{(x+3)}\cancel{(x-3)}} \cdot \frac{(x-1)\cancel{(x-3)}}{\cancel{(x+4)}(x+4)} = \frac{x-1}{x+4}$$

Answer:

$\frac{x-1}{x+4}$; the variable restrictions are $x
eq -4, x
eq -3, x
eq 3$ (third option)