Question

multiplying rational expressions quiz
algebra 2 a ae teacher (acc ed) / module 6: rational functions and relations

1. multiply \\(\frac{x^2 + 8x + 7}{x^2 + 9x + 14} \cdot \frac{x^2 + 5x + 6}{x^2 - 5x - 6}\\). completely simplify your answer and state any variable restrictions.

\\(\frac{x + 3}{x - 6}\\); the variable restrictions are \\(x \
eq -7\\), \\(x \
eq -2\\), \\(x \
eq -1\\), \\(x \
eq 6\\)
\\(\frac{x + 3}{x + 6}\\); the variable restrictions are \\(x \
eq -7\\), \\(x \
eq -2\\), \\(x \
eq -1\\), \\(x \
eq 6\\)
\\(\frac{3x}{x + 6}\\); the variable restrictions are \\(x \
eq -7\\), \\(x \
eq -2\\), \\(x \
eq -1\\), \\(x \
eq 6\\)
\\(\frac{x - 3}{x - 6}\\); the variable restrictions are \\(x \
eq -7\\), \\(x \
eq -2\\), \\(x \
eq -1\\), \\(x \
eq 6\\)

Explanation:

Step1: Factor numerators/denominators

$x^2+8x+7=(x+1)(x+7)$; $x^2+9x+14=(x+2)(x+7)$; $x^2+5x+6=(x+2)(x+3)$; $x^2-5x-6=(x-6)(x+1)$

Step2: Write product of factored forms

$\frac{(x+1)(x+7)}{(x+2)(x+7)} \cdot \frac{(x+2)(x+3)}{(x-6)(x+1)}$

Step3: Cancel common factors

Cancel $(x+1),(x+7),(x+2)$ → $\frac{x+3}{x-6}$

Step4: Find restrictions

Denominators≠0: $x≠-7,-2,-1,6$

Answer:

$\frac{x+3}{x-6}$; the variable restrictions are $x
eq -7$, $x
eq -2$, $x
eq -1$, $x
eq 6$