Question

hw 12.1_multiply & divide rational expressions
score: 10/11 answered: 10/11
question 11
simplify each. be sure to put numerator and denominator in parenthesis as needed.
\\(\frac{x^2 - 7x + 12}{x^2 - 3x - 4} \cdot \frac{4x + 4}{x^2 - 8x + 15} = \\)
\\(\frac{3x - 15}{x^2 + x - 2} \div \frac{x^2 - 4x - 5}{x^2 + 3x + 2} = \\)
question help: video

Explanation:

Step1: Factor first rational product numerators/denominators

$\frac{x^2-7x+12}{x^2-3x-4} \cdot \frac{4x+4}{x^2-8x+15} = \frac{(x-3)(x-4)}{(x-4)(x+1)} \cdot \frac{4(x+1)}{(x-3)(x-5)}$

Step2: Cancel common factors

Cancel $(x-3)$, $(x-4)$, $(x+1)$ from numerator/denominator:
$\frac{\cancel{(x-3)}\cancel{(x-4)}}{\cancel{(x-4)}\cancel{(x+1)}} \cdot \frac{4\cancel{(x+1)}}{\cancel{(x-3)}(x-5)} = \frac{4}{x-5}$

Step3: Rewrite division as multiplication by reciprocal

$\frac{3x-15}{x^2+x-2} \div \frac{x^2-4x-5}{x^2+3x+2} = \frac{3x-15}{x^2+x-2} \cdot \frac{x^2+3x+2}{x^2-4x-5}$

Step4: Factor all numerators/denominators

$= \frac{3(x-5)}{(x+2)(x-1)} \cdot \frac{(x+2)(x+1)}{(x-5)(x+1)}$

Step5: Cancel common factors

Cancel $(x-5)$, $(x+2)$, $(x+1)$ from numerator/denominator:
$\frac{3\cancel{(x-5)}}{\cancel{(x+2)}(x-1)} \cdot \frac{\cancel{(x+2)}\cancel{(x+1)}}{\cancel{(x-5)}\cancel{(x+1)}} = \frac{3}{x-1}$

Answer:

$\frac{4}{x-5}$
$\frac{3}{x-1}$