How do you multiply (x^2-x-6)/(x^2+4x+3) *( x^2-x-12)/(x^2-2x-8)?

Apr 5, 2016

$\frac{{x}^{2} - x - 6}{{x}^{2} + 4 x + 3} \cdot \frac{{x}^{2} - x - 12}{{x}^{2} - x - 8} = \frac{x - 3}{x + 1}$

Explanation:

To solve the multiplication, we should first factorize each quadratic polynomial. Hence,

${x}^{2} - x - 6 = {x}^{2} - 3 x + 2 x - 6 = x \left(x - 3\right) + 2 \left(x - 3\right) = \left(x + 2\right) \left(x - 3\right)$

${x}^{2} + 4 x + 3 = {x}^{2} + 3 x + x + 3 = x \left(x + 3\right) + 1 \left(x + 3\right) = \left(x + 1\right) \left(x + 3\right)$

${x}^{2} - x - 12 = {x}^{2} - 4 x + 3 x - 12 = x \left(x - 4\right) + 3 \left(x - 4\right) = \left(x + 3\right) \left(x - 4\right)$

${x}^{2} - x - 8 = {x}^{2} - 4 x + 2 x - 8 = x \left(x - 4\right) + 2 \left(x - 4\right) = \left(x + 2\right) \left(x - 4\right)$

Hence $\frac{{x}^{2} - x - 6}{{x}^{2} + 4 x + 3} \cdot \frac{{x}^{2} - x - 12}{{x}^{2} - x - 8}$

= $\frac{\left(x + 2\right) \left(x - 3\right)}{\left(x + 1\right) \left(x + 3\right)} \cdot \frac{\left(x + 3\right) \left(x - 4\right)}{\left(x + 2\right) \left(x - 4\right)}$

= $\frac{\left(\cancel{x + 2}\right) \left(x - 3\right)}{\left(x + 1\right) \left(\cancel{x + 3}\right)} \cdot \frac{\left(\cancel{x + 3}\right) \left(\cancel{x - 4}\right)}{\cancel{\left(x + 2\right)} \left(\cancel{x - 4}\right)}$

=$\frac{x - 3}{x + 1}$