# How do you find the product (x^2-3x-10)/(x^2+2x-35)*(x^2+4x-21)/(x^2+9x+14)?

Jan 24, 2017

$\frac{{x}^{2} - 3 x - 10}{{x}^{2} + 2 x - 35} \cdot \frac{{x}^{2} + 4 x - 21}{{x}^{2} + 9 x + 14} =$

#### Explanation:

To solve $\frac{{x}^{2} - 3 x - 10}{{x}^{2} + 2 x - 35} \cdot \frac{{x}^{2} + 4 x - 21}{{x}^{2} + 9 x + 14} = \frac{x - 3}{x + 7}$,

we just need to factorize each of the quadratic polynomial.

${x}^{2} - 3 x - 10 = {x}^{2} - 5 x + 2 x - 10 = x \left(x - 5\right) + 2 \left(x - 5\right) = \left(x + 2\right) \left(x - 5\right)$

${x}^{2} + 2 x - 35 = {x}^{2} + 7 x - 5 x - 35 = x \left(x + 7\right) - 5 \left(x + 7\right) = \left(x - 5\right) \left(x + 7\right)$

${x}^{2} + 4 x - 21 = {x}^{2} + 7 x - 3 x - 21 = x \left(x + 7\right) - 3 \left(x + 7\right) = \left(x - 3\right) \left(x + 7\right)$

${x}^{2} + 9 x + 14 = {x}^{2} + 7 x + 2 x + 14 = x \left(x + 7\right) + 2 \left(x + 7\right) = \left(x + 2\right) \left(x + 7\right)$

Hence $\frac{{x}^{2} - 3 x - 10}{{x}^{2} + 2 x - 35} \cdot \frac{{x}^{2} + 4 x - 21}{{x}^{2} + 9 x + 14}$

= $\frac{\left(x + 2\right) \left(x - 5\right)}{\left(x - 5\right) \left(x + 7\right)} \times \frac{\left(x - 3\right) \left(x + 7\right)}{\left(x + 2\right) \left(x + 7\right)}$

= $\frac{\cancel{\left(x + 2\right)} \cancel{\left(x - 5\right)}}{\cancel{\left(x - 5\right)} \cancel{\left(x + 7\right)}} \times \frac{\left(x - 3\right) \cancel{\left(x + 7\right)}}{\cancel{\left(x + 2\right)} \left(x + 7\right)}$

= $\frac{x - 3}{x + 7}$