# How do you multiply and simplify \frac { x ^ { 2} - 7x - 30} { x ^ { 2} - 4x - 21} \cdot \frac { 14- 2x } { x ^ { 2} - 19x + 90}?

Jun 22, 2018

$\frac{2 \left(7 - x\right)}{\left(x - 7\right) \left(x - 9\right)}$

#### Explanation:

$\setminus \frac{{x}^{2} - 7 x - 30}{{x}^{2} - 4 x - 21} \setminus \cdot \setminus \frac{14 - 2 x}{{x}^{2} - 19 x + 90}$

first factor each term:

$\frac{\left(x + 3\right) \left(x - 10\right)}{\left(x + 3\right) \left(x - 7\right)} \cdot \frac{2 \left(7 - x\right)}{\left(x - 9\right) \left(x - 10\right)}$

Cancel terms:

$\frac{\cancel{\left(x + 3\right)} \cancel{\left(x - 10\right)}}{\cancel{\left(x + 3\right)} \left(x - 7\right)} \cdot \frac{2 \left(7 - x\right)}{\left(x - 9\right) \cancel{\left(x - 10\right)}}$

$\frac{2 \left(7 - x\right)}{\left(x - 7\right) \left(x - 9\right)}$

Jun 22, 2018

$- \frac{2}{x - 9}$

#### Explanation:

For the second-degree terms, we need to think of two numbers that sum up to the middle term and have a product of the last term. These will become our factors.

$\left(\frac{\textcolor{b l u e}{{x}^{2} - 7 x - 30}}{\textcolor{\mathrm{da} r k v i o \le t}{{x}^{2} - 4 x - 21}}\right) \cdot \left(\frac{14 - 2 x}{\textcolor{\lim e}{{x}^{2} - 19 x + 90}}\right)$

$\frac{\textcolor{b l u e}{\left(x - 10\right) \left(x + 3\right)}}{\textcolor{\mathrm{da} r k v i o \le t}{\left(x - 7\right) \left(x + 3\right)}} \cdot \frac{14 - 2 x}{\textcolor{\lim e}{\left(x - 10\right) \left(x - 9\right)}}$

We can factor a $2$ out of the black term, which will leave us with

$\frac{\left(x - 10\right) \left(x + 3\right)}{\left(x - 7\right) \left(x + 3\right)} \cdot \frac{2 \left(7 - x\right)}{\left(x - 10\right) \left(x - 9\right)}$

Like terms on the top and bottom will cancel. We're left with

$\frac{\cancel{\left(x - 10\right)} \cancel{\left(x + 3\right)}}{\left(x - 7\right) \cancel{\left(x + 3\right)}} \cdot \frac{2 \left(7 - x\right)}{\cancel{\left(x - 10\right)} \left(x - 9\right)}$

$\frac{1}{x - 7} \cdot \frac{2 \left(7 - x\right)}{x - 9}$

Which simplifies to

$\frac{2 \textcolor{\mathmr{and} a n \ge}{\left(7 - x\right)}}{\left(x - 7\right) \left(x - 9\right)}$

$\frac{2 \cdot \textcolor{\mathmr{and} a n \ge}{- 1 \left(x - 7\right)}}{\left(x - 7\right) \left(x - 9\right)}$

$\frac{2 \cdot \textcolor{\mathmr{and} a n \ge}{- 1 \cancel{\left(x - 7\right)}}}{\cancel{\left(x - 7\right)} \left(x - 9\right)}$

$\implies - \frac{2}{x - 9}$