# Question #85711

Jan 18, 2017

$r = \frac{23 \cdot 5}{x + 7} = \frac{115}{x + 7} \implies {r}^{2} = {115}^{2} / {\left(x + 7\right)}^{2} = \frac{13225}{{x}^{2} + 14 x + 49}$
$= \frac{{23}^{2} \cdot {5}^{2}}{x + 7} ^ 2$

#### Explanation:

$r = \frac{23 \cdot 5}{x + 7} \implies {r}^{2} = {\left(\frac{23 \cdot 5}{x + 7}\right)}^{2} = {\left(23 \cdot 5\right)}^{2} / {\left(x + 7\right)}^{2} = \frac{{\left(23\right)}^{2} {\left(5\right)}^{2}}{\left(x + 7\right) \left(x + 7\right)}$

$= \frac{{\left(20 + 3\right)}^{2} \left(25\right)}{x \cdot x + 7 x + 7 x + 7 \left(7\right)}$

$= \frac{\left(\left(20\right) \left(20\right) + \left(3\right) \left(20\right) + \left(3\right) \left(20\right) + \left(3\right) \left(3\right)\right) \left(25\right)}{{x}^{2} + 14 x + 49}$

$= \frac{\left(400 + 60 + 60 + 9\right) \left(25\right)}{{x}^{2} + 14 x + 49}$

$= \frac{\left(400 + 120 + 9\right) \left(25\right)}{{x}^{2} + 14 x + 49} = \frac{\left(529\right) \left(25\right)}{{x}^{2} + 14 x + 49}$

$= \frac{\left(529\right) \left(20 + 5\right)}{{x}^{2} + 14 x + 49} = \frac{529 \left(20\right) + 529 \left(5\right)}{{x}^{2} + 14 x + 49}$

$= \frac{2 \left(5290\right) + 5290 \left(\frac{1}{2}\right)}{{x}^{2} + 14 x + 49} = \frac{10580 + 2645}{{x}^{2} + 14 x + 49}$

$\frac{13225}{{x}^{2} + 14 x + 49}$