# How do you multiply and simplify \frac { x ^ { 2} + 5x - 14} { x ^ { 2} - 9x + 20} \cdot \frac { x ^ { 2} - 3x - 10} { x ^ { 2} + 9x + 14}?

Apr 7, 2017

$\textcolor{red}{\frac{x - 2}{x + 4}}$

#### Explanation:

This answer may contain some information which you already know but I have added it for the sake of completeness. This has also caused the solution to APPEAR very long although the solution to the problem itself is quite easy and concise . So do give it a read.

Suggestions are welcome.

# Let us factorize all the four polynomials to see what we get.

To factorize a polynomial of type $a {x}^{2} + b x + c$, where $a , b , c$ are constants, try to find two numbers, say $d$ & $e$ (by trial and error) such that:-

$i .$ $b = d + e$ &
$i i .$ $d \times e = a \times c$

then $a {x}^{2} + b x + c$ can be written as;
$a {x}^{2} + \left(d + e\right) x + c = a {x}^{2} + \mathrm{dx} + e x + c$

$= a x \left(x + \frac{d}{a}\right) + e \left(x + \frac{c}{e}\right)$

[but from condition $i i .$, $d \times e = a \times c \implies \frac{d}{a} = \frac{c}{e}$]

$= a x \left(x + \frac{d}{a}\right) + e \left(x + \frac{d}{a}\right)$ OR $a x \left(x + \frac{c}{e}\right) + e \left(x + \frac{c}{e}\right)$

$= \textcolor{red}{\left(x + \frac{d}{a}\right) \left(a x + e\right)}$ OR $\textcolor{red}{\left(x + \frac{c}{e}\right) \left(a x + e\right)}$

1. ${x}^{2} + 5 x - 14$

[Here, $a = 1 , b = 5 , c = - 14$; also observe that $7 + \left(- 2\right) = 7 - 2 = 5$ & $7 \times \left(- 2\right) = - 14 = 1 \times \left(- 14\right)$]

So, ${x}^{2} + 5 x - 14 = {x}^{2} + \left(7 - 2\right) x - 14$
$= {x}^{2} + 7 x - 2 x - 14$
$= x \left(x + 7\right) - 2 \left(x + 7\right)$
$= \textcolor{red}{\left(x - 2\right) \left(x + 7\right)}$

2. ${x}^{2} - 9 x + 20$

[Here, $a = 1 , b = - 9 , c = 20$; and $- 5 + \left(- 4\right) = - 5 - 4 = - 9$ & $\left(- 5\right) \times \left(- 4\right) = 20 = 1 \times 20$]

So, ${x}^{2} - 9 x + 20 = {x}^{2} + \left(- 5 - 4\right) x + 20$
$= {x}^{2} - 5 x - 4 x + 20$
$= x \left(x - 5\right) - 4 \left(x - 20\right)$
$= \textcolor{red}{\left(x - 4\right) \left(x - 5\right)}$

Following the similar procedure we can factorize rest two.

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

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

Substituting all the values in the given equation;

$\frac{{x}^{2} + 5 x - 14}{{x}^{2} - 9 x + 20} \cdot \frac{{x}^{2} - 3 x - 10}{{x}^{2} + 9 x + 14}$

$= \frac{\left(x - 2\right) \cancel{\left(x + 7\right)}}{\left(x - 4\right) \cancel{\left(x - 5\right)}} \cdot \frac{\cancel{\left(x + 2\right)} \cancel{\left(x - 5\right)}}{\cancel{\left(x + 2\right)} \cancel{\left(x + 7\right)}}$