# How do you simplify (2x^2-2)/(x^2-6x-7)*(x^2-10x+21)?

Aug 10, 2016

$= 2 \left(x - 1\right) \left(x - 3\right)$

#### Explanation:

$\frac{2 {x}^{2} - 2}{{x}^{2} - 6 x - 7} \times \left({x}^{2} - 10 x + 21\right)$

$= \frac{2 \left({x}^{2} - 1\right)}{{x}^{2} - 7 x + x - 7} \times \left({x}^{2} - 7 x - 3 x + 21\right)$

$= \frac{2 \left(x + 1\right) \left(x - 1\right)}{x \left(x - 7\right) + 1 \left(x - 7\right)} \times \left(x \left(x - 7\right) - 3 \left(x - 7\right)\right)$

$= \frac{2 \left(x + 1\right) \left(x - 1\right)}{\left(x - 7\right) \left(x + 1\right)} \times \left(x - 7\right) \left(x - 3\right)$

$= 2 \left(x - 1\right) \left(x - 3\right)$

Aug 10, 2016

$2 \left(x - 1\right) \left(x - 3\right)$

#### Explanation:

To simplify this expression, we require to $\textcolor{b l u e}{\text{factorise}}$

Let us begin with $2 {x}^{2} - 2$ which has a common factor of 2 in both terms.
$\textcolor{b l u e}{\text{-------------------------------------------------}}$

$\Rightarrow 2 {x}^{2} - 2 = 2 \left({x}^{2} - 1\right) \ldots \ldots . . \left(A\right)$

Now ${x}^{2} - 1$ is a $\textcolor{red}{\text{difference of squares}}$ which, in general factorises as follows.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here ${x}^{2} = {\left(x\right)}^{2} \text{ and " 1=(1)^2rArra=x" and } b = 1$

$\Rightarrow {x}^{2} - 1 = \left(x - 1\right) \left(x + 1\right)$

substitute back into (A)

$\Rightarrow 2 {x}^{2} - 2 = 2 \left(x - 1\right) \left(x + 1\right)$
$\textcolor{b l u e}{\text{------------------------------------------------------}}$

The standard form of a quadratic is $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{a {x}^{2} + b x + c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and to factorise consider the factors of the product ac, which must sum to give b.
$\textcolor{b l u e}{\text{-----------------------------------------------------}}$

For ${x}^{2} - 6 x - 7 , a c = - 7 \text{ and } b = - 6$

The required values the are -7 and +1

$\Rightarrow {x}^{2} - 6 x - 7 = \left(x - 7\right) \left(x + 1\right)$

and for ${x}^{2} - 10 x + 21 , a c = 21 \text{ and } b = - 10$

the required values are -7 and -3

$\Rightarrow {x}^{2} - 10 x + 21 = \left(x - 7\right) \left(x - 3\right)$
$\textcolor{b l u e}{\text{----------------------------------------------------------}}$

Substitute all of these factors back into the original expression.

$\Rightarrow \frac{2 \left(x - 1\right) \left(x + 1\right)}{\left(x - 7\right) \left(x + 1\right)} \times \frac{\left(x - 7\right) \left(x - 3\right)}{1}$

and 'cancelling' common factors between numerator/denominator

$= \frac{2 \left(x - 1\right) \cancel{\left(x + 1\right)}}{\cancel{\left(x - 7\right)} \cancel{\left(x + 1\right)}} \times \frac{\cancel{\left(x - 7\right)} \left(x - 3\right)}{1}$

=2(x-1)(x-3)
$\textcolor{b l u e}{\text{------------}}$