How do you simplify the expression $\frac{\frac{x^{2} - 4 x - 32}{x + 1}}{\frac{x^{2} + 6 x + 8}{x^{2} - 1}}$ $((x^2-4x-32)/(x+1))/((x^2+6x+8)/(x^2-1))$ ?

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How do you simplify the expression $\frac{\frac{x^{2} - 4 x - 32}{x + 1}}{\frac{x^{2} + 6 x + 8}{x^{2} - 1}}$ $((x^2-4x-32)/(x+1))/((x^2+6x+8)/(x^2-1))$ ?

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Answer 1 · 2016-12-26T14:38:22

The answer is $= \frac{(x - 1) (x - 8)}{x + 2}$ $=((x-1)(x-8))/(x+2)$

Explanation:

Let's do some factorisations

$x^{2} - 4 x - 32 = (x + 4) (x - 8)$ $x^2-4x-32=(x+4)(x-8)$

$x^{2} + 6 x + 8 = (x + 2) (x + 4)$ $x^2+6x+8=(x+2)(x+4)$

$x^{2} - 1 = (x + 1) (x - 1)$ $x^2-1=(x+1)(x-1)$

Therefore,

$\frac{\frac{x^{2} - 4 x - 32}{x + 1}}{\frac{x^{2} + 6 x + 8}{x^{2} - 1}} = \frac{\frac{(x + 4) (x - 8)}{(x + 1)}}{\frac{(x + 2) (x + 4)}{(x + 1) (x - 1)}}$ $((x^2-4x-32)/(x+1))/((x^2+6x+8)/(x^2-1))=(((x+4)(x-8))/((x+1)))/(((x+2)(x+4))/((x+1)(x-1))$

$=(cancel(x+4)(x-8))/(cancel(x+1))*(cancel(x+1)(x-1))/((x+2)cancel(x+4))$

$=((x-1)(x-8))/(x+2)$

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How do you simplify the expression $\frac{\frac{x^{2} - 4 x - 32}{x + 1}}{\frac{x^{2} + 6 x + 8}{x^{2} - 1}}$ $((x^2-4x-32)/(x+1))/((x^2+6x+8)/(x^2-1))$ ?

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How do you simplify the expression $\frac{\frac{x^{2} - 4 x - 32}{x + 1}}{\frac{x^{2} + 6 x + 8}{x^{2} - 1}}$ $((x^2-4x-32)/(x+1))/((x^2+6x+8)/(x^2-1))$ ?