How do you simplify $\frac{x^{2} - x - 6}{x^{2} - 4}$ $(x^2 - x - 6 )/( x^2 - 4)$ ?

Question

17114 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-01T07:01:19

$\frac{x^{2} - x - 6}{x^{2} - 4} = \frac{x - 3}{x - 2}$ $(x^2-x-6 )/(x^2-4)=(x-3)/(x-2)$

To simplify $\frac{x^{2} - x - 6}{x^{2} - 4}$ $(x^2-x-6 )/(x^2-4)$ , one needs to factorize numerator and denominator.

$x^{2} - x - 6 = x^{2} - 3 x + 2 x - 6 = x (x - 3) + 2 (x - 3) = (x + 2) (x - 3)$ $x^2-x-6=x^2-3x+2x-6=x(x-3)+2(x-3)=(x+2)(x-3)$

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

Hence $\frac{x^{2} - x - 6}{x^{2} - 4} = \frac{(x + 2) (x - 3)}{(x + 2) (x - 2)}$ $(x^2-x-6 )/(x^2-4)=((x+2)(x-3))/((x+2)(x-2))$

= $\frac{x - 3}{x - 2}$ $(x-3)/(x-2)$

How do you simplify x2−x−6x2−4(x^2 - x - 6 )/( x^2 - 4)?