How do you simplify $[(x^2+2x)(x^2+2x-3)]/[(x^2+x-2)(x^2+3x)]$ ?

Question

1598 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-31T20:27:14

1

Since

$x^2+2x=x(x+2)$
$x^2+2x-3=(x+3)(x-1)$
$x^2+x-2=(x+2)(x-1)$
$x^2+3x=x(x+3)$

you would substitute in the given fraction and symplify:

$((x^2+2x)(x^2+2x-3))/((x^2+x-2)(x^2+3x))=(x(x+2)(x+3)(x-1))/((x+2)(x-1)x(x+3))=1$

How do you simplify [(x^2+2x)(x^2+2x-3)]/[(x^2+x-2)(x^2+3x)][(x^2+2x)(x^2+2x-3)]/[(x^2+x-2)(x^2+3x)]?