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

Jan 31, 2017

1

Explanation:

Since

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

you would substitute in the given fraction and symplify:

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