How do you simplify $\frac{x^{2} + 5 x + 6}{x + 1} \cdot \frac{x^{2} - 1}{x + 3}$ $(x^2+5x+6)/(x+1) * (x^2-1)/(x+3)$ ?

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Answer 1 · 2018-05-12T00:28:42

$x^{2} + x - 3$ $x^2 + x - 3$

$\frac{x^{2} + 5 x + 6}{x + 1}$ $(x^2 + 5x + 6)/ (x+1)$ • $\frac{x^{2} - 1}{x + 3}$ $(x^2 - 1)/ (x+3)$

Factor out the numerators:

$\frac{(x + 3) (x + 2)}{x + 1}$ $((x+3)(x+2))/(x+1)$ • $\frac{(x + 1) (x - 1)}{x + 3}$ $((x+1)(x-1))/(x+3)$

Cross simplify to get:

$\frac{x + 2}{1}$ $(x+2)/1$ • $\frac{x - 1}{1}$ $(x-1)/1$

Make it into factoring:

$(x + 2)$ $(x+2)$ • $(x - 1)$ $(x-1)$

You get:

$x^{2} + x - 3$ $x^2 + x - 3$

How do you simplify x2+5x+6x+1⋅x2−1x+3(x^2+5x+6)/(x+1) * (x^2-1)/(x+3)?