Question

Divide:$\frac{2n^2+10n}{n-1}÷\frac{n^2+10n+24}{n^2+8n-9}·\frac{n+4}{8n^2+12n}$.

Short answer

The required answer is$\frac{\left(n+5\right)\left(n+9\right)}{2\left(n+6\right)\left(2n+3\right)}$.

Step-by-step solution

Step 1. Given information.

The given expression is:

$\frac{2n^2+10n}{n-1}÷\frac{n^2+10n+24}{n^2+8n-9}·\frac{n+4}{8n^2+12n}$

Step 2. Simplify the expression.

$\frac{2n^2+10n}{n-1}÷\frac{n^2+10n+24}{n^2+8n-9}·\frac{n+4}{8n^2+12n}$

Rewrite the division as multiplication by reciprocal,

$\frac{2n^2+10n}{n-1}·\frac{n^2+8n-9}{n^2+10n+24}·\frac{n+4}{8n^2+12n}$

Factor the numerators and denominators and then multiply.

$\frac{2n\left(n+5\right)\left(n-1\right)\left(n+9\right)\left(n+4\right)}{4n\left(n-1\right)\left(n+4\right)\left(n+6\right)\left(2n+3\right)}$

Simplify by dividing out common factors.

$\frac{\left(n+5\right)\left(n+9\right)}{2\left(n+6\right)\left(2n+3\right)}$