Simplify ((n^2-9n-10)/(n+1))÷((n+3)/(n+10))

$\frac{n^{2} - 9 n - 10}{n + 1} \div \frac{n + 3}{n + 10}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{n^{2} - 9 n - 10}{n + 1} \cdot \frac{n + 10}{n + 3}$

Step 2

Factor $n^{2} - 9 n - 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 10$ and whose sum is $- 9$ .

$- 10, 1$

Write the factored form using these integers.

$\frac{(n - 10) (n + 1)}{n + 1} \cdot \frac{n + 10}{n + 3}$

Step 3

Simplify terms.

Cancel the common factor of $n + 1$ .

Cancel the common factor.

$\frac{(n - 10) (n + 1)}{n + 1} \cdot \frac{n + 10}{n + 3}$

Divide $n - 10$ by $1$ .

$(n - 10) \frac{n + 10}{n + 3}$

Multiply $n - 10$ by $\frac{n + 10}{n + 3}$ .

$\frac{(n - 10) (n + 10)}{n + 3}$

Simplify ((n^2-9n-10)/(n+1))÷((n+3)/(n+10))

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