Simplify (48-3n^2)/(n^2+n-20)

$\frac{48 - 3 n^{2}}{n^{2} + n - 20}$

Simplify the numerator.

Factor $3$ out of $48 - 3 n^{2}$ .

Factor $3$ out of $48$ .

$\frac{3 (16) - 3 n^{2}}{n^{2} + n - 20}$

Factor $3$ out of $- 3 n^{2}$ .

$\frac{3 (16) + 3 (- n^{2})}{n^{2} + n - 20}$

Factor $3$ out of $3 (16) + 3 (- n^{2})$ .

$\frac{3 (16 - n^{2})}{n^{2} + n - 20}$

Rewrite $16$ as $4^{2}$ .

$\frac{3 (4^{2} - n^{2})}{n^{2} + n - 20}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = n$ .

$\frac{3 (4 + n) (4 - n)}{n^{2} + n - 20}$

Factor $n^{2} + n - 20$ 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 $- 20$ and whose sum is $1$ .

$- 4, 5$

Write the factored form using these integers.

$\frac{3 (4 + n) (4 - n)}{(n - 4) (n + 5)}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $4 - n$ and $n - 4$ .

Rewrite $4$ as $- 1 (- 4)$ .

$\frac{3 (4 + n) (- 1 (- 4) - n)}{(n - 4) (n + 5)}$

Factor $- 1$ out of $- n$ .

$\frac{3 (4 + n) (- 1 (- 4) - (n))}{(n - 4) (n + 5)}$

Factor $- 1$ out of $- 1 (- 4) - (n)$ .

$\frac{3 (4 + n) (- 1 (- 4 + n))}{(n - 4) (n + 5)}$

Reorder terms.

$\frac{3 (4 + n) (- 1 (n - 4))}{(n - 4) (n + 5)}$

Cancel the common factor.

$\frac{3 (4 + n) (- 1 (n - 4))}{(n - 4) (n + 5)}$

Rewrite the expression.

$\frac{3 (4 + n) \cdot (- 1)}{n + 5}$

Simplify the expression.

Multiply $- 1$ by $3$ .

$\frac{- 3 (4 + n)}{n + 5}$

Move the negative in front of the fraction.

$- \frac{3 (4 + n)}{n + 5}$

Simplify (48-3n^2)/(n^2+n-20)

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