Basic Math Examples

$\frac{- 3 n - 27}{n - 3} \cdot \frac{n^{3} + n^{2} - 12 n}{n^{3} + 4 n^{2} - 45 n}$

Step 1

Factor $3$ out of $- 3 n - 27$ .

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Step 1.1

Factor $3$ out of $- 3 n$ .

$\frac{3 (- n) - 27}{n - 3} \cdot \frac{n^{3} + n^{2} - 12 n}{n^{3} + 4 n^{2} - 45 n}$

Step 1.2

Factor $3$ out of $- 27$ .

$\frac{3 (- n) + 3 (- 9)}{n - 3} \cdot \frac{n^{3} + n^{2} - 12 n}{n^{3} + 4 n^{2} - 45 n}$

Step 1.3

Factor $3$ out of $3 (- n) + 3 (- 9)$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n^{3} + n^{2} - 12 n}{n^{3} + 4 n^{2} - 45 n}$

Step 2

Simplify the numerator.

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Step 2.1

Factor $n$ out of $n^{3} + n^{2} - 12 n$ .

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Step 2.1.1

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

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n \cdot n^{2} + n^{2} - 12 n}{n^{3} + 4 n^{2} - 45 n}$

Step 2.1.2

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

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n \cdot n^{2} + n \cdot n - 12 n}{n^{3} + 4 n^{2} - 45 n}$

Step 2.1.3

Factor $n$ out of $- 12 n$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n \cdot n^{2} + n \cdot n + n \cdot - 12}{n^{3} + 4 n^{2} - 45 n}$

Step 2.1.4

Factor $n$ out of $n \cdot n^{2} + n \cdot n$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n^{2} + n) + n \cdot - 12}{n^{3} + 4 n^{2} - 45 n}$

Step 2.1.5

Factor $n$ out of $n (n^{2} + n) + n \cdot - 12$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n^{2} + n - 12)}{n^{3} + 4 n^{2} - 45 n}$

Step 2.2

Factor $n^{2} + n - 12$ using the AC method.

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Step 2.2.1

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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Step 2.2.2

Write the factored form using these integers.

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n ((n - 3) (n + 4))}{n^{3} + 4 n^{2} - 45 n}$

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n^{3} + 4 n^{2} - 45 n}$

Step 3

Simplify the denominator.

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Step 3.1

Factor $n$ out of $n^{3} + 4 n^{2} - 45 n$ .

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Step 3.1.1

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

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n \cdot n^{2} + 4 n^{2} - 45 n}$

Step 3.1.2

Factor $n$ out of $4 n^{2}$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n \cdot n^{2} + n (4 n) - 45 n}$

Step 3.1.3

Factor $n$ out of $- 45 n$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n \cdot n^{2} + n (4 n) + n \cdot - 45}$

Step 3.1.4

Factor $n$ out of $n \cdot n^{2} + n (4 n)$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n (n^{2} + 4 n) + n \cdot - 45}$

Step 3.1.5

Factor $n$ out of $n (n^{2} + 4 n) + n \cdot - 45$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n (n^{2} + 4 n - 45)}$

Step 3.2

Factor $n^{2} + 4 n - 45$ using the AC method.

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Step 3.2.1

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 $- 45$ and whose sum is $4$ .

$- 5, 9$

Step 3.2.2

Write the factored form using these integers.

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n ((n - 5) (n + 9))}$

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{n (n - 3) (n + 4)}{n (n - 5) (n + 9)}$

Step 4

Simplify terms.

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Step 4.1

Cancel the common factor of $n - 3$ .

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Step 4.1.1

Factor $n - 3$ out of $n (n - 3) (n + 4)$ .

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{(n - 3) (n (n + 4))}{n (n - 5) (n + 9)}$

Step 4.1.2

Cancel the common factor.

$\frac{3 (- n - 9)}{n - 3} \cdot \frac{(n - 3) (n (n + 4))}{n (n - 5) (n + 9)}$

Step 4.1.3

Rewrite the expression.

$3 (- n - 9) \cdot \frac{n (n + 4)}{n (n - 5) (n + 9)}$

Step 4.2

Combine $\frac{n (n + 4)}{n (n - 5) (n + 9)}$ and $3$ .

$\frac{n (n + 4) \cdot 3}{n (n - 5) (n + 9)} (- n - 9)$

Step 4.3

Cancel the common factor of $n$ .

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Step 4.3.1

Cancel the common factor.

$\frac{n (n + 4) \cdot 3}{n (n - 5) (n + 9)} (- n - 9)$

Step 4.3.2

Rewrite the expression.

$\frac{(n + 4) \cdot 3}{(n - 5) (n + 9)} (- n - 9)$

Step 4.4

Move $3$ to the left of $n + 4$ .

$\frac{3 \cdot (n + 4)}{(n - 5) (n + 9)} (- n - 9)$

Step 4.5

Multiply $\frac{3 (n + 4)}{(n - 5) (n + 9)}$ by $- n - 9$ .

$\frac{3 (n + 4) (- n - 9)}{(n - 5) (n + 9)}$

Step 4.6

Cancel the common factor of $- n - 9$ and $n + 9$ .

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Step 4.6.1

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

$\frac{3 (n + 4) (- (n) - 9)}{(n - 5) (n + 9)}$

Step 4.6.2

Rewrite $- 9$ as $- 1 (9)$ .

$\frac{3 (n + 4) (- (n) - 1 (9))}{(n - 5) (n + 9)}$

Step 4.6.3

Factor $- 1$ out of $- (n) - 1 (9)$ .

$\frac{3 (n + 4) (- (n + 9))}{(n - 5) (n + 9)}$

Step 4.6.4

Rewrite $- (n + 9)$ as $- 1 (n + 9)$ .

$\frac{3 (n + 4) (- 1 (n + 9))}{(n - 5) (n + 9)}$

Step 4.6.5

Cancel the common factor.

$\frac{3 (n + 4) (- 1 (n + 9))}{(n - 5) (n + 9)}$

Step 4.6.6

Rewrite the expression.

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

Step 4.7

Simplify the expression.

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Step 4.7.1

Multiply $- 1$ by $3$ .

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

Step 4.7.2

Move the negative in front of the fraction.

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