Trigonometry Examples

$\frac{k^{2} + 5 k + 6}{k^{2} + 11 k + 8} \cdot \frac{k^{2} + 9 k}{k^{2} - 2 k - 15}$

Step 1

Factor $k^{2} + 5 k + 6$ using the AC method.

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Step 1.1

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 $6$ and whose sum is $5$ .

$2, 3$

Step 1.2

Write the factored form using these integers.

$\frac{(k + 2) (k + 3)}{k^{2} + 11 k + 8} \cdot \frac{k^{2} + 9 k}{k^{2} - 2 k - 15}$

Step 2

Factor $k$ out of $k^{2} + 9 k$ .

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Step 2.1

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

$\frac{(k + 2) (k + 3)}{k^{2} + 11 k + 8} \cdot \frac{k \cdot k + 9 k}{k^{2} - 2 k - 15}$

Step 2.2

Factor $k$ out of $9 k$ .

$\frac{(k + 2) (k + 3)}{k^{2} + 11 k + 8} \cdot \frac{k \cdot k + k \cdot 9}{k^{2} - 2 k - 15}$

Step 2.3

Factor $k$ out of $k \cdot k + k \cdot 9$ .

$\frac{(k + 2) (k + 3)}{k^{2} + 11 k + 8} \cdot \frac{k (k + 9)}{k^{2} - 2 k - 15}$

Step 3

Factor $k^{2} - 2 k - 15$ using the AC method.

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Step 3.1

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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Step 3.2

Write the factored form using these integers.

$\frac{(k + 2) (k + 3)}{k^{2} + 11 k + 8} \cdot \frac{k (k + 9)}{(k - 5) (k + 3)}$

Step 4

Simplify terms.

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Step 4.1

Cancel the common factor of $k + 3$ .

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Step 4.1.1

Factor $k + 3$ out of $(k + 2) (k + 3)$ .

$\frac{(k + 3) (k + 2)}{k^{2} + 11 k + 8} \cdot \frac{k (k + 9)}{(k - 5) (k + 3)}$

Step 4.1.2

Factor $k + 3$ out of $(k - 5) (k + 3)$ .

$\frac{(k + 3) (k + 2)}{k^{2} + 11 k + 8} \cdot \frac{k (k + 9)}{(k + 3) (k - 5)}$

Step 4.1.3

Cancel the common factor.

$\frac{(k + 3) (k + 2)}{k^{2} + 11 k + 8} \cdot \frac{k (k + 9)}{(k + 3) (k - 5)}$

Step 4.1.4

Rewrite the expression.

$\frac{k + 2}{k^{2} + 11 k + 8} \cdot \frac{k (k + 9)}{k - 5}$

Step 4.2

Multiply $\frac{k + 2}{k^{2} + 11 k + 8}$ by $\frac{k (k + 9)}{k - 5}$ .

$\frac{(k + 2) (k (k + 9))}{(k^{2} + 11 k + 8) (k - 5)}$

Step 4.3

Reorder factors in $\frac{(k + 2) k (k + 9)}{(k^{2} + 11 k + 8) (k - 5)}$ .

$\frac{k (k + 2) (k + 9)}{(k^{2} + 11 k + 8) (k - 5)}$