Algebra Examples

$\frac{4 k + 2}{k^{2} - 4} \cdot \frac{k - 2}{2 k + 1}$

Step 1

Factor $2$ out of $4 k + 2$ .

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Step 1.1

Factor $2$ out of $4 k$ .

$\frac{2 (2 k) + 2}{k^{2} - 4} \cdot \frac{k - 2}{2 k + 1}$

Step 1.2

Factor $2$ out of $2$ .

$\frac{2 (2 k) + 2 (1)}{k^{2} - 4} \cdot \frac{k - 2}{2 k + 1}$

Step 1.3

Factor $2$ out of $2 (2 k) + 2 (1)$ .

$\frac{2 (2 k + 1)}{k^{2} - 4} \cdot \frac{k - 2}{2 k + 1}$

Step 2

Simplify the denominator.

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Step 2.1

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

$\frac{2 (2 k + 1)}{k^{2} - 2^{2}} \cdot \frac{k - 2}{2 k + 1}$

Step 2.2

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

$\frac{2 (2 k + 1)}{(k + 2) (k - 2)} \cdot \frac{k - 2}{2 k + 1}$

Step 3

Reduce the expression by cancelling the common factors.

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Step 3.1

Cancel the common factor of $2 k + 1$ .

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Step 3.1.1

Factor $2 k + 1$ out of $2 (2 k + 1)$ .

$\frac{(2 k + 1) \cdot 2}{(k + 2) (k - 2)} \cdot \frac{k - 2}{2 k + 1}$

Step 3.1.2

Cancel the common factor.

$\frac{(2 k + 1) \cdot 2}{(k + 2) (k - 2)} \cdot \frac{k - 2}{2 k + 1}$

Step 3.1.3

Rewrite the expression.

$\frac{2}{(k + 2) (k - 2)} \cdot (k - 2)$

Step 3.2

Cancel the common factor of $k - 2$ .

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Step 3.2.1

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

$\frac{2}{(k - 2) (k + 2)} \cdot (k - 2)$

Step 3.2.2

Cancel the common factor.

$\frac{2}{(k - 2) (k + 2)} \cdot (k - 2)$

Step 3.2.3

Rewrite the expression.

$\frac{2}{k + 2}$