Simplify (k^2+9k-10)/34*2/(k^2-1)

$\frac{k^{2} + 9 k - 10}{34} \cdot \frac{2}{k^{2} - 1}$

Step 1

Factor $k^{2} + 9 k - 10$ 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 $9$ .

$- 1, 10$

Write the factored form using these integers.

$\frac{(k - 1) (k + 10)}{34} \cdot \frac{2}{k^{2} - 1}$

Step 2

Simplify the denominator.

Rewrite $1$ as $1^{2}$ .

$\frac{(k - 1) (k + 10)}{34} \cdot \frac{2}{k^{2} - 1^{2}}$

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 $b = 1$ .

$\frac{(k - 1) (k + 10)}{34} \cdot \frac{2}{(k + 1) (k - 1)}$

Step 3

Simplify terms.

Cancel the common factor of $k - 1$ .

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

$\frac{(k - 1) (k + 10)}{34} \cdot \frac{2}{(k - 1) (k + 1)}$

Cancel the common factor.

$\frac{(k - 1) (k + 10)}{34} \cdot \frac{2}{(k - 1) (k + 1)}$

Rewrite the expression.

$\frac{k + 10}{34} \cdot \frac{2}{k + 1}$

Cancel the common factor of $2$ .

Factor $2$ out of $34$ .

$\frac{k + 10}{2 (17)} \cdot \frac{2}{k + 1}$

Cancel the common factor.

$\frac{k + 10}{2 \cdot 17} \cdot \frac{2}{k + 1}$

Rewrite the expression.

$\frac{k + 10}{17} \cdot \frac{1}{k + 1}$

Multiply $\frac{k + 10}{17}$ by $\frac{1}{k + 1}$ .

$\frac{k + 10}{17 (k + 1)}$

Simplify (k^2+9k-10)/34*2/(k^2-1)

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