Divide ((5k^2-4k-1)/(5k^2+16k+11))÷((25k^2-1)/(5k^2+6k-11))

$\frac{5 k^{2} - 4 k - 1}{5 k^{2} + 16 k + 11} \div \frac{25 k^{2} - 1}{5 k^{2} + 6 k - 11}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{5 k^{2} - 4 k - 1}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 1 = - 5$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 k$ .

$\frac{5 k^{2} - 4 (k) - 1}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Rewrite $- 4$ as $1$ plus $- 5$

$\frac{5 k^{2} + (1 - 5) k - 1}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Apply the distributive property.

$\frac{5 k^{2} + 1 k - 5 k - 1}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Multiply $k$ by $1$ .

$\frac{5 k^{2} + k - 5 k - 1}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(5 k^{2} + k) - 5 k - 1}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Factor out the greatest common factor (GCF) from each group.

$\frac{k (5 k + 1) - (5 k + 1)}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Factor the polynomial by factoring out the greatest common factor, $5 k + 1$ .

$\frac{(5 k + 1) (k - 1)}{5 k^{2} + 16 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Step 3

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot 11 = 55$ and whose sum is $b = 16$ .

Factor $16$ out of $16 k$ .

$\frac{(5 k + 1) (k - 1)}{5 k^{2} + 16 (k) + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Rewrite $16$ as $5$ plus $11$

$\frac{(5 k + 1) (k - 1)}{5 k^{2} + (5 + 11) k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Apply the distributive property.

$\frac{(5 k + 1) (k - 1)}{5 k^{2} + 5 k + 11 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(5 k + 1) (k - 1)}{(5 k^{2} + 5 k) + 11 k + 11} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Factor out the greatest common factor (GCF) from each group.

$\frac{(5 k + 1) (k - 1)}{5 k (k + 1) + 11 (k + 1)} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Factor the polynomial by factoring out the greatest common factor, $k + 1$ .

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{5 k^{2} + 6 k - 11}{25 k^{2} - 1}$

Step 4

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 11 = - 55$ and whose sum is $b = 6$ .

Factor $6$ out of $6 k$ .

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{5 k^{2} + 6 (k) - 11}{25 k^{2} - 1}$

Rewrite $6$ as $- 5$ plus $11$

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{5 k^{2} + (- 5 + 11) k - 11}{25 k^{2} - 1}$

Apply the distributive property.

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{5 k^{2} - 5 k + 11 k - 11}{25 k^{2} - 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{(5 k^{2} - 5 k) + 11 k - 11}{25 k^{2} - 1}$

Factor out the greatest common factor (GCF) from each group.

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{5 k (k - 1) + 11 (k - 1)}{25 k^{2} - 1}$

Factor the polynomial by factoring out the greatest common factor, $k - 1$ .

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{(k - 1) (5 k + 11)}{25 k^{2} - 1}$

Step 5

Simplify the denominator.

Rewrite $25 k^{2}$ as ${(5 k)}^{2}$ .

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{(k - 1) (5 k + 11)}{{(5 k)}^{2} - 1}$

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

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{(k - 1) (5 k + 11)}{{(5 k)}^{2} - 1^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5 k$ and $b = 1$ .

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{(k - 1) (5 k + 11)}{(5 k + 1) (5 k - 1)}$

Step 6

Cancel the common factor of $5 k + 1$ .

Cancel the common factor.

$\frac{(5 k + 1) (k - 1)}{(k + 1) (5 k + 11)} \cdot \frac{(k - 1) (5 k + 11)}{(5 k + 1) (5 k - 1)}$

Rewrite the expression.

$\frac{k - 1}{(k + 1) (5 k + 11)} \cdot \frac{(k - 1) (5 k + 11)}{5 k - 1}$

Step 7

Cancel the common factor of $5 k + 11$ .

Factor $5 k + 11$ out of $(k + 1) (5 k + 11)$ .

$\frac{k - 1}{(5 k + 11) (k + 1)} \cdot \frac{(k - 1) (5 k + 11)}{5 k - 1}$

Factor $5 k + 11$ out of $(k - 1) (5 k + 11)$ .

$\frac{k - 1}{(5 k + 11) (k + 1)} \cdot \frac{(5 k + 11) (k - 1)}{5 k - 1}$

Cancel the common factor.

$\frac{k - 1}{(5 k + 11) (k + 1)} \cdot \frac{(5 k + 11) (k - 1)}{5 k - 1}$

Rewrite the expression.

$\frac{k - 1}{k + 1} \cdot \frac{k - 1}{5 k - 1}$

Step 8

Multiply $\frac{k - 1}{k + 1}$ by $\frac{k - 1}{5 k - 1}$ .

$\frac{(k - 1) (k - 1)}{(k + 1) (5 k - 1)}$

Step 9

Raise $k - 1$ to the power of $1$ .

$\frac{{(k - 1)}^{1} (k - 1)}{(k + 1) (5 k - 1)}$

Step 10

Raise $k - 1$ to the power of $1$ .

$\frac{{(k - 1)}^{1} {(k - 1)}^{1}}{(k + 1) (5 k - 1)}$

Step 11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{{(k - 1)}^{1 + 1}}{(k + 1) (5 k - 1)}$

Step 12

Add $1$ and $1$ .

$\frac{{(k - 1)}^{2}}{(k + 1) (5 k - 1)}$

Divide ((5k^2-4k-1)/(5k^2+16k+11))÷((25k^2-1)/(5k^2+6k-11))

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