Pre-Algebra Examples

$\frac{4 k^{2} - 3 k - 1}{4 k^{2} + 9 k + 5} \div \frac{16 k^{2} - 1}{4 k^{2} + 1 k - 5}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{4 k^{2} - 3 k - 1}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 2

Factor by grouping.

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Step 2.1

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 $a \cdot c = 4 \cdot - 1 = - 4$ and whose sum is $b = - 3$ .

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Step 2.1.1

Factor $- 3$ out of $- 3 k$ .

$\frac{4 k^{2} - 3 (k) - 1}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 2.1.2

Rewrite $- 3$ as $1$ plus $- 4$

$\frac{4 k^{2} + (1 - 4) k - 1}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 2.1.3

Apply the distributive property.

$\frac{4 k^{2} + 1 k - 4 k - 1}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 2.1.4

Multiply $k$ by $1$ .

$\frac{4 k^{2} + k - 4 k - 1}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 2.2

Factor out the greatest common factor from each group.

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Step 2.2.1

Group the first two terms and the last two terms.

$\frac{(4 k^{2} + k) - 4 k - 1}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 2.2.2

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

$\frac{k (4 k + 1) - (4 k + 1)}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $4 k + 1$ .

$\frac{(4 k + 1) (k - 1)}{4 k^{2} + 9 k + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 3

Factor by grouping.

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Step 3.1

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 $a \cdot c = 4 \cdot 5 = 20$ and whose sum is $b = 9$ .

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Step 3.1.1

Factor $9$ out of $9 k$ .

$\frac{(4 k + 1) (k - 1)}{4 k^{2} + 9 (k) + 5} \cdot \frac{4 k^{2} + 1 k - 5}{16 k^{2} - 1}$

Step 3.1.2

Rewrite $9$ as $4$ plus $5$

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

Step 3.1.3

Apply the distributive property.

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

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

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

Step 3.2.2

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

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

Step 3.3

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

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

Step 4

Factor by grouping.

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Step 4.1

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 $a \cdot c = 4 \cdot - 5 = - 20$ and whose sum is $b = 1$ .

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Step 4.1.1

Factor $1$ out of $1 k$ .

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

Step 4.1.2

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

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

Step 4.1.3

Apply the distributive property.

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

Step 4.2

Factor out the greatest common factor from each group.

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Step 4.2.1

Group the first two terms and the last two terms.

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

Step 4.2.2

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

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

Step 4.3

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

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

Step 5

Simplify the denominator.

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Step 5.1

Rewrite $16 k^{2}$ as ${(4 k)}^{2}$ .

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

Step 5.2

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

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

Step 5.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 k$ and $b = 1$ .

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

Step 6

Simplify terms.

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Step 6.1

Cancel the common factor of $4 k + 1$ .

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Step 6.1.1

Cancel the common factor.

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

Step 6.1.2

Rewrite the expression.

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

Step 6.2

Cancel the common factor of $4 k + 5$ .

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Step 6.2.1

Factor $4 k + 5$ out of $(k + 1) (4 k + 5)$ .

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

Step 6.2.2

Factor $4 k + 5$ out of $(k - 1) (4 k + 5)$ .

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

Step 6.2.3

Cancel the common factor.

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

Step 6.2.4

Rewrite the expression.

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

Step 6.3

Multiply $\frac{k - 1}{k + 1}$ by $\frac{k - 1}{4 k - 1}$ .

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

Step 7

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

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

Step 8

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

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

Step 9

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

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

Step 10

Add $1$ and $1$ .

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