Pre-Algebra Examples

$\frac{3 y^{2} \cdot (13 y) + 4}{y^{2} - 16} \div \frac{4 y^{2} - 1}{2 y^{2} - 9 y + 4}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{3 y^{2} \cdot (13 y) + 4}{y^{2} - 16} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 2

Simplify the numerator.

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Step 2.1

Rewrite using the commutative property of multiplication.

$\frac{3 \cdot 13 y^{2} y + 4}{y^{2} - 16} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 2.2

Multiply $y^{2}$ by $y$ by adding the exponents.

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Step 2.2.1

Move $y$ .

$\frac{3 \cdot 13 (y \cdot y^{2}) + 4}{y^{2} - 16} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 2.2.2

Multiply $y$ by $y^{2}$ .

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Step 2.2.2.1

Raise $y$ to the power of $1$ .

$\frac{3 \cdot 13 (y^{1} y^{2}) + 4}{y^{2} - 16} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 2.2.2.2

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

$\frac{3 \cdot 13 y^{1 + 2} + 4}{y^{2} - 16} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 2.2.3

Add $1$ and $2$ .

$\frac{3 \cdot 13 y^{3} + 4}{y^{2} - 16} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 2.3

Multiply $3$ by $13$ .

$\frac{39 y^{3} + 4}{y^{2} - 16} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 3

Simplify the denominator.

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Step 3.1

Rewrite $16$ as $4^{2}$ .

$\frac{39 y^{3} + 4}{y^{2} - 4^{2}} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

Step 3.2

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

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{2 y^{2} - 9 y + 4}{4 y^{2} - 1}$

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 $a \cdot c = 2 \cdot 4 = 8$ and whose sum is $b = - 9$ .

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Step 4.1.1

Factor $- 9$ out of $- 9 y$ .

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{2 y^{2} - 9 (y) + 4}{4 y^{2} - 1}$

Step 4.1.2

Rewrite $- 9$ as $- 1$ plus $- 8$

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{2 y^{2} + (- 1 - 8) y + 4}{4 y^{2} - 1}$

Step 4.1.3

Apply the distributive property.

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{2 y^{2} - 1 y - 8 y + 4}{4 y^{2} - 1}$

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.

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{(2 y^{2} - 1 y) - 8 y + 4}{4 y^{2} - 1}$

Step 4.2.2

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

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{y (2 y - 1) - 4 (2 y - 1)}{4 y^{2} - 1}$

Step 4.3

Factor the polynomial by factoring out the greatest common factor, $2 y - 1$ .

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{(2 y - 1) (y - 4)}{4 y^{2} - 1}$

Step 5

Simplify the denominator.

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Step 5.1

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{(2 y - 1) (y - 4)}{{(2 y)}^{2} - 1}$

Step 5.2

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

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{(2 y - 1) (y - 4)}{{(2 y)}^{2} - 1^{2}}$

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 $a = 2 y$ and $b = 1$ .

$\frac{39 y^{3} + 4}{(y + 4) (y - 4)} \cdot \frac{(2 y - 1) (y - 4)}{(2 y + 1) (2 y - 1)}$

Step 6

Simplify terms.

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Step 6.1

Cancel the common factor of $y - 4$ .

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Step 6.1.1

Factor $y - 4$ out of $(y + 4) (y - 4)$ .

$\frac{39 y^{3} + 4}{(y - 4) (y + 4)} \cdot \frac{(2 y - 1) (y - 4)}{(2 y + 1) (2 y - 1)}$

Step 6.1.2

Factor $y - 4$ out of $(2 y - 1) (y - 4)$ .

$\frac{39 y^{3} + 4}{(y - 4) (y + 4)} \cdot \frac{(y - 4) (2 y - 1)}{(2 y + 1) (2 y - 1)}$

Step 6.1.3

Cancel the common factor.

$\frac{39 y^{3} + 4}{(y - 4) (y + 4)} \cdot \frac{(y - 4) (2 y - 1)}{(2 y + 1) (2 y - 1)}$

Step 6.1.4

Rewrite the expression.

$\frac{39 y^{3} + 4}{y + 4} \cdot \frac{2 y - 1}{(2 y + 1) (2 y - 1)}$

Step 6.2

Multiply $\frac{39 y^{3} + 4}{y + 4}$ by $\frac{2 y - 1}{(2 y + 1) (2 y - 1)}$ .

$\frac{(39 y^{3} + 4) (2 y - 1)}{(y + 4) ((2 y + 1) (2 y - 1))}$

Step 6.3

Cancel the common factor of $2 y - 1$ .

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Step 6.3.1

Cancel the common factor.

$\frac{(39 y^{3} + 4) (2 y - 1)}{(y + 4) ((2 y + 1) (2 y - 1))}$

Step 6.3.2

Rewrite the expression.

$\frac{39 y^{3} + 4}{(y + 4) (2 y + 1)}$