Algebra Examples

$\frac{25 - 4 x^{2}}{6 x^{2} + 9 x - 15} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $25$ as $5^{2}$ .

$\frac{5^{2} - 4 x^{2}}{6 x^{2} + 9 x - 15} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 1.2

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

$\frac{5^{2} - {(2 x)}^{2}}{6 x^{2} + 9 x - 15} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 1.3

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

$\frac{(5 + 2 x) (5 - (2 x))}{6 x^{2} + 9 x - 15} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 1.4

Multiply $2$ by $- 1$ .

$\frac{(5 + 2 x) (5 - 2 x)}{6 x^{2} + 9 x - 15} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2

Simplify the denominator.

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Step 2.1

Factor $3$ out of $6 x^{2} + 9 x - 15$ .

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Step 2.1.1

Factor $3$ out of $6 x^{2}$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2}) + 9 x - 15} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.1.2

Factor $3$ out of $9 x$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2}) + 3 (3 x) - 15} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.1.3

Factor $3$ out of $- 15$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2}) + 3 (3 x) + 3 (- 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.1.4

Factor $3$ out of $3 (2 x^{2}) + 3 (3 x)$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2} + 3 x) + 3 (- 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.1.5

Factor $3$ out of $3 (2 x^{2} + 3 x) + 3 (- 5)$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2} + 3 x - 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.2

Factor by grouping.

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Step 2.2.1

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 3$ .

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Step 2.2.1.1

Factor $3$ out of $3 x$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2} + 3 (x) - 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.2.1.2

Rewrite $3$ as $- 2$ plus $5$

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2} + (- 2 + 5) x - 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.2.1.3

Apply the distributive property.

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x^{2} - 2 x + 5 x - 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$\frac{(5 + 2 x) (5 - 2 x)}{3 ((2 x^{2} - 2 x) + 5 x - 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.2.2.2

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

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x (x - 1) + 5 (x - 1))} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 2.2.3

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

$\frac{(5 + 2 x) (5 - 2 x)}{3 ((x - 1) (2 x + 5))} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{6 x^{2} - 2 x - 4}{2 x^{2} - x - 10}$

Step 3

Simplify the numerator.

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Step 3.1

Factor $2$ out of $6 x^{2} - 2 x - 4$ .

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Step 3.1.1

Factor $2$ out of $6 x^{2}$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2}) - 2 x - 4}{2 x^{2} - x - 10}$

Step 3.1.2

Factor $2$ out of $- 2 x$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2}) + 2 (- x) - 4}{2 x^{2} - x - 10}$

Step 3.1.3

Factor $2$ out of $- 4$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2}) + 2 (- x) + 2 (- 2)}{2 x^{2} - x - 10}$

Step 3.1.4

Factor $2$ out of $2 (3 x^{2}) + 2 (- x)$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2} - x) + 2 (- 2)}{2 x^{2} - x - 10}$

Step 3.1.5

Factor $2$ out of $2 (3 x^{2} - x) + 2 (- 2)$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2} - x - 2)}{2 x^{2} - x - 10}$

Step 3.2

Factor by grouping.

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Step 3.2.1

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 $a \cdot c = 3 \cdot - 2 = - 6$ and whose sum is $b = - 1$ .

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Step 3.2.1.1

Factor $- 1$ out of $- x$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2} - (x) - 2)}{2 x^{2} - x - 10}$

Step 3.2.1.2

Rewrite $- 1$ as $2$ plus $- 3$

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2} + (2 - 3) x - 2)}{2 x^{2} - x - 10}$

Step 3.2.1.3

Apply the distributive property.

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x^{2} + 2 x - 3 x - 2)}{2 x^{2} - x - 10}$

Step 3.2.2

Factor out the greatest common factor from each group.

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Step 3.2.2.1

Group the first two terms and the last two terms.

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 ((3 x^{2} + 2 x) - 3 x - 2)}{2 x^{2} - x - 10}$

Step 3.2.2.2

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

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (x (3 x + 2) - (3 x + 2))}{2 x^{2} - x - 10}$

Step 3.2.3

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 ((3 x + 2) (x - 1))}{2 x^{2} - x - 10}$

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x + 2) (x - 1)}{2 x^{2} - x - 10}$

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 $a \cdot c = 2 \cdot - 10 = - 20$ and whose sum is $b = - 1$ .

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Step 4.1.1

Factor $- 1$ out of $- x$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x + 2) (x - 1)}{2 x^{2} - (x) - 10}$

Step 4.1.2

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

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x + 2) (x - 1)}{2 x^{2} + (4 - 5) x - 10}$

Step 4.1.3

Apply the distributive property.

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x + 2) (x - 1)}{2 x^{2} + 4 x - 5 x - 10}$

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.

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x + 2) (x - 1)}{(2 x^{2} + 4 x) - 5 x - 10}$

Step 4.2.2

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

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x + 2) (x - 1)}{2 x (x + 2) - 5 (x + 2)}$

Step 4.3

Factor the polynomial by factoring out the greatest common factor, $x + 2$ .

$\frac{(5 + 2 x) (5 - 2 x)}{3 (x - 1) (2 x + 5)} \cdot \frac{2 (3 x + 2) (x - 1)}{(x + 2) (2 x - 5)}$

Step 5

Cancel the common factor of $x - 1$ .

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Step 5.1

Factor $x - 1$ out of $3 (x - 1) (2 x + 5)$ .

$\frac{(5 + 2 x) (5 - 2 x)}{(x - 1) (3 (2 x + 5))} \cdot \frac{2 (3 x + 2) (x - 1)}{(x + 2) (2 x - 5)}$

Step 5.2

Factor $x - 1$ out of $2 (3 x + 2) (x - 1)$ .

$\frac{(5 + 2 x) (5 - 2 x)}{(x - 1) (3 (2 x + 5))} \cdot \frac{(x - 1) (2 (3 x + 2))}{(x + 2) (2 x - 5)}$

Step 5.3

Cancel the common factor.

$\frac{(5 + 2 x) (5 - 2 x)}{(x - 1) (3 (2 x + 5))} \cdot \frac{(x - 1) (2 (3 x + 2))}{(x + 2) (2 x - 5)}$

Step 5.4

Rewrite the expression.

$\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x + 5)} \cdot \frac{2 (3 x + 2)}{(x + 2) (2 x - 5)}$

Step 6

Multiply $\frac{(5 + 2 x) (5 - 2 x)}{3 (2 x + 5)}$ by $\frac{2 (3 x + 2)}{(x + 2) (2 x - 5)}$ .

$\frac{(5 + 2 x) (5 - 2 x) (2 (3 x + 2))}{3 (2 x + 5) ((x + 2) (2 x - 5))}$

Step 7

Cancel the common factor of $5 + 2 x$ and $2 x + 5$ .

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Step 7.1

Reorder terms.

$\frac{(2 x + 5) (5 - 2 x) (2 (3 x + 2))}{3 (2 x + 5) ((x + 2) (2 x - 5))}$

Step 7.2

Cancel the common factor.

$\frac{(2 x + 5) (5 - 2 x) (2 (3 x + 2))}{3 (2 x + 5) ((x + 2) (2 x - 5))}$

Step 7.3

Rewrite the expression.

$\frac{(5 - 2 x) (2 (3 x + 2))}{3 ((x + 2) (2 x - 5))}$

Step 8

Cancel the common factor of $5 - 2 x$ and $2 x - 5$ .

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Step 8.1

Rewrite $5$ as $- 1 (- 5)$ .

$\frac{(- 1 (- 5) - 2 x) (2 (3 x + 2))}{3 ((x + 2) (2 x - 5))}$

Step 8.2

Factor $- 1$ out of $- 2 x$ .

$\frac{(- 1 (- 5) - (2 x)) (2 (3 x + 2))}{3 ((x + 2) (2 x - 5))}$

Step 8.3

Factor $- 1$ out of $- 1 (- 5) - (2 x)$ .

$\frac{- 1 (- 5 + 2 x) (2 (3 x + 2))}{3 ((x + 2) (2 x - 5))}$

Step 8.4

Reorder terms.

$\frac{- 1 (2 x - 5) (2 (3 x + 2))}{3 (x + 2) (2 x - 5)}$

Step 8.5

Cancel the common factor.

$\frac{- 1 (2 x - 5) (2 (3 x + 2))}{3 (x + 2) (2 x - 5)}$

Step 8.6

Rewrite the expression.

$\frac{- 1 (2 (3 x + 2))}{3 (x + 2)}$

Step 9

Simplify the expression.

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Step 9.1

Multiply $- 1$ by $2$ .

$\frac{- 2 (3 x + 2)}{3 (x + 2)}$

Step 9.2

Move the negative in front of the fraction.

$- \frac{2 (3 x + 2)}{3 (x + 2)}$