Algebra Examples

$\frac{\frac{\frac{x^{2} - 3 x - 10}{x + 3}}{2 x^{2} - 7 x - 15}}{x^{2} - 9}$

Step 1

Multiply the numerator by the reciprocal of the denominator.

$\frac{\frac{x^{2} - 3 x - 10}{x + 3}}{2 x^{2} - 7 x - 15} \cdot \frac{1}{x^{2} - 9}$

Step 2

Multiply the numerator by the reciprocal of the denominator.

$\frac{x^{2} - 3 x - 10}{x + 3} \cdot \frac{1}{2 x^{2} - 7 x - 15} \frac{1}{x^{2} - 9}$

Step 3

Factor $x^{2} - 3 x - 10$ using the AC method.

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Step 3.1

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 $- 3$ .

$- 5, 2$

Step 3.2

Write the factored form using these integers.

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{1}{2 x^{2} - 7 x - 15} \frac{1}{x^{2} - 9}$

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 $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 7$ .

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Step 4.1.1

Factor $- 7$ out of $- 7 x$ .

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{1}{2 x^{2} - 7 (x) - 15} \frac{1}{x^{2} - 9}$

Step 4.1.2

Rewrite $- 7$ as $3$ plus $- 10$

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{1}{2 x^{2} + (3 - 10) x - 15} \frac{1}{x^{2} - 9}$

Step 4.1.3

Apply the distributive property.

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{1}{2 x^{2} + 3 x - 10 x - 15} \frac{1}{x^{2} - 9}$

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.

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{1}{(2 x^{2} + 3 x) - 10 x - 15} \frac{1}{x^{2} - 9}$

Step 4.2.2

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

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

Step 4.3

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

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

Step 5

Simplify terms.

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Step 5.1

Cancel the common factor of $x - 5$ .

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Step 5.1.1

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

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

Step 5.1.2

Cancel the common factor.

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

Step 5.1.3

Rewrite the expression.

$\frac{x + 2}{x + 3} \cdot \frac{1}{2 x + 3} \frac{1}{x^{2} - 9}$

Step 5.2

Multiply $\frac{x + 2}{x + 3}$ by $\frac{1}{2 x + 3}$ .

$\frac{x + 2}{(x + 3) (2 x + 3)} \cdot \frac{1}{x^{2} - 9}$

Step 6

Simplify the denominator.

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Step 6.1

Rewrite $9$ as $3^{2}$ .

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

Step 6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

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

Step 7

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

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Step 7.1

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

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

Step 7.2

Raise $x + 3$ to the power of $1$ .

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

Step 7.3

Raise $x + 3$ to the power of $1$ .

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

Step 7.4

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

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

Step 7.5

Add $1$ and $1$ .

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