Precalculus Examples

$\frac{x^{2} - 2 x - 35}{2 x^{3} - 3 x^{2}} \cdot \frac{4 x^{3} - 9 x}{7 x - 49}$

Step 1

Factor $x^{2} - 2 x - 35$ using the AC method.

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Step 1.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 1.2

Write the factored form using these integers.

$\frac{(x - 7) (x + 5)}{2 x^{3} - 3 x^{2}} \cdot \frac{4 x^{3} - 9 x}{7 x - 49}$

Step 2

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

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Step 2.1

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

$\frac{(x - 7) (x + 5)}{x^{2} (2 x) - 3 x^{2}} \cdot \frac{4 x^{3} - 9 x}{7 x - 49}$

Step 2.2

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

$\frac{(x - 7) (x + 5)}{x^{2} (2 x) + x^{2} \cdot - 3} \cdot \frac{4 x^{3} - 9 x}{7 x - 49}$

Step 2.3

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

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{4 x^{3} - 9 x}{7 x - 49}$

Step 3

Simplify the numerator.

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Step 3.1

Factor $x$ out of $4 x^{3} - 9 x$ .

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Step 3.1.1

Factor $x$ out of $4 x^{3}$ .

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{x (4 x^{2}) - 9 x}{7 x - 49}$

Step 3.1.2

Factor $x$ out of $- 9 x$ .

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{x (4 x^{2}) + x \cdot - 9}{7 x - 49}$

Step 3.1.3

Factor $x$ out of $x (4 x^{2}) + x \cdot - 9$ .

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{x (4 x^{2} - 9)}{7 x - 49}$

Step 3.2

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

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{x ({(2 x)}^{2} - 9)}{7 x - 49}$

Step 3.3

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

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{x ({(2 x)}^{2} - 3^{2})}{7 x - 49}$

Step 3.4

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

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{x (2 x + 3) (2 x - 3)}{7 x - 49}$

Step 4

Simplify terms.

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Step 4.1

Factor $7$ out of $7 x - 49$ .

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Step 4.1.1

Factor $7$ out of $7 x$ .

$\frac{(x - 7) (x + 5)}{x^{2} (2 x - 3)} \cdot \frac{x (2 x + 3) (2 x - 3)}{7 (x) - 49}$

Step 4.1.2

Factor $7$ out of $- 49$ .

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

Step 4.1.3

Factor $7$ out of $7 x + 7 \cdot - 7$ .

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

Step 4.2

Combine.

$\frac{(x - 7) (x + 5) (x (2 x + 3) (2 x - 3))}{x^{2} (2 x - 3) (7 (x - 7))}$

Step 4.3

Cancel the common factor of $x - 7$ .

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Step 4.3.1

Cancel the common factor.

$\frac{(x - 7) (x + 5) (x (2 x + 3) (2 x - 3))}{x^{2} (2 x - 3) (7 (x - 7))}$

Step 4.3.2

Rewrite the expression.

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

Step 4.4

Cancel the common factor of $x$ and $x^{2}$ .

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Step 4.4.1

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

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

Step 4.4.2

Cancel the common factors.

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Step 4.4.2.1

Factor $x$ out of $x^{2} (2 x - 3) \cdot (7)$ .

$\frac{x ((x + 5) ((2 x + 3) (2 x - 3)))}{x ((x (2 x - 3)) \cdot 7)}$

Step 4.4.2.2

Cancel the common factor.

$\frac{x ((x + 5) ((2 x + 3) (2 x - 3)))}{x ((x (2 x - 3)) \cdot 7)}$

Step 4.4.2.3

Rewrite the expression.

$\frac{(x + 5) ((2 x + 3) (2 x - 3))}{(x (2 x - 3)) \cdot 7}$

Step 4.5

Cancel the common factor of $2 x - 3$ .

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Step 4.5.1

Cancel the common factor.

$\frac{(x + 5) ((2 x + 3) (2 x - 3))}{x (2 x - 3) \cdot 7}$

Step 4.5.2

Rewrite the expression.

$\frac{(x + 5) (2 x + 3)}{(x) \cdot 7}$

Step 4.6

Move $7$ to the left of $x$ .

$\frac{(x + 5) (2 x + 3)}{7 x}$