Pre-Algebra Examples

$\frac{2 x^{2} + 3 x + 1}{x^{2} + 2 x - 15} \div \frac{x^{2} + 6 x + 5}{2 x^{2} - 7 x + 3}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{2 x^{2} + 3 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

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

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Step 2.1.1

Factor $3$ out of $3 x$ .

$\frac{2 x^{2} + 3 (x) + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Step 2.1.2

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

$\frac{2 x^{2} + (1 + 2) x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Step 2.1.3

Apply the distributive property.

$\frac{2 x^{2} + 1 x + 2 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Step 2.1.4

Multiply $x$ by $1$ .

$\frac{2 x^{2} + x + 2 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

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.

$\frac{(2 x^{2} + x) + 2 x + 1}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Step 2.2.2

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

$\frac{x (2 x + 1) + 1 (2 x + 1)}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$\frac{(2 x + 1) (x + 1)}{x^{2} + 2 x - 15} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

Step 3

Factor $x^{2} + 2 x - 15$ 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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 3.2

Write the factored form using these integers.

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} + 6 x + 5}$

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 7$ .

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Step 4.1.1

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

$\frac{(2 x + 1) (x + 1)}{(x - 3) (x + 5)} \cdot \frac{2 x^{2} - 7 (x) + 3}{x^{2} + 6 x + 5}$

Step 4.1.2

Rewrite $- 7$ as $- 1$ plus $- 6$

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

Step 4.1.3

Apply the distributive property.

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

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.

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

Step 4.2.2

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

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

Step 4.3

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

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

Step 5

Factor $x^{2} + 6 x + 5$ using the AC method.

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Step 5.1

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 $5$ and whose sum is $6$ .

$1, 5$

Step 5.2

Write the factored form using these integers.

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

Step 6

Cancel the common factor of $x + 1$ .

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Step 6.1

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

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

Step 6.2

Cancel the common factor.

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

Step 6.3

Rewrite the expression.

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

Step 7

Cancel the common factor of $x - 3$ .

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Step 7.1

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

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

Step 7.2

Cancel the common factor.

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

Step 7.3

Rewrite the expression.

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

Step 8

Multiply $\frac{2 x + 1}{x + 5}$ by $\frac{2 x - 1}{x + 5}$ .

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

Step 9

Raise $x + 5$ to the power of $1$ .

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

Step 10

Raise $x + 5$ to the power of $1$ .

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

Step 11

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

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

Step 12

Add $1$ and $1$ .

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

Step 13

Expand $(2 x + 1) (2 x - 1)$ using the FOIL Method.

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Step 13.1

Apply the distributive property.

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

Step 13.2

Apply the distributive property.

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

Step 13.3

Apply the distributive property.

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

Step 14

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

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Step 14.1

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

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

Step 14.2

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

$\frac{2 x (2 x) + 0 + 1 \cdot - 1}{{(x + 5)}^{2}}$

Step 14.3

Add $2 x (2 x)$ and $0$ .

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

Step 15

Simplify each term.

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Step 15.1

Rewrite using the commutative property of multiplication.

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

Step 15.2

Multiply $x$ by $x$ by adding the exponents.

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Step 15.2.1

Move $x$ .

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

Step 15.2.2

Multiply $x$ by $x$ .

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

Step 15.3

Multiply $2$ by $2$ .

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

Step 15.4

Multiply $- 1$ by $1$ .

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

Step 16

Split the fraction $\frac{4 x^{2} - 1}{{(x + 5)}^{2}}$ into two fractions.

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

Step 17

Move the negative in front of the fraction.

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