Algebra Examples

$\frac{- 2 d^{2} + d + 15}{9 - d^{2}} \div \frac{4 d + 1}{2 d^{2} + 11 d + 15}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{- 2 d^{2} + d + 15}{9 - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

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 $a \cdot c = - 2 \cdot 15 = - 30$ and whose sum is $b = 1$ .

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Step 2.1.1

Multiply by $1$ .

$\frac{- 2 d^{2} + 1 d + 15}{9 - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 2.1.2

Rewrite $1$ as $- 5$ plus $6$

$\frac{- 2 d^{2} + (- 5 + 6) d + 15}{9 - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 2.1.3

Apply the distributive property.

$\frac{- 2 d^{2} - 5 d + 6 d + 15}{9 - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

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.

$\frac{(- 2 d^{2} - 5 d) + 6 d + 15}{9 - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 2.2.2

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

$\frac{d (- 2 d - 5) - 3 (- 2 d - 5)}{9 - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $- 2 d - 5$ .

$\frac{(- 2 d - 5) (d - 3)}{9 - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 3

Simplify the denominator.

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Step 3.1

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

$\frac{(- 2 d - 5) (d - 3)}{3^{2} - d^{2}} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

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 $a = 3$ and $b = d$ .

$\frac{(- 2 d - 5) (d - 3)}{(3 + d) (3 - d)} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4

Reduce the expression by cancelling the common factors.

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Step 4.1

Cancel the common factor of $d - 3$ and $3 - d$ .

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Step 4.1.1

Factor $- 1$ out of $d$ .

$\frac{(- 2 d - 5) (- 1 (- d) - 3)}{(3 + d) (3 - d)} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4.1.2

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{(- 2 d - 5) (- 1 (- d) - 1 (3))}{(3 + d) (3 - d)} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4.1.3

Factor $- 1$ out of $- 1 (- d) - 1 (3)$ .

$\frac{(- 2 d - 5) (- 1 (- d + 3))}{(3 + d) (3 - d)} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4.1.4

Reorder terms.

$\frac{(- 2 d - 5) (- 1 (- d + 3))}{(3 + d) (- d + 3)} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4.1.5

Cancel the common factor.

$\frac{(- 2 d - 5) (- 1 (- d + 3))}{(3 + d) (- d + 3)} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4.1.6

Rewrite the expression.

$\frac{(- 2 d - 5) \cdot (- 1)}{3 + d} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4.2

Simplify the expression.

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Step 4.2.1

Move $- 1$ to the left of $- 2 d - 5$ .

$\frac{- 1 \cdot (- 2 d - 5)}{3 + d} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 4.2.2

Move the negative in front of the fraction.

$- \frac{- 2 d - 5}{3 + d} \cdot \frac{2 d^{2} + 11 d + 15}{4 d + 1}$

Step 5

Factor by grouping.

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Step 5.1

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 $b = 11$ .

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Step 5.1.1

Factor $11$ out of $11 d$ .

$- \frac{- 2 d - 5}{3 + d} \cdot \frac{2 d^{2} + 11 (d) + 15}{4 d + 1}$

Step 5.1.2

Rewrite $11$ as $5$ plus $6$

$- \frac{- 2 d - 5}{3 + d} \cdot \frac{2 d^{2} + (5 + 6) d + 15}{4 d + 1}$

Step 5.1.3

Apply the distributive property.

$- \frac{- 2 d - 5}{3 + d} \cdot \frac{2 d^{2} + 5 d + 6 d + 15}{4 d + 1}$

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

$- \frac{- 2 d - 5}{3 + d} \cdot \frac{(2 d^{2} + 5 d) + 6 d + 15}{4 d + 1}$

Step 5.2.2

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

$- \frac{- 2 d - 5}{3 + d} \cdot \frac{d (2 d + 5) + 3 (2 d + 5)}{4 d + 1}$

Step 5.3

Factor the polynomial by factoring out the greatest common factor, $2 d + 5$ .

$- \frac{- 2 d - 5}{3 + d} \cdot \frac{(2 d + 5) (d + 3)}{4 d + 1}$

Step 6

Multiply $- \frac{- 2 d - 5}{3 + d} \cdot \frac{(2 d + 5) (d + 3)}{4 d + 1}$ .

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Step 6.1

Multiply $\frac{(2 d + 5) (d + 3)}{4 d + 1}$ by $\frac{- 2 d - 5}{3 + d}$ .

$- \frac{(2 d + 5) (d + 3) (- 2 d - 5)}{(4 d + 1) (3 + d)}$

Step 6.2

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

$- \frac{(- (- 2 d) + 5) (d + 3) (- 2 d - 5)}{(4 d + 1) (3 + d)}$

Step 6.3

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

$- \frac{(- (- 2 d) - 1 (- 5)) (d + 3) (- 2 d - 5)}{(4 d + 1) (3 + d)}$

Step 6.4

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

$- \frac{- (- 2 d - 5) (d + 3) (- 2 d - 5)}{(4 d + 1) (3 + d)}$

Step 6.5

Rewrite $- (- 2 d - 5)$ as $- 1 (- 2 d - 5)$ .

$- \frac{- 1 (- 2 d - 5) (d + 3) (- 2 d - 5)}{(4 d + 1) (3 + d)}$

Step 6.6

Raise $- 2 d - 5$ to the power of $1$ .

$- \frac{- 1 ({(- 2 d - 5)}^{1} (- 2 d - 5)) (d + 3)}{(4 d + 1) (3 + d)}$

Step 6.7

Raise $- 2 d - 5$ to the power of $1$ .

$- \frac{- 1 ({(- 2 d - 5)}^{1} {(- 2 d - 5)}^{1}) (d + 3)}{(4 d + 1) (3 + d)}$

Step 6.8

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

$- \frac{- 1 {(- 2 d - 5)}^{1 + 1} (d + 3)}{(4 d + 1) (3 + d)}$

Step 6.9

Add $1$ and $1$ .

$- \frac{- 1 {(- 2 d - 5)}^{2} (d + 3)}{(4 d + 1) (3 + d)}$

Step 7

Cancel the common factor of $d + 3$ and $3 + d$ .

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Step 7.1

Reorder terms.

$- \frac{- 1 {(- 2 d - 5)}^{2} (d + 3)}{(4 d + 1) (d + 3)}$

Step 7.2

Cancel the common factor.

$- \frac{- 1 {(- 2 d - 5)}^{2} (d + 3)}{(4 d + 1) (d + 3)}$

Step 7.3

Rewrite the expression.

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

Step 8

Move the negative in front of the fraction.

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

Step 9

Multiply $- - \frac{{(- 2 d - 5)}^{2}}{4 d + 1}$ .

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Step 9.1

Multiply $- 1$ by $- 1$ .

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

Step 9.2

Multiply $\frac{{(- 2 d - 5)}^{2}}{4 d + 1}$ by $1$ .

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