Basic Math Examples

$\frac{15 - 22 b - 5}{25 b^{2} + 20 b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 1

Simplify the numerator.

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Step 1.1

Subtract $5$ from $15$ .

$\frac{- 22 b + 10}{25 b^{2} + 20 b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 1.2

Factor $2$ out of $- 22 b + 10$ .

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Step 1.2.1

Factor $2$ out of $- 22 b$ .

$\frac{2 (- 11 b) + 10}{25 b^{2} + 20 b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 1.2.2

Factor $2$ out of $10$ .

$\frac{2 (- 11 b) + 2 (5)}{25 b^{2} + 20 b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 1.2.3

Factor $2$ out of $2 (- 11 b) + 2 (5)$ .

$\frac{2 (- 11 b + 5)}{25 b^{2} + 20 b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

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 $a \cdot c = 25 \cdot 3 = 75$ and whose sum is $b = 20$ .

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Step 2.1.1

Factor $20$ out of $20 b$ .

$\frac{2 (- 11 b + 5)}{25 b^{2} + 20 (b) + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 2.1.2

Rewrite $20$ as $5$ plus $15$

$\frac{2 (- 11 b + 5)}{25 b^{2} + (5 + 15) b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 2.1.3

Apply the distributive property.

$\frac{2 (- 11 b + 5)}{25 b^{2} + 5 b + 15 b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

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.

$\frac{2 (- 11 b + 5)}{(25 b^{2} + 5 b) + 15 b + 3} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 2.2.2

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

$\frac{2 (- 11 b + 5)}{5 b (5 b + 1) + 3 (5 b + 1)} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $5 b + 1$ .

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{20 b^{2} - 13 b - 15}{5 + 17 b - 12 b^{2}}$

Step 3

Factor by grouping.

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Step 3.1

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 $a \cdot c = 20 \cdot - 15 = - 300$ and whose sum is $b = - 13$ .

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Step 3.1.1

Factor $- 13$ out of $- 13 b$ .

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{20 b^{2} - 13 (b) - 15}{5 + 17 b - 12 b^{2}}$

Step 3.1.2

Rewrite $- 13$ as $12$ plus $- 25$

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{20 b^{2} + (12 - 25) b - 15}{5 + 17 b - 12 b^{2}}$

Step 3.1.3

Apply the distributive property.

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{20 b^{2} + 12 b - 25 b - 15}{5 + 17 b - 12 b^{2}}$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{(20 b^{2} + 12 b) - 25 b - 15}{5 + 17 b - 12 b^{2}}$

Step 3.2.2

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

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{4 b (5 b + 3) - 5 (5 b + 3)}{5 + 17 b - 12 b^{2}}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $5 b + 3$ .

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{(5 b + 3) (4 b - 5)}{5 + 17 b - 12 b^{2}}$

Step 4

Factor by grouping.

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Step 4.1

Reorder terms.

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{(5 b + 3) (4 b - 5)}{- 12 b^{2} + 17 b + 5}$

Step 4.2

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 $a \cdot c = - 12 \cdot 5 = - 60$ and whose sum is $b = 17$ .

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Step 4.2.1

Factor $17$ out of $17 b$ .

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{(5 b + 3) (4 b - 5)}{- 12 b^{2} + 17 (b) + 5}$

Step 4.2.2

Rewrite $17$ as $- 3$ plus $20$

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{(5 b + 3) (4 b - 5)}{- 12 b^{2} + (- 3 + 20) b + 5}$

Step 4.2.3

Apply the distributive property.

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{(5 b + 3) (4 b - 5)}{- 12 b^{2} - 3 b + 20 b + 5}$

Step 4.3

Factor out the greatest common factor from each group.

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Step 4.3.1

Group the first two terms and the last two terms.

$\frac{2 (- 11 b + 5)}{(5 b + 1) (5 b + 3)} \cdot \frac{(5 b + 3) (4 b - 5)}{(- 12 b^{2} - 3 b) + 20 b + 5}$

Step 4.3.2

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

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

Step 4.4

Factor the polynomial by factoring out the greatest common factor, $- 4 b - 1$ .

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

Step 5

Cancel the common factor of $5 b + 3$ .

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Step 5.1

Factor $5 b + 3$ out of $(5 b + 1) (5 b + 3)$ .

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

Step 5.2

Cancel the common factor.

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

Step 5.3

Rewrite the expression.

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

Step 6

Multiply $\frac{2 (- 11 b + 5)}{5 b + 1}$ by $\frac{4 b - 5}{(- 4 b - 1) (3 b - 5)}$ .

$\frac{2 (- 11 b + 5) (4 b - 5)}{(5 b + 1) ((- 4 b - 1) (3 b - 5))}$

Step 7

Factor $- 1$ out of $- 11 b$ .

$\frac{2 (- (11 b) + 5) (4 b - 5)}{(5 b + 1) (- 4 b - 1) (3 b - 5)}$

Step 8

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

$\frac{2 (- (11 b) - 1 (- 5)) (4 b - 5)}{(5 b + 1) (- 4 b - 1) (3 b - 5)}$

Step 9

Factor $- 1$ out of $- (11 b) - 1 (- 5)$ .

$\frac{2 (- (11 b - 5)) (4 b - 5)}{(5 b + 1) (- 4 b - 1) (3 b - 5)}$

Step 10

Rewrite $- (11 b - 5)$ as $- 1 (11 b - 5)$ .

$\frac{2 (- 1 (11 b - 5)) (4 b - 5)}{(5 b + 1) (- 4 b - 1) (3 b - 5)}$

Step 11

Factor $- 1$ out of $- 4 b$ .

$\frac{2 (- 1 (11 b - 5)) (4 b - 5)}{(5 b + 1) (- (4 b) - 1) (3 b - 5)}$

Step 12

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

$\frac{2 (- 1 (11 b - 5)) (4 b - 5)}{(5 b + 1) (- (4 b) - 1 (1)) (3 b - 5)}$

Step 13

Factor $- 1$ out of $- (4 b) - 1 (1)$ .

$\frac{2 (- 1 (11 b - 5)) (4 b - 5)}{(5 b + 1) (- (4 b + 1)) (3 b - 5)}$

Step 14

Rewrite $- (4 b + 1)$ as $- 1 (4 b + 1)$ .

$\frac{2 (- 1 (11 b - 5)) (4 b - 5)}{(5 b + 1) (- 1 (4 b + 1)) (3 b - 5)}$

Step 15

Cancel the common factor.

$\frac{2 (- 1 (11 b - 5)) (4 b - 5)}{(5 b + 1) (- 1 (4 b + 1)) (3 b - 5)}$

Step 16

Rewrite the expression.

$\frac{2 (11 b - 5) (4 b - 5)}{((5 b + 1) (4 b + 1)) (3 b - 5)}$

Step 17

Reorder factors in $\frac{2 (11 b - 5) (4 b - 5)}{((5 b + 1) (4 b + 1)) (3 b - 5)}$ .

$\frac{2 (11 b - 5) (4 b - 5)}{(5 b + 1) (4 b + 1) (3 b - 5)}$