Basic Math Examples

$\frac{10^{2} + 42 b + 36}{6 b^{2} - 2 b - 60} \div \frac{40 b + 48}{3 b^{2} - 13 b + 10}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{10^{2} + 42 b + 36}{6 b^{2} - 2 b - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 2

Simplify the numerator.

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Step 2.1

Let $u = b$ . Substitute $u$ for all occurrences of $b$ .

$\frac{42 u + 136}{6 b^{2} - 2 b - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 2.2

Factor $2$ out of $42 u + 136$ .

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Step 2.2.1

Factor $2$ out of $42 u$ .

$\frac{2 (21 u) + 136}{6 b^{2} - 2 b - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 2.2.2

Factor $2$ out of $136$ .

$\frac{2 (21 u) + 2 (68)}{6 b^{2} - 2 b - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 2.2.3

Factor $2$ out of $2 (21 u) + 2 (68)$ .

$\frac{2 (21 u + 68)}{6 b^{2} - 2 b - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 2.3

Replace all occurrences of $u$ with $b$ .

$\frac{2 (21 b + 68)}{6 b^{2} - 2 b - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3

Simplify the denominator.

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Step 3.1

Factor $2$ out of $6 b^{2} - 2 b - 60$ .

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Step 3.1.1

Factor $2$ out of $6 b^{2}$ .

$\frac{2 (21 b + 68)}{2 (3 b^{2}) - 2 b - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.1.2

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

$\frac{2 (21 b + 68)}{2 (3 b^{2}) + 2 (- b) - 60} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.1.3

Factor $2$ out of $- 60$ .

$\frac{2 (21 b + 68)}{2 (3 b^{2}) + 2 (- b) + 2 (- 30)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.1.4

Factor $2$ out of $2 (3 b^{2}) + 2 (- b)$ .

$\frac{2 (21 b + 68)}{2 (3 b^{2} - b) + 2 (- 30)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.1.5

Factor $2$ out of $2 (3 b^{2} - b) + 2 (- 30)$ .

$\frac{2 (21 b + 68)}{2 (3 b^{2} - b - 30)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.2

Factor by grouping.

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Step 3.2.1

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 $a \cdot c = 3 \cdot - 30 = - 90$ and whose sum is $b = - 1$ .

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Step 3.2.1.1

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

$\frac{2 (21 b + 68)}{2 (3 b^{2} - (b) - 30)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.2.1.2

Rewrite $- 1$ as $9$ plus $- 10$

$\frac{2 (21 b + 68)}{2 (3 b^{2} + (9 - 10) b - 30)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.2.1.3

Apply the distributive property.

$\frac{2 (21 b + 68)}{2 (3 b^{2} + 9 b - 10 b - 30)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.2.2

Factor out the greatest common factor from each group.

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Step 3.2.2.1

Group the first two terms and the last two terms.

$\frac{2 (21 b + 68)}{2 ((3 b^{2} + 9 b) - 10 b - 30)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.2.2.2

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

$\frac{2 (21 b + 68)}{2 (3 b (b + 3) - 10 (b + 3))} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 3.2.3

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

$\frac{2 (21 b + 68)}{2 ((b + 3) (3 b - 10))} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

$\frac{2 (21 b + 68)}{2 (b + 3) (3 b - 10)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 4

Cancel the common factor of $2$ .

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Step 4.1

Cancel the common factor.

$\frac{2 (21 b + 68)}{2 (b + 3) (3 b - 10)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

Step 4.2

Rewrite the expression.

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{3 b^{2} - 13 b + 10}{40 b + 48}$

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 $a \cdot c = 3 \cdot 10 = 30$ and whose sum is $b = - 13$ .

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Step 5.1.1

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

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{3 b^{2} - 13 (b) + 10}{40 b + 48}$

Step 5.1.2

Rewrite $- 13$ as $- 3$ plus $- 10$

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{3 b^{2} + (- 3 - 10) b + 10}{40 b + 48}$

Step 5.1.3

Apply the distributive property.

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{3 b^{2} - 3 b - 10 b + 10}{40 b + 48}$

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.

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{(3 b^{2} - 3 b) - 10 b + 10}{40 b + 48}$

Step 5.2.2

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

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{3 b (b - 1) - 10 (b - 1)}{40 b + 48}$

Step 5.3

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

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{(b - 1) (3 b - 10)}{40 b + 48}$

Step 6

Factor $8$ out of $40 b + 48$ .

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Step 6.1

Factor $8$ out of $40 b$ .

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{(b - 1) (3 b - 10)}{8 (5 b) + 48}$

Step 6.2

Factor $8$ out of $48$ .

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{(b - 1) (3 b - 10)}{8 (5 b) + 8 (6)}$

Step 6.3

Factor $8$ out of $8 (5 b) + 8 (6)$ .

$\frac{21 b + 68}{(b + 3) (3 b - 10)} \cdot \frac{(b - 1) (3 b - 10)}{8 (5 b + 6)}$

Step 7

Cancel the common factor of $3 b - 10$ .

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Step 7.1

Factor $3 b - 10$ out of $(b + 3) (3 b - 10)$ .

$\frac{21 b + 68}{(3 b - 10) (b + 3)} \cdot \frac{(b - 1) (3 b - 10)}{8 (5 b + 6)}$

Step 7.2

Factor $3 b - 10$ out of $(b - 1) (3 b - 10)$ .

$\frac{21 b + 68}{(3 b - 10) (b + 3)} \cdot \frac{(3 b - 10) (b - 1)}{8 (5 b + 6)}$

Step 7.3

Cancel the common factor.

$\frac{21 b + 68}{(3 b - 10) (b + 3)} \cdot \frac{(3 b - 10) (b - 1)}{8 (5 b + 6)}$

Step 7.4

Rewrite the expression.

$\frac{21 b + 68}{b + 3} \cdot \frac{b - 1}{8 (5 b + 6)}$

Step 8

Multiply $\frac{21 b + 68}{b + 3}$ by $\frac{b - 1}{8 (5 b + 6)}$ .

$\frac{(21 b + 68) (b - 1)}{(b + 3) (8 (5 b + 6))}$

Step 9

Move $8$ to the left of $b + 3$ .

$\frac{(21 b + 68) (b - 1)}{8 (b + 3) (5 b + 6)}$