Basic Math Examples

$(\frac{7 a + 14 a^{2}}{9 a^{2} + 6 a + 1}) (\frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1})$

Step 1

Factor $7 a$ out of $7 a + 14 a^{2}$ .

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Step 1.1

Factor $7 a$ out of $7 a$ .

$\frac{7 a (1) + 14 a^{2}}{9 a^{2} + 6 a + 1} \cdot \frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1}$

Step 1.2

Factor $7 a$ out of $14 a^{2}$ .

$\frac{7 a (1) + 7 a (2 a)}{9 a^{2} + 6 a + 1} \cdot \frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1}$

Step 1.3

Factor $7 a$ out of $7 a (1) + 7 a (2 a)$ .

$\frac{7 a (1 + 2 a)}{9 a^{2} + 6 a + 1} \cdot \frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1}$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $9 a^{2}$ as ${(3 a)}^{2}$ .

$\frac{7 a (1 + 2 a)}{{(3 a)}^{2} + 6 a + 1} \cdot \frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1}$

Step 2.2

Rewrite $1$ as $1^{2}$ .

$\frac{7 a (1 + 2 a)}{{(3 a)}^{2} + 6 a + 1^{2}} \cdot \frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1}$

Step 2.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 a = 2 \cdot (3 a) \cdot 1$

Step 2.4

Rewrite the polynomial.

$\frac{7 a (1 + 2 a)}{{(3 a)}^{2} + 2 \cdot (3 a) \cdot 1 + 1^{2}} \cdot \frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1}$

Step 2.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 3 a$ and $b = 1$ .

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{6 a^{2} + 5 a + 1}{4 a^{2} - 1}$

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 $a \cdot c = 6 \cdot 1 = 6$ and whose sum is $b = 5$ .

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Step 3.1.1

Factor $5$ out of $5 a$ .

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{6 a^{2} + 5 (a) + 1}{4 a^{2} - 1}$

Step 3.1.2

Rewrite $5$ as $2$ plus $3$

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{6 a^{2} + (2 + 3) a + 1}{4 a^{2} - 1}$

Step 3.1.3

Apply the distributive property.

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{6 a^{2} + 2 a + 3 a + 1}{4 a^{2} - 1}$

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.

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{(6 a^{2} + 2 a) + 3 a + 1}{4 a^{2} - 1}$

Step 3.2.2

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

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{2 a (3 a + 1) + 1 (3 a + 1)}{4 a^{2} - 1}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $3 a + 1$ .

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{(3 a + 1) (2 a + 1)}{4 a^{2} - 1}$

Step 4

Simplify the denominator.

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Step 4.1

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

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{(3 a + 1) (2 a + 1)}{{(2 a)}^{2} - 1}$

Step 4.2

Rewrite $1$ as $1^{2}$ .

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{(3 a + 1) (2 a + 1)}{{(2 a)}^{2} - 1^{2}}$

Step 4.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 a$ and $b = 1$ .

$\frac{7 a (1 + 2 a)}{{(3 a + 1)}^{2}} \cdot \frac{(3 a + 1) (2 a + 1)}{(2 a + 1) (2 a - 1)}$

Step 5

Simplify terms.

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Step 5.1

Combine.

$\frac{7 a (1 + 2 a) ((3 a + 1) (2 a + 1))}{{(3 a + 1)}^{2} ((2 a + 1) (2 a - 1))}$

Step 5.2

Cancel the common factor of $3 a + 1$ and ${(3 a + 1)}^{2}$ .

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Step 5.2.1

Factor $3 a + 1$ out of $7 a (1 + 2 a) ((3 a + 1) (2 a + 1))$ .

$\frac{(3 a + 1) (7 a (1 + 2 a) (2 a + 1))}{{(3 a + 1)}^{2} ((2 a + 1) (2 a - 1))}$

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Factor $3 a + 1$ out of ${(3 a + 1)}^{2} ((2 a + 1) (2 a - 1))$ .

$\frac{(3 a + 1) (7 a (1 + 2 a) (2 a + 1))}{(3 a + 1) ((3 a + 1) ((2 a + 1) (2 a - 1)))}$

Step 5.2.2.2

Cancel the common factor.

$\frac{(3 a + 1) (7 a (1 + 2 a) (2 a + 1))}{(3 a + 1) ((3 a + 1) ((2 a + 1) (2 a - 1)))}$

Step 5.2.2.3

Rewrite the expression.

$\frac{7 a (1 + 2 a) (2 a + 1)}{(3 a + 1) ((2 a + 1) (2 a - 1))}$

Step 5.3

Cancel the common factor of $2 a + 1$ .

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Step 5.3.1

Cancel the common factor.

$\frac{7 a (1 + 2 a) (2 a + 1)}{(3 a + 1) ((2 a + 1) (2 a - 1))}$

Step 5.3.2

Rewrite the expression.

$\frac{7 a (1 + 2 a)}{(3 a + 1) (2 a - 1)}$