Basic Math Examples

$\frac{2 a b + 6 c b + a q + 3 c q}{a^{2} + 6 a c + 9 c^{2}}$

Step 1

Simplify the numerator.

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Step 1.1

Factor out the greatest common factor from each group.

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Step 1.1.1

Group the first two terms and the last two terms.

$\frac{(2 a b + 6 c b) + a q + 3 c q}{a^{2} + 6 a c + 9 c^{2}}$

Step 1.1.2

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

$\frac{2 b (a + 3 c) + q (a + 3 c)}{a^{2} + 6 a c + 9 c^{2}}$

Step 1.2

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

$\frac{(a + 3 c) (2 b + q)}{a^{2} + 6 a c + 9 c^{2}}$

Step 2

Factor using the perfect square rule.

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Step 2.1

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

$\frac{(a + 3 c) (2 b + q)}{a^{2} + 6 a c + {(3 c)}^{2}}$

Step 2.2

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

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

Step 2.3

Rewrite the polynomial.

$\frac{(a + 3 c) (2 b + q)}{a^{2} + 2 \cdot a \cdot (3 c) + {(3 c)}^{2}}$

Step 2.4

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

$\frac{(a + 3 c) (2 b + q)}{{(a + 3 c)}^{2}}$

Step 3

Cancel the common factors.

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Step 3.1

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

$\frac{(a + 3 c) (2 b + q)}{(a + 3 c) (a + 3 c)}$

Step 3.2

Cancel the common factor.

$\frac{(a + 3 c) (2 b + q)}{(a + 3 c) (a + 3 c)}$

Step 3.3

Rewrite the expression.

$\frac{2 b + q}{a + 3 c}$