Basic Math Examples

$\frac{b - 2 a}{b^{2} + 4 b a + 4 a^{2}} \div \frac{b^{2} - 4 a^{2}}{b^{2} - 7 b a + 10 a^{2}}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{b - 2 a}{b^{2} + 4 b a + 4 a^{2}} \cdot \frac{b^{2} - 7 b a + 10 a^{2}}{b^{2} - 4 a^{2}}$

Step 2

Factor using the perfect square rule.

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Step 2.1

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

$\frac{b - 2 a}{b^{2} + 4 b a + {(2 a)}^{2}} \cdot \frac{b^{2} - 7 b a + 10 a^{2}}{b^{2} - 4 a^{2}}$

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.

$4 b a = 2 \cdot b \cdot (2 a)$

Step 2.3

Rewrite the polynomial.

$\frac{b - 2 a}{b^{2} + 2 \cdot b \cdot (2 a) + {(2 a)}^{2}} \cdot \frac{b^{2} - 7 b a + 10 a^{2}}{b^{2} - 4 a^{2}}$

Step 2.4

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

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{b^{2} - 7 b a + 10 a^{2}}{b^{2} - 4 a^{2}}$

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 $a \cdot c = 10 \cdot 1 = 10$ and whose sum is $b = - 7$ .

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Step 3.1.1

Reorder terms.

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{10 a^{2} + b^{2} - 7 a b}{b^{2} - 4 a^{2}}$

Step 3.1.2

Reorder $b^{2}$ and $- 7 a b$ .

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{10 a^{2} - 7 a b + b^{2}}{b^{2} - 4 a^{2}}$

Step 3.1.3

Factor $- 7$ out of $- 7 a b$ .

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{10 a^{2} - 7 (a b) + b^{2}}{b^{2} - 4 a^{2}}$

Step 3.1.4

Rewrite $- 7$ as $- 2$ plus $- 5$

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{10 a^{2} + (- 2 - 5) (a b) + b^{2}}{b^{2} - 4 a^{2}}$

Step 3.1.5

Apply the distributive property.

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{10 a^{2} - 2 (a b) - 5 (a b) + b^{2}}{b^{2} - 4 a^{2}}$

Step 3.1.6

Move parentheses.

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{10 a^{2} - 2 a b - 5 a b + b^{2}}{b^{2} - 4 a^{2}}$

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.

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{(10 a^{2} - 2 a b) - 5 a b + b^{2}}{b^{2} - 4 a^{2}}$

Step 3.2.2

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

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{2 a (5 a - b) - b (5 a - b)}{b^{2} - 4 a^{2}}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $5 a - b$ .

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{(5 a - b) (2 a - b)}{b^{2} - 4 a^{2}}$

Step 4

Simplify the denominator.

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Step 4.1

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

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{(5 a - b) (2 a - b)}{b^{2} - {(2 a)}^{2}}$

Step 4.2

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

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{(5 a - b) (2 a - b)}{(b + 2 a) (b - (2 a))}$

Step 4.3

Multiply $2$ by $- 1$ .

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{(5 a - b) (2 a - b)}{(b + 2 a) (b - 2 a)}$

Step 5

Simplify terms.

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Step 5.1

Cancel the common factor of $b - 2 a$ .

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Step 5.1.1

Factor $b - 2 a$ out of $(b + 2 a) (b - 2 a)$ .

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{(5 a - b) (2 a - b)}{(b - 2 a) (b + 2 a)}$

Step 5.1.2

Cancel the common factor.

$\frac{b - 2 a}{{(b + 2 a)}^{2}} \cdot \frac{(5 a - b) (2 a - b)}{(b - 2 a) (b + 2 a)}$

Step 5.1.3

Rewrite the expression.

$\frac{1}{{(b + 2 a)}^{2}} \cdot \frac{(5 a - b) (2 a - b)}{b + 2 a}$

Step 5.2

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

$\frac{(5 a - b) (2 a - b)}{{(b + 2 a)}^{2} (b + 2 a)}$

Step 6

Multiply ${(b + 2 a)}^{2}$ by $b + 2 a$ by adding the exponents.

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Step 6.1

Multiply ${(b + 2 a)}^{2}$ by $b + 2 a$ .

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Step 6.1.1

Raise $b + 2 a$ to the power of $1$ .

$\frac{(5 a - b) (2 a - b)}{{(b + 2 a)}^{2} {(b + 2 a)}^{1}}$

Step 6.1.2

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

$\frac{(5 a - b) (2 a - b)}{{(b + 2 a)}^{2 + 1}}$

Step 6.2

Add $2$ and $1$ .

$\frac{(5 a - b) (2 a - b)}{{(b + 2 a)}^{3}}$