Basic Math Examples

\frac{a^{2} + 7}{a^{2} - 25} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{a^{2} - 25} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}$

Step 1

Simplify the denominator.

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Step 1.1

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

\frac{a^{2} + 7}{a^{2} - 5^{2}} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{a^{2} - 5^{2}} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}$

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = a

$a = a$ and

b = 5

$b = 5$ .

\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}$

\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{3} - 3 a^{2} + 7 a - 21}{a^{2} + 2 a - 15}$

Step 2

Simplify the numerator.

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Step 2.1

Factor out the greatest common factor from each group.

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Step 2.1.1

Group the first two terms and the last two terms.

\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{(a^{3} - 3 a^{2}) + 7 a - 21}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{(a^{3} - 3 a^{2}) + 7 a - 21}{a^{2} + 2 a - 15}$

Step 2.1.2

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

\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{2} (a - 3) + 7 (a - 3)}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{2} (a - 3) + 7 (a - 3)}{a^{2} + 2 a - 15}$

\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{2} (a - 3) + 7 (a - 3)}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{2} (a - 3) + 7 (a - 3)}{a^{2} + 2 a - 15}$

Step 2.2

Factor the polynomial by factoring out the greatest common factor,

a - 3

$a - 3$ .

\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{(a - 3) (a^{2} + 7)}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{(a - 3) (a^{2} + 7)}{a^{2} + 2 a - 15}$

\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{(a - 3) (a^{2} + 7)}{a^{2} + 2 a - 15}

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{(a - 3) (a^{2} + 7)}{a^{2} + 2 a - 15}$

Step 3

Factor

$a^{2} + 2 a - 15$ using the AC method.

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Step 3.1

Consider the form

$x^{2} + b x + c$ . Find a pair of integers whose product is

$c$ and whose sum is

$b$ . In this case, whose product is

$- 15$ and whose sum is

$2$ .

$- 3, 5$

Step 3.2

Write the factored form using these integers.

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

Step 4

Cancel the common factor of

$a - 3$ .

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Step 4.1

Cancel the common factor.

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

Step 4.2

Rewrite the expression.

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{2} + 7}{a + 5}$

Step 5

Multiply

$\frac{a^{2} + 7}{(a + 5) (a - 5)} \cdot \frac{a^{2} + 7}{a + 5}$ .

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Step 5.1

Multiply

$\frac{a^{2} + 7}{(a + 5) (a - 5)}$ by

$\frac{a^{2} + 7}{a + 5}$ .

$\frac{(a^{2} + 7) (a^{2} + 7)}{(a + 5) (a - 5) (a + 5)}$

Step 5.2

Raise

$a^{2} + 7$ to the power of

$1$ .

$\frac{{(a^{2} + 7)}^{1} (a^{2} + 7)}{(a + 5) (a - 5) (a + 5)}$

Step 5.3

Raise

$a^{2} + 7$ to the power of

$1$ .

$\frac{{(a^{2} + 7)}^{1} {(a^{2} + 7)}^{1}}{(a + 5) (a - 5) (a + 5)}$

Step 5.4

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{{(a^{2} + 7)}^{1 + 1}}{(a + 5) (a - 5) (a + 5)}$

Step 5.5

Add

$1$ and

$1$ .

$\frac{{(a^{2} + 7)}^{2}}{(a + 5) (a - 5) (a + 5)}$

Step 5.6

Raise

$a + 5$ to the power of

$1$ .

$\frac{{(a^{2} + 7)}^{2}}{{(a + 5)}^{1} (a + 5) (a - 5)}$

Step 5.7

Raise

$a + 5$ to the power of

$1$ .

$\frac{{(a^{2} + 7)}^{2}}{{(a + 5)}^{1} {(a + 5)}^{1} (a - 5)}$

Step 5.8

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{{(a^{2} + 7)}^{2}}{{(a + 5)}^{1 + 1} (a - 5)}$

Step 5.9

Add

$1$ and

$1$ .

$\frac{{(a^{2} + 7)}^{2}}{{(a + 5)}^{2} (a - 5)}$