Pre-Algebra Examples

$(25 n^{2} + 0 x^{2} + 0 x - 81) \div (5 n + 9)$

Step 1

Rewrite the division as a fraction.

$\frac{25 n^{2} + 0 x^{2} + 0 x - 81}{5 n + 9}$

Step 2

Simplify the numerator.

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Step 2.1

Factor using the perfect square rule.

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Step 2.1.1

Rearrange terms.

$\frac{25 n^{2} + 0 x + 0 x^{2} - 81}{5 n + 9}$

Step 2.1.2

Rewrite $25 n^{2}$ as ${(5 n)}^{2}$ .

$\frac{{(5 n)}^{2} + 0 x + 0 x^{2} - 81}{5 n + 9}$

Step 2.1.3

Rewrite $0 x^{2}$ as ${(0 x)}^{2}$ .

$\frac{{(5 n)}^{2} + 0 x + {(0 x)}^{2} - 81}{5 n + 9}$

Step 2.1.4

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

$0 x = 2 \cdot (5 n) \cdot (0 x)$

Step 2.1.5

Rewrite the polynomial.

$\frac{{(5 n)}^{2} + 2 \cdot (5 n) \cdot (0 x) + {(0 x)}^{2} - 81}{5 n + 9}$

Step 2.1.6

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 5 n$ and $b = 0 x$ .

$\frac{{(5 n + 0 x)}^{2} - 81}{5 n + 9}$

Step 2.2

Rewrite $81$ as $9^{2}$ .

$\frac{{(5 n + 0 x)}^{2} - 9^{2}}{5 n + 9}$

Step 2.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5 n + 0 x$ and $b = 9$ .

$\frac{(5 n + 0 x + 9) (5 n + 0 x - 9)}{5 n + 9}$

Step 2.4

Simplify.

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Step 2.4.1

Multiply $0$ by $x$ .

$\frac{(5 n + 0 + 9) (5 n + 0 x - 9)}{5 n + 9}$

Step 2.4.2

Add $5 n$ and $0$ .

$\frac{(5 n + 9) (5 n + 0 x - 9)}{5 n + 9}$

Step 2.4.3

Multiply $0$ by $x$ .

$\frac{(5 n + 9) (5 n + 0 - 9)}{5 n + 9}$

Step 2.4.4

Add $5 n$ and $0$ .

$\frac{(5 n + 9) (5 n - 9)}{5 n + 9}$

Step 3

Cancel the common factor of $5 n + 9$ .

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Step 3.1

Cancel the common factor.

$\frac{(5 n + 9) (5 n - 9)}{5 n + 9}$

Step 3.2

Divide $5 n - 9$ by $1$ .

$5 n - 9$