Pre-Algebra Examples

$\frac{1}{x + 3} + \frac{3 x^{2}}{x^{2} - 9}$

Step 1

Simplify the denominator.

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Step 1.1

Rewrite $9$ as $3^{2}$ .

$\frac{1}{x + 3} + \frac{3 x^{2}}{x^{2} - 3^{2}}$

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

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

Step 2

To write $\frac{1}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

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

Step 3

Multiply $\frac{1}{x + 3}$ by $\frac{x - 3}{x - 3}$ .

$\frac{x - 3}{(x + 3) (x - 3)} + \frac{3 x^{2}}{(x + 3) (x - 3)}$

Step 4

Combine the numerators over the common denominator.

$\frac{x - 3 + 3 x^{2}}{(x + 3) (x - 3)}$

Step 5

Reorder terms.

$\frac{3 x^{2} + x - 3}{(x + 3) (x - 3)}$