Pre-Algebra Examples

$\frac{3 x^{4}}{5} - \frac{16 x^{2}}{15}$

Step 1

Factor $\frac{1}{15}$ out of each term.

$\frac{1}{15} (9 x^{4} - (16 x^{2}))$

Step 2

Remove parentheses.

$\frac{1}{15} \cdot (9 x^{4} - 1 \cdot 16 x^{2})$

Step 3

Factor $x^{2}$ out of $9 x^{4} - 1 \cdot 16 x^{2}$ .

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Step 3.1

Factor $x^{2}$ out of $9 x^{4}$ .

$\frac{1}{15} \cdot (x^{2} (9 x^{2}) - 1 \cdot 16 x^{2})$

Step 3.2

Factor $x^{2}$ out of $- 1 \cdot 16 x^{2}$ .

$\frac{1}{15} \cdot (x^{2} (9 x^{2}) + x^{2} (- 1 \cdot 16))$

Step 3.3

Factor $x^{2}$ out of $x^{2} (9 x^{2}) + x^{2} (- 1 \cdot 16)$ .

$\frac{1}{15} \cdot (x^{2} (9 x^{2} - 1 \cdot 16))$

Step 4

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

$\frac{1}{15} \cdot (x^{2} ({(3 x)}^{2} - 1 \cdot 16))$

Step 5

Rewrite $16$ as $4^{2}$ .

$\frac{1}{15} \cdot (x^{2} ({(3 x)}^{2} - 4^{2}))$

Step 6

Factor.

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Step 6.1

Factor.

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Step 6.1.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 4$ .

$\frac{1}{15} \cdot (x^{2} ((3 x + 4) (3 x - 4)))$

Step 6.1.2

Remove unnecessary parentheses.

$\frac{1}{15} \cdot (x^{2} (3 x + 4) (3 x - 4))$

Step 6.2

Remove unnecessary parentheses.

$\frac{1}{15} \cdot x^{2} (3 x + 4) (3 x - 4)$

Step 7

Combine $\frac{1}{15}$ and $x^{2}$ .

$\frac{x^{2}}{15} (3 x + 4) (3 x - 4)$