Pre-Algebra Examples

$\frac{5 x^{4}}{7} - \frac{36 x^{2}}{35}$

Step 1

Factor $\frac{1}{35}$ out of each term.

$\frac{1}{35} (25 x^{4} - (36 x^{2}))$

Step 2

Remove parentheses.

$\frac{1}{35} \cdot (25 x^{4} - 1 \cdot 36 x^{2})$

Step 3

Factor $x^{2}$ out of $25 x^{4} - 1 \cdot 36 x^{2}$ .

Tap for more steps...

Step 3.1

Factor $x^{2}$ out of $25 x^{4}$ .

$\frac{1}{35} \cdot (x^{2} (25 x^{2}) - 1 \cdot 36 x^{2})$

Step 3.2

Factor $x^{2}$ out of $- 1 \cdot 36 x^{2}$ .

$\frac{1}{35} \cdot (x^{2} (25 x^{2}) + x^{2} (- 1 \cdot 36))$

Step 3.3

Factor $x^{2}$ out of $x^{2} (25 x^{2}) + x^{2} (- 1 \cdot 36)$ .

$\frac{1}{35} \cdot (x^{2} (25 x^{2} - 1 \cdot 36))$

Step 4

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

$\frac{1}{35} \cdot (x^{2} ({(5 x)}^{2} - 1 \cdot 36))$

Step 5

Rewrite $36$ as $6^{2}$ .

$\frac{1}{35} \cdot (x^{2} ({(5 x)}^{2} - 6^{2}))$

Step 6

Factor.

Tap for more steps...

Step 6.1

Factor.

Tap for more steps...

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 $a = 5 x$ and $b = 6$ .

$\frac{1}{35} \cdot (x^{2} ((5 x + 6) (5 x - 6)))$

Step 6.1.2

Remove unnecessary parentheses.

$\frac{1}{35} \cdot (x^{2} (5 x + 6) (5 x - 6))$

Step 6.2

Remove unnecessary parentheses.

$\frac{1}{35} \cdot x^{2} (5 x + 6) (5 x - 6)$

Step 7

Combine $\frac{1}{35}$ and $x^{2}$ .

$\frac{x^{2}}{35} (5 x + 6) (5 x - 6)$