Pre-Algebra Examples

$5 p^{2} q^{2} + 20 p q - 60$

Step 1

Factor $5$ out of $5 p^{2} q^{2} + 20 p q - 60$ .

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Step 1.1

Factor $5$ out of $5 p^{2} q^{2}$ .

$5 (p^{2} q^{2}) + 20 p q - 60$

Step 1.2

Factor $5$ out of $20 p q$ .

$5 (p^{2} q^{2}) + 5 (4 p q) - 60$

Step 1.3

Factor $5$ out of $- 60$ .

$5 (p^{2} q^{2}) + 5 (4 p q) + 5 \cdot - 12$

Step 1.4

Factor $5$ out of $5 (p^{2} q^{2}) + 5 (4 p q)$ .

$5 (p^{2} q^{2} + 4 p q) + 5 \cdot - 12$

Step 1.5

Factor $5$ out of $5 (p^{2} q^{2} + 4 p q) + 5 \cdot - 12$ .

$5 (p^{2} q^{2} + 4 p q - 12)$

Step 2

Rewrite $p^{2} q^{2}$ as ${(p q)}^{2}$ .

$5 ({(p q)}^{2} + 4 (p q) - 12)$

Step 3

Let $u = p q$ . Substitute $u$ for all occurrences of $p q$ .

$5 (u^{2} + 4 u - 12)$

Step 4

Factor $u^{2} + 4 u - 12$ using the AC method.

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Step 4.1

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 $- 12$ and whose sum is $4$ .

$- 2, 6$

Step 4.2

Write the factored form using these integers.

$5 ((u - 2) (u + 6))$

Step 5

Factor.

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Step 5.1

Replace all occurrences of $u$ with $p q$ .

$5 ((p q - 2) (p q + 6))$

Step 5.2

Remove unnecessary parentheses.

$5 (p q - 2) (p q + 6)$