Factor m^6-7m^4-18m^2

$m^{6} - 7 m^{4} - 18 m^{2}$

Step 1

Factor $m^{2}$ out of $m^{6} - 7 m^{4} - 18 m^{2}$ .

Factor $m^{2}$ out of $m^{6}$ .

$m^{2} m^{4} - 7 m^{4} - 18 m^{2}$

Factor $m^{2}$ out of $- 7 m^{4}$ .

$m^{2} m^{4} + m^{2} (- 7 m^{2}) - 18 m^{2}$

Factor $m^{2}$ out of $- 18 m^{2}$ .

$m^{2} m^{4} + m^{2} (- 7 m^{2}) + m^{2} \cdot - 18$

Factor $m^{2}$ out of $m^{2} m^{4} + m^{2} (- 7 m^{2})$ .

$m^{2} (m^{4} - 7 m^{2}) + m^{2} \cdot - 18$

Factor $m^{2}$ out of $m^{2} (m^{4} - 7 m^{2}) + m^{2} \cdot - 18$ .

$m^{2} (m^{4} - 7 m^{2} - 18)$

Step 2

Rewrite $m^{4}$ as ${(m^{2})}^{2}$ .

$m^{2} ({(m^{2})}^{2} - 7 m^{2} - 18)$

Step 3

Let $u = m^{2}$ . Substitute $u$ for all occurrences of $m^{2}$ .

$m^{2} (u^{2} - 7 u - 18)$

Step 4

Factor $u^{2} - 7 u - 18$ using the AC method.

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 $- 18$ and whose sum is $- 7$ .

$- 9, 2$

Write the factored form using these integers.

$m^{2} ((u - 9) (u + 2))$

Step 5

Replace all occurrences of $u$ with $m^{2}$ .

$m^{2} ((m^{2} - 9) (m^{2} + 2))$

Step 6

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

$m^{2} ((m^{2} - 3^{2}) (m^{2} + 2))$

Step 7

Factor.

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

$m^{2} ((m + 3) (m - 3) (m^{2} + 2))$

Remove unnecessary parentheses.

$m^{2} (m + 3) (m - 3) (m^{2} + 2)$

Factor m^6-7m^4-18m^2

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