Algebra Examples

$8 b^{2} m^{2} + 24 b^{2} m n + 18 b^{2} n^{2}$

Step 1

Factor $2 b^{2}$ out of $8 b^{2} m^{2} + 24 b^{2} m n + 18 b^{2} n^{2}$ .

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Step 1.1

Factor $2 b^{2}$ out of $8 b^{2} m^{2}$ .

$2 b^{2} (4 m^{2}) + 24 b^{2} m n + 18 b^{2} n^{2}$

Step 1.2

Factor $2 b^{2}$ out of $24 b^{2} m n$ .

$2 b^{2} (4 m^{2}) + 2 b^{2} (12 m n) + 18 b^{2} n^{2}$

Step 1.3

Factor $2 b^{2}$ out of $18 b^{2} n^{2}$ .

$2 b^{2} (4 m^{2}) + 2 b^{2} (12 m n) + 2 b^{2} (9 n^{2})$

Step 1.4

Factor $2 b^{2}$ out of $2 b^{2} (4 m^{2}) + 2 b^{2} (12 m n)$ .

$2 b^{2} (4 m^{2} + 12 m n) + 2 b^{2} (9 n^{2})$

Step 1.5

Factor $2 b^{2}$ out of $2 b^{2} (4 m^{2} + 12 m n) + 2 b^{2} (9 n^{2})$ .

$2 b^{2} (4 m^{2} + 12 m n + 9 n^{2})$

Step 2

Factor using the perfect square rule.

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Step 2.1

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

$2 b^{2} ({(2 m)}^{2} + 12 m n + 9 n^{2})$

Step 2.2

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

$2 b^{2} ({(2 m)}^{2} + 12 m n + {(3 n)}^{2})$

Step 2.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 m n = 2 \cdot (2 m) \cdot (3 n)$

Step 2.4

Rewrite the polynomial.

$2 b^{2} ({(2 m)}^{2} + 2 \cdot (2 m) \cdot (3 n) + {(3 n)}^{2})$

Step 2.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 m$ and $b = 3 n$ .

$2 b^{2} {(2 m + 3 n)}^{2}$