Basic Math Examples

$a^{2} - 4 b^{2} + 12 b - 9$

Step 1

Regroup terms.

$a^{2} - 4 b^{2} + 12 b - 9$

Step 2

Factor $- 1$ out of $- 4 b^{2} + 12 b - 9$ .

Tap for more steps...

Step 2.1

Factor $- 1$ out of $- 4 b^{2}$ .

$a^{2} - (4 b^{2}) + 12 b - 9$

Step 2.2

Factor $- 1$ out of $12 b$ .

$a^{2} - (4 b^{2}) - (- 12 b) - 9$

Step 2.3

Rewrite $- 9$ as $- 1 (9)$ .

$a^{2} - (4 b^{2}) - (- 12 b) - 1 (9)$

Step 2.4

Factor $- 1$ out of $- (4 b^{2}) - (- 12 b)$ .

$a^{2} - (4 b^{2} - 12 b) - 1 (9)$

Step 2.5

Factor $- 1$ out of $- (4 b^{2} - 12 b) - 1 (9)$ .

$a^{2} - (4 b^{2} - 12 b + 9)$

Step 3

Factor using the perfect square rule.

Tap for more steps...

Step 3.1

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

$a^{2} - ({(2 b)}^{2} - 12 b + 9)$

Step 3.2

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

$a^{2} - ({(2 b)}^{2} - 12 b + 3^{2})$

Step 3.3

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

$12 b = 2 \cdot (2 b) \cdot 3$

Step 3.4

Rewrite the polynomial.

$a^{2} - ({(2 b)}^{2} - 2 \cdot (2 b) \cdot 3 + 3^{2})$

Step 3.5

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

$a^{2} - {(2 b - 3)}^{2}$

Step 4

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

$(a + 2 b - 3) (a - (2 b - 3))$

Step 5

Simplify.

Tap for more steps...

Step 5.1

Apply the distributive property.

$(a + 2 b - 3) (a - (2 b) - - 3)$

Step 5.2

Multiply $2$ by $- 1$ .

$(a + 2 b - 3) (a - 2 b - - 3)$

Step 5.3

Multiply $- 1$ by $- 3$ .

$(a + 2 b - 3) (a - 2 b + 3)$