Algebra Examples

$4 x^{2} - y^{2} + 6 y - 9$

Step 1

Regroup terms.

$4 x^{2} - y^{2} + 6 y - 9$

Step 2

Factor $- 1$ out of $- y^{2} + 6 y - 9$ .

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Step 2.1

Factor $- 1$ out of $- y^{2}$ .

$4 x^{2} - (y^{2}) + 6 y - 9$

Step 2.2

Factor $- 1$ out of $6 y$ .

$4 x^{2} - (y^{2}) - (- 6 y) - 9$

Step 2.3

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

$4 x^{2} - (y^{2}) - (- 6 y) - 1 (9)$

Step 2.4

Factor $- 1$ out of $- (y^{2}) - (- 6 y)$ .

$4 x^{2} - (y^{2} - 6 y) - 1 (9)$

Step 2.5

Factor $- 1$ out of $- (y^{2} - 6 y) - 1 (9)$ .

$4 x^{2} - (y^{2} - 6 y + 9)$

Step 3

Factor using the perfect square rule.

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Step 3.1

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

$4 x^{2} - (y^{2} - 6 y + 3^{2})$

Step 3.2

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

$6 y = 2 \cdot y \cdot 3$

Step 3.3

Rewrite the polynomial.

$4 x^{2} - (y^{2} - 2 \cdot y \cdot 3 + 3^{2})$

Step 3.4

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

$4 x^{2} - {(y - 3)}^{2}$

Step 4

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

${(2 x)}^{2} - {(y - 3)}^{2}$

Step 5

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

$(2 x + y - 3) (2 x - (y - 3))$

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

$(2 x + y - 3) (2 x - y - - 3)$

Step 6.2

Multiply $- 1$ by $- 3$ .

$(2 x + y - 3) (2 x - y + 3)$