Factor 49y^2-x^2+6x-9

$49 y^{2} - x^{2} + 6 x - 9$

Step 1

Regroup terms.

$49 y^{2} - x^{2} + 6 x - 9$

Step 2

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

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

$49 y^{2} - (x^{2}) + 6 x - 9$

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

$49 y^{2} - (x^{2}) - (- 6 x) - 9$

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

$49 y^{2} - (x^{2}) - (- 6 x) - 1 (9)$

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

$49 y^{2} - (x^{2} - 6 x) - 1 (9)$

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

$49 y^{2} - (x^{2} - 6 x + 9)$

Step 3

Factor using the perfect square rule.

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

$49 y^{2} - (x^{2} - 6 x + 3^{2})$

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

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$49 y^{2} - (x^{2} - 2 \cdot x \cdot 3 + 3^{2})$

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

$49 y^{2} - {(x - 3)}^{2}$

Step 4

Rewrite $49 y^{2}$ as ${(7 y)}^{2}$ .

${(7 y)}^{2} - {(x - 3)}^{2}$

Step 5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7 y$ and $b = x - 3$ .

$(7 y + x - 3) (7 y - (x - 3))$

Step 6

Simplify.

Apply the distributive property.

$(7 y + x - 3) (7 y - x - - 3)$

Multiply $- 1$ by $- 3$ .

$(7 y + x - 3) (7 y - x + 3)$

Factor 49y^2-x^2+6x-9

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