Factor (x^2-4)(x^2+6x+9)

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

Step 1

Rewrite $4$ as $2^{2}$ .

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

Step 2

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

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

Step 3

Factor using the perfect square rule.

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

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

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.

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

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

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