factor completely.\n$(x^{2}-4)(x^{2}+6x + 9)=$

Explanation:

Step1: Factor difference - of - squares

Use the formula $a^{2}-b^{2}=(a + b)(a - b)$. For $x^{2}-4$, where $a = x$ and $b = 2$, we have $x^{2}-4=(x + 2)(x - 2)$.

Step2: Factor perfect - square trinomial

Use the formula $a^{2}+2ab + b^{2}=(a + b)^{2}$. For $x^{2}+6x + 9$, where $a=x$ and $b = 3$, we have $x^{2}+6x + 9=(x + 3)^{2}$.

Step3: Write the factored form

Substitute the factored forms back into the original expression: $(x^{2}-4)(x^{2}+6x + 9)=(x + 2)(x - 2)(x + 3)^{2}$.

Answer:

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