Factor completely. (x^(2)-4)(x^(2)+6x+9)=

Factorize first term: Factor the first term (x^2 - 4). The term (x^2 - 4) is a difference of squares, which can be factored into (x - 2)(x + 2).Factorize second term: Factor the second term (x^2 + 6x + 9). The term (x^2 + 6x + 9) is a perfect square trinomial, which can be factored into (x + 3)(x + 3) or (x + 3)^2.Combine factored forms: Combine the factored forms of both terms. Now we have the factored forms of both terms: (x - 2)(x + 2) and (x + 3)^2. The completely factored form of the original expression is the product of these factored forms.Write final expression: Write the final factored expression. The final factored expression is (x - 2)(x + 2)(x + 3)^2.

Home

Math Problems

Algebra 2

Factor sums and differences of cubes

Full solution

Q. Factor completely.

\newline

\left(x^{2}-4\right)\left(x^{2}+6 x+9\right)=

Factorize first term: Factor the first term $(x^2 - 4)$ . $\newline$ The term $(x^2 - 4)$ is a difference of squares, which can be factored into $(x - 2)(x + 2)$ .
Factorize second term: Factor the second term $(x^2 + 6x + 9)$ . $\newline$ The term $(x^2 + 6x + 9)$ is a perfect square trinomial, which can be factored into $(x + 3)(x + 3)$ or $(x + 3)^2$ .
Combine factored forms: Combine the factored forms of both terms. $\newline$ Now we have the factored forms of both terms: $(x - 2)(x + 2)$ and $(x + 3)^2$ . The completely factored form of the original expression is the product of these factored forms.
Write final expression: Write the final factored expression. $\newline$ The final factored expression is $(x - 2)(x + 2)(x + 3)^2$ .