Question

the polynomial $x^3 + 4x^2 - 9x - 36$ has 4 terms.
use factoring by grouping to find the correct
factorization.
$(x^2 + 9)(x + 4)$
$(x^2 - 9)(x - 4)$
$(x^2 + 9)(x - 4)$
$(x^2 - 9)(x + 4)$
complete
$x + 4$ is prime
$x^2 - 9$ can be factored using the
formula.
done
intro

Explanation:

Step1: Group polynomial terms

Group first two and last two terms:
$(x^3 + 4x^2) + (-9x - 36)$

Step2: Factor out GCF from groups

Factor $x^2$ from first group, $-9$ from second:
$x^2(x + 4) - 9(x + 4)$

Step3: Factor common binomial

Factor out $(x+4)$ from both terms:
$(x^2 - 9)(x + 4)$

Step4: Identify factoring formula for $x^2-9$

$x^2 - 9$ is a difference of squares, $a^2-b^2=(a-b)(a+b)$

Answer:

1. The full factorization steps confirm the correct initial factorization is $\boldsymbol{(x^2 - 9)(x + 4)}$
2. The formula to factor $x^2 - 9$ is the difference of squares formula, which follows $\boldsymbol{a^2 - b^2 = (a - b)(a + b)}$ (here $a=x$, $b=3$, so $x^2-9=(x-3)(x+3)$)