Question

factor completely.
$7q^{3}+7q^{2}-9q - 9$

Explanation:

Step1: Group terms into pairs

Group the first two and last two terms:
$(7q^3 + 7q^2) + (-9q - 9)$

Step2: Factor out GCF from each group

Factor $7q^2$ from first group, $-9$ from second:
$7q^2(q + 1) - 9(q + 1)$

Step3: Factor out common binomial

Factor $(q + 1)$ from both terms:
$(q + 1)(7q^2 - 9)$

Step4: Check for further factoring

$7q^2 - 9$ is a difference of squares: $7q^2 - 9 = (\sqrt{7}q)^2 - 3^2$, so factor it:
$(q + 1)(\sqrt{7}q - 3)(\sqrt{7}q + 3)$
(Note: If restricting to integer factors, stop at $(q + 1)(7q^2 - 9)$; full real factorization is above.)

Answer:

$(q + 1)(\sqrt{7}q - 3)(\sqrt{7}q + 3)$
Or for integer factorization: $(q + 1)(7q^2 - 9)$