Question

question
factor completely.
-2k³ + 6k² + 4k - 12

Explanation:

Step1: Group terms

Group the first two terms and the last two terms: $(-2k^3 + 6k^2) + (4k - 12)$

Step2: Factor out GCF from each group

Factor out $-2k^2$ from the first group and $4$ from the second group: $-2k^2(k - 3) + 4(k - 3)$

Step3: Factor out common binomial

Factor out $(k - 3)$: $(k - 3)(-2k^2 + 4)$

Step4: Factor out GCF from the second factor

Factor out $-2$ from $-2k^2 + 4$: $(k - 3)(-2)(k^2 - 2)$ or $-2(k - 3)(k^2 - 2)$ (both are correct, but we can also write it as $2(3 - k)(k^2 - 2)$ or other equivalent forms, but the standard factored form is usually with positive leading coefficient for the quadratic, so $-2(k - 3)(k^2 - 2)$ or $2(3 - k)(k^2 - 2)$)

Answer:

$-2(k - 3)(k^2 - 2)$ (or equivalent forms like $2(3 - k)(k^2 - 2)$)