High School

Factor the expression completely, or state that the polynomial is prime.

[tex]\[ x^3 + 3x^2 - 16x - 48 \][/tex]

Select the correct choice below and, if necessary, fill in the answer box to complete your choice:

A. [tex]\[ x^3 + 3x^2 - 16x - 48 = \square \][/tex]

B. The polynomial is prime.

Answer :

To factor the polynomial [tex]\(x^3 + 3x^2 - 16x - 48\)[/tex] completely, follow these steps:

1. Look for common factors: First, check if there is any common factor in all the terms. Here, there is no common factor in all the terms of the polynomial.

2. Try factoring by grouping: Group the terms in pairs to see if you can factor them easily:

[tex]\[
(x^3 + 3x^2) + (-16x - 48)
\][/tex]

- From the first group [tex]\((x^3 + 3x^2)\)[/tex], factor out [tex]\(x^2\)[/tex]:
[tex]\[
x^2(x + 3)
\][/tex]

- From the second group [tex]\((-16x - 48)\)[/tex], factor out [tex]\(-16\)[/tex]:
[tex]\[
-16(x + 3)
\][/tex]

3. Combine the grouped factors: Notice that [tex]\((x + 3)\)[/tex] is common in both terms:
[tex]\[
x^2(x + 3) - 16(x + 3) = (x^2 - 16)(x + 3)
\][/tex]

4. Factor further if possible: The expression [tex]\((x^2 - 16)\)[/tex] is a difference of squares, which can be factored into
[tex]\[
(x - 4)(x + 4)
\][/tex]

5. Write the complete factorization: Substitute back to include all factors:
[tex]\[
(x - 4)(x + 4)(x + 3)
\][/tex]

So, the polynomial [tex]\(x^3 + 3x^2 - 16x - 48\)[/tex] can be factored completely as [tex]\((x - 4)(x + 3)(x + 4)\)[/tex]. Therefore, the correct choice is:

A. [tex]\(x^3 + 3x^2 - 16x - 48 = (x - 4)(x + 3)(x + 4)\)[/tex]