Question

question 6 of 25
what is the complete factorization of the polynomial below?
$x^3 + x^2 + 9x + 9$

a. $(x + 1)(x + 3i)(x - 3i)$
b. $(x - 1)(x + 3i)(x + 3i)$
c. $(x - 1)(x + 3i)(x - 3i)$
d. $(x + 1)(x + 3i)(x + 3i)$

Explanation:

Step1: Group terms for factoring

$x^3 + x^2 + 9x + 9 = (x^3 + x^2) + (9x + 9)$

Step2: Factor out common terms

$= x^2(x + 1) + 9(x + 1)$

Step3: Factor out $(x + 1)$

$= (x + 1)(x^2 + 9)$

Step4: Factor $x^2 + 9$ as difference of squares

$x^2 + 9 = x^2 - (-9) = (x + 3i)(x - 3i)$

Step5: Combine factors

$= (x + 1)(x + 3i)(x - 3i)$

Answer:

A. $(x+1)(x+3i)(x-3i)$