Question 14 of 25What is the complete factorization of the polynomial below?x³+x² + 4x+4O A. (x-1)(x+2)(x+2)B. (x + 1)(x + 2)(x + 2)O C. (x + 1)(x + 2)(x-2)O D. (x-1)(x + 2)(x-2)SUBMIT

GIVEN:
We are given the following polynomial;
[tex]x^3+x^2+4x+4[/tex]Required;
We are required to factorize this polynomial completely.
Step-by-step explanation;
To factorize this polynomial, we start by grouping;
[tex](x^3+x^2)+(4x+4)[/tex]We now take the common factor in each group;
[tex]\begin{gathered} x^2(x+1)+4(x+1) \\ \\ (x^2+4)(x+1) \end{gathered}[/tex]Next, we factorize the first parenthesis. To do this we set the equation equal to zero and solve for x as follows;
[tex]\begin{gathered} x^2+4=0 \\ \\ Subtract\text{ }4\text{ }from\text{ }both\text{ }sides: \\ \\ x^2=-4 \\ \\ Take\text{ }the\text{ }square\text{ }root\text{ }of\text{ }both\text{ }sides: \\ \\ x=\pm\sqrt{-4} \\ \\ x=(\pm\sqrt{-1}\times\sqrt{4}) \\ \\ x=\pm2i \end{gathered}[/tex]Therefore, the factors of the other parenthesis are;
[tex](x+2i)(x-2i)[/tex]Therefore, the complete factorization of the polynomial is;
ANSWER:
[tex](x+1)(x+2i)(x-2i)[/tex]Option C is the correct answer.