According to the Fundamental Theorem of Algebra, which polynomial function has exactly 11 roots?

A. [tex]\( f(x) = (x-1)(x+1)^{11} \)[/tex]
B. [tex]\( f(x) = (x+2)^3(x^2-7x+3)^4 \)[/tex]
C. [tex]\( f(x) = (x^5 + 7x + 14)^8 \)[/tex]
D. [tex]\( f(x) = 11x^5 + 5x + 25 \)[/tex]



Answer :

Let's analyze each polynomial function to determine the degree and find which one has exactly 11 roots.

1. First polynomial: [tex]\( f(x) = (x-1)(x+1)^{11} \)[/tex]
- The term [tex]\( (x-1) \)[/tex] has a degree of 1.
- The term [tex]\( (x+1)^{11} \)[/tex] has a degree of [tex]\( 11 \)[/tex].
- Therefore, the total degree of the polynomial is [tex]\( 1 + 11 = 12 \)[/tex].

2. Second polynomial: [tex]\( f(x) = (x+2)^3(x^2-7x+3)^4 \)[/tex]
- The term [tex]\( (x+2)^3 \)[/tex] has a degree of [tex]\( 3 \)[/tex].
- The term [tex]\( (x^2-7x+3)^4 \)[/tex] involves a quadratic polynomial [tex]\( (x^2-7x+3) \)[/tex] raised to the power of [tex]\( 4 \)[/tex], which means it has a degree of [tex]\( 2 \times 4 = 8 \)[/tex].
- Therefore, the total degree of the polynomial is [tex]\( 3 + 8 = 11 \)[/tex].

3. Third polynomial: [tex]\( f(x) = (x^5 + 7x + 14)^8 \)[/tex]
- Inside the polynomial [tex]\( (x^5 + 7x + 14) \)[/tex], the highest degree term is [tex]\( x^5 \)[/tex].
- When raised to the power of [tex]\( 8 \)[/tex], the degree becomes [tex]\( 5 \times 8 = 40 \)[/tex].
- Therefore, the total degree of the polynomial is [tex]\( 40 \)[/tex].

4. Fourth polynomial: [tex]\( f(x) = 11x^5 + 5x + 25 \)[/tex]
- The highest degree term is [tex]\( 11x^5 \)[/tex], which means it has a degree of [tex]\( 5 \)[/tex].
- Therefore, the total degree of the polynomial is [tex]\( 5 \)[/tex].

Based on the analysis, the polynomial function [tex]\( f(x) = (x+2)^3(x^2-7x+3)^4 \)[/tex] has exactly 11 roots.

Thus, the polynomial function with exactly 11 roots is:
[tex]\[ f(x) = (x+2)^3(x^2-7x+3)^4 \][/tex]