according to the fundamental theorem of algebra, which polynomial function has exactly 11 roots?\n$f(x)=(x…

according to the fundamental theorem of algebra, which polynomial function has exactly 11 roots?\n$f(x)=(x - 1)(x + 1)^{11}$\n$f(x)=(x + 2)^{3}(x^{2}-7x + 3)^{4}$\n$f(x)=(x^{5}+7x + 14)^{6}$\n$f(x)=11x^{5}+5x + 25$

according to the fundamental theorem of algebra, which polynomial function has exactly 11 roots?\n$f(x)=(x - 1)(x + 1)^{11}$\n$f(x)=(x + 2)^{3}(x^{2}-7x + 3)^{4}$\n$f(x)=(x^{5}+7x + 14)^{6}$\n$f(x)=11x^{5}+5x + 25$

Answer

Explanation:

Step1: Recall the Fundamental Theorem of Algebra

The number of roots of a polynomial is equal to its degree.

Step2: Calculate the degree of (f(x)=(x - 1)(x + 1)^{11})

Using the rule of exponents in polynomial - multiplication, the degree is (1+11 = 12).

Step3: Calculate the degree of (f(x)=(x + 2)^{3}(x^{2}-7x + 3)^{4})

The degree of ((x + 2)^{3}) is 3 and the degree of ((x^{2}-7x + 3)^{4}) is (2\times4=8). The degree of the product is (3 + 8=11).

Step4: Calculate the degree of (f(x)=(x^{5}+7x + 14)^{6})

The degree is (5\times6 = 30).

Step5: Calculate the degree of (f(x)=11x^{5}+5x + 25)

The degree is 5.

Answer:

(f(x)=(x + 2)^{3}(x^{2}-7x + 3)^{4})