according to the fundamental theorem of algebra, which polynomial function has exactly 8 roots?\n…

Explanation:

Step1: Recall the Fundamental Theorem of Algebra

The number of roots of a non - zero polynomial is equal to its degree. For a product of polynomials $P(x)=f(x)g(x)$, the degree of $P(x)$ is the sum of the degrees of $f(x)$ and $g(x)$. For a polynomial of the form $(ax^n + bx^{n - 1}+\cdots+ c)^m$, the degree is $mn$.

Step2: Calculate the degree of $f(x)=(3x^{2}-4x - 5)(2x^{6}-5)$

The degree of $3x^{2}-4x - 5$ is $2$ and the degree of $2x^{6}-5$ is $6$. The degree of $f(x)$ is $2 + 6=8$.

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

Using the power - rule for polynomials, if $y=(3x^{4}+2x)^{4}$, the degree is $4\times4 = 16$.

Step4: Calculate the degree of $f(x)=(4x^{2}-7)^{3}$

The degree of $4x^{2}-7$ is $2$, so the degree of $(4x^{2}-7)^{3}$ is $3\times2=6$.

Step5: Calculate the degree of $f(x)=(6x^{8}-4x^{5}-1)(3x^{2}-4)$

The degree of $6x^{8}-4x^{5}-1$ is $8$ and the degree of $3x^{2}-4$ is $2$. The degree of $f(x)$ is $8 + 2=10$.

Answer:

$f(x)=(3x^{2}-4x - 5)(2x^{6}-5)$