College

Using the Factor Theorem, which of the polynomial functions has the zeros [tex]4, \sqrt{7}[/tex], and [tex]-\sqrt{7}[/tex]?

A. [tex]f(x) = x^3 - 4x^2 - 7x + 28[/tex]
B. [tex]f(x) = x^3 + 4x^2 - 7x - 28[/tex]
C. [tex]f(x) = x^3 - 4x^2 + 7x + 28[/tex]
D. [tex]f(x) = x^3 + 4x^2 - 7x + 28[/tex]

Answer :

Sure! Let's use the Factor Theorem to determine which polynomial function has the zeros [tex]\(4\)[/tex], [tex]\(\sqrt{7}\)[/tex], and [tex]\(-\sqrt{7}\)[/tex].

The Factor Theorem states that if [tex]\(c\)[/tex] is a zero of a polynomial [tex]\(f(x)\)[/tex], then [tex]\((x - c)\)[/tex] is a factor of [tex]\(f(x)\)[/tex].

Given the zeros are [tex]\(4\)[/tex], [tex]\(\sqrt{7}\)[/tex], and [tex]\(-\sqrt{7}\)[/tex], the polynomial can be expressed as:

[tex]\[ f(x) = (x - 4)(x - \sqrt{7})(x + \sqrt{7}) \][/tex]

First, let's simplify the factors involving the square roots:

[tex]\[ (x - \sqrt{7})(x + \sqrt{7}) = x^2 - (\sqrt{7})^2 = x^2 - 7 \][/tex]

So, the polynomial becomes:

[tex]\[ f(x) = (x - 4)(x^2 - 7) \][/tex]

Next, we need to expand this product:

[tex]\[
\begin{align*}
f(x) & = (x - 4)(x^2 - 7) \\
& = x \cdot (x^2 - 7) - 4 \cdot (x^2 - 7) \\
& = x^3 - 7x - 4x^2 + 28 \\
& = x^3 - 4x^2 - 7x + 28
\end{align*}
\][/tex]

Therefore, the polynomial function [tex]\(f(x)\)[/tex] is:

[tex]\[ f(x) = x^3 - 4x^2 - 7x + 28 \][/tex]

Comparing this with the provided options:
- [tex]\(f(x)=x^3 - 4 x^2 - 7 x + 28\)[/tex]

So, the correct option is:

[tex]\[ f(x)=x^3 - 4x^2 - 7x + 28 \][/tex]

Therefore, the correct answer is:

[tex]\[ f(x)=x^3 - 4 x^2 - 7 x + 28 \][/tex]