High School

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 :

To determine which polynomial function has the zeros [tex]\( 4 \)[/tex], [tex]\( \sqrt{7} \)[/tex], and [tex]\( -\sqrt{7} \)[/tex], we use the Factor Theorem. The Factor Theorem states that if [tex]\( r \)[/tex] is a root of the polynomial [tex]\( f(x) \)[/tex], then [tex]\( (x - r) \)[/tex] is a factor of [tex]\( f(x) \)[/tex].

Given the zeros [tex]\( 4 \)[/tex], [tex]\( \sqrt{7} \)[/tex], and [tex]\( -\sqrt{7} \)[/tex], the polynomial can be written in factored form as:

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

Let's expand this product:

First, notice that [tex]\((x - \sqrt{7})(x + \sqrt{7})\)[/tex] is a difference of squares:

[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]

Now we expand this product:

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

Therefore, the polynomial [tex]\( f(x) = x^3 - 4x^2 - 7x + 28 \)[/tex] is the one that has the zeros [tex]\( 4 \)[/tex], [tex]\( \sqrt{7} \)[/tex], and [tex]\( -\sqrt{7} \)[/tex].

So, the correct polynomial function is:

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