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

A. [tex]$f(x) = x^3 - 4x^2 + 7x + 28[tex]$[/tex]
B. [tex]$[/tex]f(x) = x^3 - 4x^2 - 7x + 28$[/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 of the given polynomial functions has the specified zeros [tex]\( 4, \sqrt{7}, \)[/tex] and [tex]\( -\sqrt{7} \)[/tex], we will use the Factor Theorem. The Factor Theorem states that for a polynomial [tex]\( f(x) \)[/tex], [tex]\( x=c \)[/tex] is a zero of the polynomial if and only if [tex]\( f(c) = 0 \)[/tex].

Let's test each polynomial to see which one satisfies [tex]\( f(4) = 0 \)[/tex], [tex]\( f(\sqrt{7}) = 0 \)[/tex], and [tex]\( f(-\sqrt{7}) = 0 \)[/tex].

### Polynomial 1: [tex]\( f(x) = x^3 - 4x^2 + 7x + 28 \)[/tex]
1. [tex]\( f(4) = 4^3 - 4 \cdot 4^2 + 7 \cdot 4 + 28 = 64 - 64 + 28 + 28 = 56 \neq 0 \)[/tex]

Since [tex]\( f(4) \neq 0 \)[/tex], [tex]\( 4 \)[/tex] is not a zero of this polynomial.

### Polynomial 2: [tex]\( f(x) = x^3 - 4x^2 - 7x + 28 \)[/tex]
1. [tex]\( f(4) = 4^3 - 4 \cdot 4^2 - 7 \cdot 4 + 28 = 64 - 64 - 28 + 28 = 0 \)[/tex]
- This suggests [tex]\( f(4) \)[/tex] might be a zero. Next, check the other zeros.

2. [tex]\( f(\sqrt{7}) = (\sqrt{7})^3 - 4(\sqrt{7})^2 - 7(\sqrt{7}) + 28 = 7\sqrt{7} - 4 \cdot 7 - 7\sqrt{7} + 28 = 0 \)[/tex]

3. [tex]\( f(-\sqrt{7}) = (-\sqrt{7})^3 - 4(-\sqrt{7})^2 - 7(-\sqrt{7}) + 28 = -7\sqrt{7} - 4 \cdot 7 + 7\sqrt{7} + 28 = 0 \)[/tex]

Thus, [tex]\( f(\sqrt{7}) = 0 \)[/tex] and [tex]\( f(-\sqrt{7}) = 0 \)[/tex], confirming that all specified zeros satisfy this polynomial.

### Polynomial 3: [tex]\( f(x) = x^3 + 4x^2 - 7x + 28 \)[/tex]
1. [tex]\( f(4) = 4^3 + 4 \cdot 4^2 - 7 \cdot 4 + 28 = 64 + 64 - 28 + 28 = 128 \neq 0 \)[/tex]

Since [tex]\( f(4) \neq 0 \)[/tex], [tex]\( 4 \)[/tex] is not a zero of this polynomial.

### Polynomial 4: [tex]\( f(x) = x^3 + 4x^2 - 7x - 28 \)[/tex]
1. [tex]\( f(4) = 4^3 + 4 \cdot 4^2 - 7 \cdot 4 - 28 = 64 + 64 - 28 - 28 = 72 \neq 0 \)[/tex]

Since [tex]\( f(4) \neq 0 \)[/tex], [tex]\( 4 \)[/tex] is not a zero of this polynomial.

After evaluating the polynomials, we find that the second polynomial [tex]\( f(x) = x^3 - 4x^2 - 7x + 28 \)[/tex] satisfies [tex]\( f(4) = 0 \)[/tex], [tex]\( f(\sqrt{7}) = 0 \)[/tex], and [tex]\( f(-\sqrt{7}) = 0 \)[/tex].

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