College

Write a polynomial function with zeros at [tex]-5[/tex], [tex]-3[/tex], and [tex]3[/tex].

A. [tex]f(x) = x^3 + 5x^2 - 9x + 12[/tex]
B. [tex]f(x) = x^3 - 45x^2 + 5x - 9[/tex]
C. [tex]f(x) = x^3 + 5x^2 - 9x - 45[/tex]
D. [tex]f(x) = x^3 - 4x^2 - 360x + 12[/tex]

Answer :

To find a polynomial function with zeros at [tex]$-5$[/tex], [tex]$-3$[/tex], and [tex]$3$[/tex], we start by writing its factors corresponding to each of these zeros. If [tex]$a$[/tex] is a zero of the polynomial, then [tex]$(x - a)$[/tex] is a factor. Therefore, the factors are:

[tex]$$
(x + 5),\quad (x + 3),\quad \text{and} \quad (x - 3).
$$[/tex]

The polynomial function can be written as the product of these factors:

[tex]$$
f(x) = (x + 5)(x + 3)(x - 3).
$$[/tex]

Next, we simplify the product. Notice that the factors [tex]$(x + 3)$[/tex] and [tex]$(x - 3)$[/tex] form a difference of squares:

[tex]$$
(x + 3)(x - 3) = x^2 - 9.
$$[/tex]

Now, substitute this result back into the expression for [tex]$f(x)$[/tex]:

[tex]$$
f(x) = (x + 5)(x^2 - 9).
$$[/tex]

To obtain the expanded form of the polynomial, distribute [tex]$(x + 5)$[/tex] over [tex]$x^2 - 9$[/tex]:

[tex]\[
\begin{aligned}
f(x) &= x(x^2 - 9) + 5(x^2 - 9) \\
&= x^3 - 9x + 5x^2 - 45 \\
&= x^3 + 5x^2 - 9x - 45.
\end{aligned}
\][/tex]

Thus, the polynomial function with zeros at [tex]$-5$[/tex], [tex]$-3$[/tex], and [tex]$3$[/tex] is

[tex]$$
f(x) = x^3 + 5x^2 - 9x - 45.
$$[/tex]

Among the multiple choices provided, this corresponds to the third option.