High School

Given the function [tex]f(x)=2x^5+15x^4+60x^3+30x^2-182x+75[/tex], and knowing that [tex]-3-4i[/tex] and [tex]\frac{1}{2}[/tex] are zeros:

(a) Find all the zeros of the function. Write your answers in exact form.

If there is more than one answer, separate them with commas. Select "None" if applicable.

The zeros of [tex]f(x)[/tex] are: [tex]\square[/tex].

Answer :

We start with the polynomial
[tex]$$
f(x)=2x^5+15x^4+60x^3+30x^2-182x+75.
$$[/tex]

We are given that [tex]$\frac{1}{2}$[/tex] and [tex]$-3-4i$[/tex] are zeros. Since the coefficients of [tex]$f(x)$[/tex] are real, the complex conjugate of [tex]$-3-4i$[/tex], namely [tex]$-3+4i$[/tex], is also a zero.

This information tells us that the polynomial has the following factors corresponding to these zeros:
- Since [tex]$x=\frac{1}{2}$[/tex] is a zero, the factor is
[tex]$$
2x-1.
$$[/tex]

- The zeros [tex]$x=-3-4i$[/tex] and [tex]$x=-3+4i$[/tex] together give the quadratic factor. Notice that
[tex]$$
(x+3-4i)(x+3+4i) = (x+3)^2 + 16.
$$[/tex]

Thus, we have two factors:
[tex]$$
2x-1 \quad \text{and} \quad (x+3)^2+16.
$$[/tex]

Next, we divide [tex]$f(x)$[/tex] by the product of these two factors. That is, we write
[tex]$$
f(x)=(2x-1)\Big[(x+3)^2+16\Big]\cdot Q(x),
$$[/tex]
where [tex]$Q(x)$[/tex] is the quotient polynomial once the known factors are removed. It turns out that the quotient is a quadratic:
[tex]$$
Q(x)=x^2+2x-3.
$$[/tex]

Now, we factor [tex]$Q(x)$[/tex]:
[tex]$$
x^2+2x-3=(x-1)(x+3).
$$[/tex]

The factors of [tex]$f(x)$[/tex] are now completely determined:
[tex]$$
f(x)=(2x-1)\Big[(x+3)^2+16\Big](x-1)(x+3).
$$[/tex]

From these factors, the zeros of the polynomial are obtained by setting each factor equal to zero:
1. From [tex]$2x-1=0$[/tex]:
[tex]$$
2x-1=0 \quad \Rightarrow \quad x=\frac{1}{2}.
$$[/tex]

2. From [tex]$(x+3)^2+16=0$[/tex]:
[tex]\[
(x+3)^2 = -16 \quad \Rightarrow \quad x+3 = \pm 4i \quad \Rightarrow \quad x=-3\pm 4i.
\][/tex]

3. From [tex]$x-1=0$[/tex]:
[tex]$$
x=1.
$$[/tex]

4. From [tex]$x+3=0$[/tex]:
[tex]$$
x=-3.
$$[/tex]

Thus, all the zeros, in exact form, are:
[tex]$$
\frac{1}{2},\quad -3-4i,\quad -3+4i,\quad -3,\quad 1.
$$[/tex]