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]
[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]
