Given:
• Degree of polynomial = 3
,
• Zeros of the polynomial: 2, 3 - 2i
Let's find the polynomial.
Since the polynomail is of degree 3, it's highest exponent will be 3.
Equate the zeros to zero:
x = 2
Subtract 2 from both sides:
x - 2 = 2 - 2
x - 2 = 0
x = (3 - 2i)
Since this root is a complex conjugate, we have the other complex root: (3 + 2i)
Hence, we have:
(x - (3 - 2i)) and (x - (3 + 2i)).
Therefore, to write the function, we have:
[tex]f(x)=(x-2)(x-(3-2i))(x-(3+2i))[/tex]
Now, simplify the expression:
[tex]\begin{gathered} f(x)=(x-2)(x-3+2i)(x-3-2i) \\ \\ f(x)=x(x-3+2i)-2(x-3+2i)(x-3-2i) \\ \\ f(x)=x^2-3x+2ix-2x+6-4i(x-3-2i) \\ \\ f(x)=x^2-5x+2ix-4i+6(x-3-2i) \end{gathered}[/tex]
Solving further:
[tex]\begin{gathered} f(x)=x(x^2-5x+2ix-4i+6)-3(x^2-5x+2ix-4i+6)-2i(x^2-5x+2ix-4i+6) \\ \\ f(x)=x^3-5x^2+2ix^2-4ix+6x-3x^2+15x-6ix+12i-18-2ix^2+10ix-4i^2x-8-12i^{} \end{gathered}[/tex]
Combine like terms:
[tex]\begin{gathered} f(x)=x^3-5x^2-3x^2-4ix-6ix+10ix+2ix^2-2ix^2+6x+15x+12i-12i-8-16 \\ \\ f(x)=x^3-8x^2+25x-26 \end{gathered}[/tex]
ANSWER:
[tex]f(x)=x^3-8x^2+25x-26[/tex]
