Given:
[tex]f(x)=2x^4-9x^3+17x^2-16x+6[/tex]We have one root: 1+ i
Let's find the remaining roots.
Given that the root is a complex root, complex roots come in conjugate pairs.
Thus, we have:
1+i and 1 - i
In the factored form we have
(x - 1+i) = 0
(x - 1 - i) = 0
Thus, we have:
Thus, we have:
[tex]\begin{gathered} (x-1+i)(x-1-i)=0 \\ \\ \text{ Expand:} \\ x^2-2x+2 \end{gathered}[/tex]Now, divide using the long division method:
[tex]\frac{2x^4-9x^3+17x^2-16x+6}{x^2-2x+2}=2x^2-5x+3[/tex]The quotient after using the long division method to divide is:
[tex]2x^2-5x+3[/tex]Factor the quotient by grouping:
[tex]\begin{gathered} 2x^2-5x+3=0 \\ \\ 2x^2-2x-3x+3=0 \\ \\ (2x^2-2x)-2x+3=0 \\ \\ Factor\text{ 2x from the first group:} \\ 2x(x-1)-3x+3=0 \\ \\ Factor\text{ 3 from the second group:} \\ 2x(x-1)-3(x-1)=0 \\ \\ (2x-3)(x-1)=0 \end{gathered}[/tex]Solve each factor for x:
[tex]\begin{gathered} 2x-3=0 \\ 2x=3 \\ x=\frac{3}{2} \\ \\ \\ x-1=0 \\ x=1 \end{gathered}[/tex]Therefore, the roots of the function are:
[tex]x=\frac{3}{2},1,1+i,1-i[/tex]ANSWER: C
The remaining roots are:
3/2, 1
