Answer:
C) f(x) = (x - 2)(x - 3)(x + 5i)(x - 5i)
Explanation:
Given the function:
[tex]f\mleft(x\mright)=x^4-5x^3+31x^2-125x+150[/tex]
First, we check for a linear factor of f(x) by using the remainder theorem.
We try 2, -2, 3 and -3.
[tex]\begin{gathered} f(2)=2^4-5(2)^3+31(2)^2-125(2)+150=0 \\ f(-2)=(-2)^4-5(-2)^3+31(-2)^2-125(-2)+150=580 \\ f(3)=3^4-5(3)^3+31(3)^2-125(3)+150=0 \\ f(-3)=(-3)^4-5(-3)^3+31(-3)^2-125(-3)+150=1020 \end{gathered}[/tex]
It means that: x-2 and x-3 are factors of f(x) since they have a remainder of zero.
We divide f(x) by (x-2)(x-3) to obtain:
[tex]\frac{x^4-5x^3+31x^2-125x+150}{(x-2)(x-3)}=x^2+25[/tex]
If we solve x²+25, we have:
[tex]\begin{gathered} x^2=-25 \\ x=\pm\sqrt[]{-25} \\ x=\pm5i \end{gathered}[/tex]
Therefore, the other roots are x+5i and x-5i.
The correct choice is:
C) f(x) = (x - 2)(x - 3)(x + 5i)(x - 5i)
