Question

Write a linear factorization of the function. f(x) = x^4-5x^3+31x^2 - 125x + 150 A) f(x) = (x - 2)(x - 3)(x^2+25) B) f(x) - (x-2)(x+5)(x - 3i)(x + 3i) C) f(x) = (x - 2)(x - 3)(x + 5i)(x - 5i) D) f(x) = (x+5)(x - 3)(x-2i)(x + 2i)

Answer 1

C) f(x) = (x - 2)(x - 3)(x + 5i)(x - 5i)

Step-by-step explanation:

Given the function:

`f\mleft(x\mright)=x^4-5x^3+31x^2-125x+150`

First, we check for a linear factor of f(x) by using the remainder theorem.

We try 2, -2, 3 and -3.

`\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}`

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:

`(x^4-5x^3+31x^2-125x+150)/((x-2)(x-3))=x^2+25`

If we solve x²+25, we have:

`\begin{gathered} x^2=-25 \\ x=\pm\sqrt[]{-25} \\ x=\pm5i \end{gathered}`

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)