find a fourth degree polynomial

[rrr1116]

I need help setting this problem up and solving it.

Find a fourth degree polynomial function f(x) having only real coefficients, -1,2, and i as zeros, and f(3)=80

 

[MarkFL]

Hello, and welcome to FMH!

According to the complex conjugate roots theorem, what must the 4th root be?

 

[Steven G]

Let r₁, r₂, r₃ and r₄ be the four roots of the polynomial.
Then f(x) = k(x-r₁)(x-r₂)(x-r₃)(x-r₄)

You need to find k, r₁, r₂, r₃ and r₄

 

[MarkFL]

To follow up, by the complex conjugate roots theorem, we know the 4th root must be $-i$, and so the quartic polynomial may be written as:

[MATH]f(x)=k(x+1)(x-2)(x-i)(x+i)=k(x^2-x-2)(x^2+1)=k\left(x^4-x^3-x^2-x-2\right)[/MATH]
Now, we can use the given point on the curve to determine the parameter $k$:

[MATH]f(3)=k\left(3^4-3^3-3^2-3-2\right)=40k=80\implies k=2[/MATH]
And so we have:

[MATH]f(x)=2\left(x^4-x^3-x^2-x-2\right)=2x^4-2x^3-2x^2-2x-4[/MATH]