Find an nth degree polynomial with real coefficients satisfying the given conditions

Question

n=4;

-1,3 and 2+4i are zeroes;

f(1)=-204

Answer 1

Since this polynomial has real coefficients and contains the complex root 2+4i it follows that the complex conjugate of that root (namely 2-4i) must be another root of the polynomial. This means that the four roots of this polynomial are -1, 3, 2+4i, 2-4i so the formula for this polynomial must be:

f(x)=a(x+1)(x-3)[x-(2+4i)][x-(2-4i)]

where "a" is some constant whose value we can determine by observing that f(1)=-204:

f(1)=a(1+1)(1-3)[1-(2+4i)][1-(2-4i)]=a*2*-2(-1-4i)(-1+4i)=a*-4*17=-68a=-204 -> a=204/68=3

Therefore, we have:

f(x)=3(x+1)(x-3)[x-(2+4i)][x-(2-4i)]=3(x²-2x-3)(x²-4x+20)=3(x⁴-6x³+25x²-28x-60)=3x⁴-18x³+75x²-84x-180

Answer 2

Hi Adbul,

I'd like you to check out this similar question: https://www.wyzant.com/resources/answers/748019/let-a-sub-n-be-the-leading-coefficient

This should help you write the function in factored form. Then, you will this need to use f(1)=-204 to solve for constants in the function.

If you need any help, please let me know.