Find the third degree polynomial function.?

Question

Find the third degree polynomial function f(x) such that f(1-2i)=0, f(1/2)=0 and satisfying the boundary condition f(2)=15. Write the answer in standard form. hint: non-real, complex zeroes occur in conjugate pairs; use the boundary condition to determine the leading coefficient.

It's a homework problem and i would want to know how to do it

alex · Accepted Answer

f(x) = a(2x-1)(x^2 - 2x +5) , find a from f(2)=15

Morningfox · Answer

I think you missed the condition that the coefficients have to be real.  Otherwise, the complex zeros do NOT have to occur in conjugate pairs.

The function is f(x) = A ( x -1+2i) (x -1 -2i) (x -0.5)
The value of A is set so that f(2) = 15.

So we have f(x) = A ( x^3 -2.5x^2 + 6x -2.5).
Just calculate the value of  ( x^3 -2.5x^2 + 6x -2.5) at x =2, and set A = 15 / (that value).

J · Answer

Unless you specify that the coefficients are real numbers, a QUADRATIC polynomial can have those properties:

f(x) = (2-4i) (x-1+2i) (x-1/2)

Trending News

Find the third degree polynomial function.?

3 Answers