Find a polynomial function​ f(x) of least degree having only real coefficients with zeros of​ 0, 2 i​, and 3+i

Answer 1

The standard method of generating a polynomial of specific zeros is to build it up as products of (x - a₁), (x - a₂), etc., and then multiplying it all out. If the zeros are real numbers, then they can be plugged in for a₁ etc. If complex numbers are involved, then you will also need their complex conjugates to be zeros. So in this case we need solutions of 0, 2i and 3 + i. This means that we also need -2i and 3 - i to be solutions.

I'm going to treat each pair of complex solutions separately, and I'll deal with ±2i first.

(x - 2i)(x + 2i) = x² + 2ix - 2ix - (2i)² = x² - 4i² = x² - 4(-1) = x² + 4.

I'm going to deal with 3 + i and 3 - i a little different. I'm going to begin with what would ordinarily be the last step of an equation that was solved:

x = 3 ± i

x - 3 = ±i

(x - 3)² = -1

x² - 6x + 9 = -1

x² - 6x + 10 = 0

So now I can multiply these two polynomials together and set the result equal to 0.

(x² + 4)(x² - 6x + 10) = 0

x⁴ - 6x³ + 10x² + 4x² - 24x - 40 = 0

x⁴ - 6x³ + 14x² - 24x + 40 = 0

Now all that remains is to make sure that 0 is a solution. This can be done by multiplying the whole equation by x.

x(x⁴ - 6x³ + 14x² - 24x + 40) = 0

Answer: x⁵ - 6x⁴ + 14x³ - 24x² + 40x = 0

Answer 2

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

= x ( x^2 + 4) ( (x-3)^2 + 1)

= x ( x^2 + 4) ( x^2 - 6x + 10)