Given a degree 5 polynomial g(x) with real coefficient 2 that has the leading coefficient t 2, and has the zeros -5i,1 and 3, and 0.

Answer 1

Polynomials wit zeros c₁, c₂, ... c_n have factors (x - c₁), (x - c₂), ..., (x - c_n)

Therefore a polynomial with zeros -5i, 1, 3, and 0 has factors (x+5i), (x -1), (x-3) and (x-0)

Because we need real coefficients, we have to include the conjugate of the -5i zero.

Therefore, 5i is another zero and another factor is (x - 5i)

So we can write a polynomial with these zeros as (x+5i)(x-5i)(x-1)(x-3)(x-0)

The x-0 can just be written as x.

So a polynomial with these factors would look like g(x) = x(x+5i)(x-5i)(x-1)(x-3)

We need a leading coefficient of 2. Right now, if we were to expand this, we would have a leading coefficient of 1, so we need to multiply this polynomial by 2.

g(x) = 2x(x+5i)(x-5i)(x-1)(x-3). This has a total of 5 complex zeros, so all the requirements are satisfied.