Write a polynomial function of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. Put your answer in standard form. 5,3,-2i

Question

this one has me really stuck

Answer 1

Because real coefficients we have one more zero 2i. So we have 4 zeros and because degree is 4.

We looking for polynomial in factored form: f(x) = (x - 5)(x - 3)(x + 2i)(x - 2i) = (x² - 8x + 15)(x² + 4) =

= x⁴ - 8x³ + 19x² - 32x + 60

Answer 2

f(x) = (x-5)(x-3)(x^2+4)

= (x^2-8x + 15)(x^2 + 4)

= x^4 - 8x^3 + 19x^2 - 32x + 60

that's the answer if you want real coefficients

you could get a 3rd degree function if you allow for imaginary coefficients'

f(x) = (x-5)(x-3)(x+2i) =

x^3 + (2i-8)x^2 + (15-16i)x + 30i

but these coefficients are not "rational" They aren't "irrational" either though. As the rational/irrational distinction applies only to real numbers, not to imaginary numbers. the real parts of the cubic polynomial include the coefficients of i, the 2 and -16, which are real and rational.