how do i write a polynomial in standard form with the given roots -6,2i, and 8

Answer 1

(x+6)(x-8)(x^2+4) =

(x^2-2x-48)(x^2+4) =

x^4 - 2x^3 - 44x^2 -8x - 192

if there's an imaginary zero, then there's also a 2nd zero, they're conjugate pairs

x-2i times x+2i = x^2 +4. take each zero, change its sign and stick an x in front of it, to create the corresponding factor. Then multiply each of the factors together to get the polynomial.

Answer 2

Assuming REAL coefficients

(X+6)(x-8)(x+2i)(x-2i) =

(x^2 -2x -48)(x^2+4)

you can finish the algebra