How to solve "Find a polynomial of degree 5 with integer coefficients that has zeros −2,√2i , i , and y-intercept of 20"?

Question

√2i is really throwing me off, help with detailed steps would be a life saver.

Answer 1

then -i*sqrt(2) and -i are also solutions

the factors are P(x) = C (x+2)(x + i*sqrt(2)) (x - i*sqrt(2)) ( x+i)(x-i)

y-intercept is (0,20)

f(0) = 20

20 = C(2)(i*sqrt(2)) ( -i*sqrt(2)) (i)(-i)

20 = C(2) (-2)(1)

20 = -4C

C=-5

P(x) = -5(x+2)(x+i*sqrt(2))(x - i*sqrt(2)) (x + i)(x-i)

= -5 ( x+2)(x^2 +2)(x^2 + 1)

you can multiply the rest out if you need