Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point.

Question

Degree 4

Zeros

-1,2,√2i

Solution Point f(1)=18

Answer 1

f(x) = k(x+1)(x-2) ( x - sqrt(2)*i)(x + sqrt(2)*i)

f(x) = k(x^2 - 2x + 1x - 2)( x^2 - 2*(-1))

f(x) = k(x^2 -x - 2)(x^2 + 2)

f(1) = 18

k*(1 - 1 - 2)(1 + 2) = 18

k*(-2)(3) = 18

k*(-6) = 18

k = -3

The polynomial function is:

y = f(x) = -3(x^2 - x - 2)(x^2 +2)

You can finish multiplying them together if you must