Form a polynomial f(x) with real coefficients with the given degree and zeros

Question

Degree 5

Zeros: - 4, -i, - 9+i

Answer 1

x=-4

x+4 is one factor

x=-i

x^2 = -1

x^2+1 is another factor

x=-9+i

x+9 = i

x^2 +18x +81 =-1

x^2 +18x +82 is another factor

multiply the 3 factors together

f(x) = (x^2+18x+82)(x^2+1)(x+4) is a 5th degree polynomial when expanded, with the zeros -4, -i, -6+i

there's actually 5 zeros, as the imaginary zeros come in conjugate pairs: -i and i are both zeros

there are also other 5th degree polynomials with the same zeros

f(x) = a(x^2+18x+82)(x^2+1)(x+4) where a=any real number or any real integer for integer polynomials

check the answer, replace x with 4

f(4) = (16+18(4)+82)(16+1)(-4+4)

= (170)(17)(0)

= 0

f(-i) = ... ((-i)^2+1)(i+4)

= ...(0)(i+4)

=0

f(-6+i) = ((-6+i)^2 +18(-6+i)+82)((-6+i)^2+1)(-6+i+4)

=0