Form a polynomial with real coefficients having the given degree and zeros.

Question

degree: 5

zeros: 8;-i;-6+i

The polynomial is

f(x)=a(?)

Answer 1

a(x-8)(x^2 +1)(x^2 +12x +37) is degree 5 with given zeros. there are 4 imaginary zeros as imaginary zeros come in conjugate pairs, if i is a zero, so is -i

a can = any real number except 0

calculate the 2nd quadratic factor

x = -6+ i

x+6 = i square both sides

x^2 +12x +36 = -1

f(x) = a(x-8)(x^2+1)(x^2 +12x +37) = a(x^3 -8x^2 +x -8)(x^2 +12x +37)

f(x) = a(x^5 +4x^4 +60x^3 -59x - 296)