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

Question

Degree 5; zeros : -1 ; -i ; -2 + i

Answer 1

Since the polynomial has degree 5, there can't be more than 5 distinct roots.

Since f(x) has real coefficients, complex roots occur in conjugate pairs. So, since -i is a root, so is i. Since -2+i is a root, so is -2-i.

f(x) = (x+1)(x - (-i))(x - i)(x - (-2+i))(x - (-2-i))

= (x+1)(x+i)(x-i)[(x+2)-i][(x+2+i]

= (x+1)(x²+1)[(x+2)² -i²]

= (x+1)(x²+1)(x²+4x+5)

= (x³ + x² + x + 1)(x² + 4x + 5)

= x⁵ + 4x⁴ + 5x³ + x⁴ + 4x³ + 5x² + x³ + 4x² + 5x + x² + 4x + 5

= x⁵ + 5x⁴ + 10x³ + 10x² + 9x + 5