Find a polynomial function f with real coefficients that satisfies the given conditions. Degree 4; zeros 0 (multiplicity 2), 2-i; f(2) =48

Answer 1

Degree of 4: highest power of 4 and 4 zeros (both real and imaginary)

zeros: x = 0, x = 0 (multiplicity of 2), x = 2-i but since imaginary numbers come in pairs the 2nd zero is x = 2+i (conjugate of 2-i)

f(2) = 48

Start with the zeros and set each of the zeros equal to zero and multiply them

x=0, x=0, x = 2-i -> x-(2-i), x = 2+i -> x-(2+i)

a(x)(x)(x-(2-i))(x-(2+i))=y

y=ax²(x²-4x+5)

y=a(x⁴-4x³+5x²)

Find a by plugging in the point

48 = a((2)⁴-4(2)³+5(2)²)

48=a(4)

a = 12

Answer (rewrite as f(x)): f(x) = (12)(x⁴-4x³+5x²)

Answer 2