Form a fifthdegree polynomial function with real coefficient

Form a fifth-degree polynomial function with real coefficients such that 3i, 1-2i, and -1 are -1 are zeros and f(0) = 90. f(x) = (Simplify your answer. Type an expression using x as the variable.)

Solution

given 3i , 1-2i , -1 are rrots

and coeeficients are real

=>conjugates of complex roots i.e, -3i, 1+2i are also the roots

the equation of form

f(x)=k(x-3i)(x+3i)(x-(1-2i))(x-(1+2i))(x+1)

f(x)=k(x²-(3i)²)(x²+x(-(1-2i)-(1+2i))+ (1-2i)(1+2i))(x+1)

f(x)=k(x²+9)(x²-2x+5)(x+1)

f(x)=k(x⁵-x⁴+12x³-4x²+27x+45)

f(0)=90

k(0⁵-0⁴+12*0³-4*0²+27*0+45)=90

45k=90

k=2

f(x)=2(x⁵-x⁴+12x³-4x²+27x+45)

f(x)=2x⁵-2x⁴+24x³ -8x²+54x +90 is required polynomial