Find a polynomial function​ f(x) of degree 3 with real coefficients that satisfies the following conditions

Question

Zero of 0 and zero of 4 having multiplicity 2; f(5)=30

Answer 1

(x-0)(x-4)^2=

x(x^2-8x+16)=0

x=0, 4, 4 as the 3 zeroes

f(5)=5

only way to get f(5) = 30 is change the zeroes, by adding a constant term

x^3 -8x^2 + 16x =-25

x^3 -8x^2 +16x +25=0

f(5) = 125 - 200 + 80 +25 = 30

but the zeroes are no longer 0, 4 and 4

f(-1) =0 one zero is now -1

the cubic polynomial factors into

(x+1)(x^2-9x+25)

x=-1 and

x=-18/2 + or - (1/2)sqr(81-100) =-9 + or - (i/2)sqr19

the other 2 zeroes are now imaginary