polynomial function

Question

Find a polynomial function of degree 3 with real coefficients that satisfies the given conditions.

Zero of

−3

having multiplicity​ 3;

​f(1​)=16.

Answer 1

f(x) = (x+3)^3

f(1) = 4^3 =64 but you want 16, so multiply by 16/64 or 1/4

f(x) =(1/4)(x+3)^3

f(x) = (1/4)(x^3 +9x^2 + 27x + 27)

f(x) = x^3/4 +9x^2/4 + 27x/4 +27/4

f(1) = 1/4 + 9/4 + 27/4 + 27/4 = (1+9+27+27)/4 =64/4 = 16