Concept:Compare degrees and coefficients to determine the polynomial.Explanation:Let f(x) be a non-zero polynomial of degree n.Then f′(x) has degree n−1, so [f′(x)]2 has degree 2n−2.Since f(x)=[f′(x)]2, degrees must be equal: n=2n−2, hence n=2.So f(x)=ax2+bx+c, where a=0.Then f′(x)=2ax+b.Therefore [f′(x)]2=(2ax+b)2=4a2x2+4abx+b2.Comparing coefficients with f(x)=ax2+bx+c gives a=4a2, so a=41, and c=b2.Thus f(x)=41x2+bx+b2=(2x+b)2.Given f(2)=0, we have (22+b)2=0, so (1+b)2=0, giving b=−1.Hence f(x)=(2x−1)2.Now f(−2)=(2−2−1)2=(−1−1)2=(−2)2=4.Answer:f(−2)=4, which is Option C.