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Linear AlgebraMediumVector Spaces & Bases

Vector Spaces & Bases · Medium problem

Relative to the basis {1, x−2, (x−2)2}\{1,\ x-2,\ (x-2)^2\} of P2P_2, a polynomial pp has coordinates (3,−1,2)(3,-1,2), meaning p(x)=3⋅1−1⋅(x−2)+2⋅(x−2)2p(x) = 3\cdot 1 - 1\cdot(x-2) + 2\cdot(x-2)^2. Find p(5)p(5).