Given a polynomial p(x) If degree is A.P(1)=P(2)=P(3).....=P(A)=1. Also P(A+1)=B Tell P(A+C).
Given a polynomial p(x) If degree is A.P(1)=P(2)=P(3).....=P(A)=1. Also P(A+1)=B Tell P(A+C).
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We know that $$$deg P = A$$$. Let $$$W(x)$$$ be $$$W(x) = P(x)-1$$$. Since $$$P(1) = P(2) = ... = P(A) = 1$$$, then $$$W(1) = W(2) = ... = W(A) = 0$$$. Let's observe that $$$deg W = deg P$$$, which means that numbers $$$1, 2, ..., A$$$ are all roots of polynomial $$$W(x)$$$. Thus,
for some real number a. Now, we know that $P(A+1) = B$ , so $$$W(A+1) = B-1 = a(A+1-1)(A+1-2)...(A+1-A) = a \times A! \implies a = (B-1)/A!$$$. So we know $$$W(x)$$$ polynomial formula — and thus, the formula for $$$P(x)$$$ (it's enough to add one). Hope I haven't made any mistake :)