Find the value of P(12)
Solution: Let f(x)=(x+1)P(x)−1, clearly f(x) is a polynomial of degree 12.
Now for x∈{0,1,2,…,11}, f(x)=(x+1)P(x)−1=x+1x+1−1=1−1=0
This shows that f(x) vanishes at the points x=0,1,2,…,11.
f(x) being of degree 12, the above statement assures that x=0,1,2,…,11. are the possible roots of f(x)
Therefore f(x)=(x+1)P(x)−1=a0(x−0)(x−1)(x−2)…(x−11)
Letting x=−1 in the above equality, we have −1=a0(−1)(−2)(−3)…(−12)⇒a0=−112!
Therefore f(x)=(x+1)P(x)−1=−112!(x−0)(x−1)(x−2)…(x−11)
Now, letting x=12 we have 13P(12)−1=−112!(12)(11)(10)…(1)=−12!12!=−1
⇒13P(12)−1=−1⇒P(12)=0
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