Let $P (x) = x^{6} + a x^{5} + b x^{4} + c x^{3} + d x^{2} + e x + f$ be a polynomial such that $P (1) = 1; P (2) = 2; P (3) = 3; P (4) = 4; P (5) = 5$ and $P (6) = 6$ , then find the value of $P (7)$ .

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Solution

Since
f(x)−x is a polynomial of degree 6 and
has 6 roots $1, 2, 3, 4, 5, 6$ by condition,
we can factorize f(x)−x as:
$f (x) - x = C (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) .$
Plug in x=0 in the above expression,
we have $3 - 0 = C \times 6!$ , hence C=36!.
Therefore,
$f (7) = 7 + (f (7) - 7) = 7 + 36! (7 - 1) (7 - 2) (7 - 3) (7 - 4) (7 - 5) (7 - 6) = 7 + 36! \times 6!$
= $10.$

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