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Factorisation of a Polynomial

Let p(x) be a...

Question

Let p(x) be a polynomial of degree 4 such that p(k) =1/k for k=1,2,3,4,5 then 12 p(6) is equal to :

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Solution

$It is given that for the polynomial p (x) of degree 4 : p (k) = \frac{1}{k} for k = 1, 2, 3, 4, 5 Therefore for k = 1, 2, 3, 4, 5 k p (k) - 1 = 0 . Now consider a polynomial q (x) such that q (x) = x p (x) - 1 As, for x = 1, 2, 3, 4, 5; q (x) = 0 . Therefore by factor theorem q (x) = c (x - 1) (x - 2) (x - 3) (x - 4) (x - 5), where c is a constant . Now, q (0) = 0 p (0) - 1 \Rightarrow c (- 1) (- 2) (- 3) (- 4) (- 5) = - 1 \Rightarrow c = \frac{- 1}{{(- 1)}^{5} 5!} = \frac{1}{5!} Thus, we can write q (x) = \frac{1}{5!} (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) q (6) = \frac{1}{5!} (- 5) (- 4) (- 3) (- 2) (- 1) = \frac{{(- 1)}^{5} 5!}{5!} = - 1 As, p (x) = \frac{q (x) + 1}{x} Hence, 12 p (6) = 12 [\frac{q (6) + 1}{6}] = 2 [1 - 1] = 0$

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Factorisation of a Polynomial

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