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Q. If $P ( x )$ is a polynomial of degree 5 with $P (1)= P (2)= P (3)= P (4)= P (5)=1$, then the value of $\frac{ P (7)-1}{ P (6)-1}$ is equal to

Complex Numbers and Quadratic Equations

Solution:

Clearly, $P(x)=a(x-1)(x-2)(x-3)(x-4)(x-5)+1$
$\therefore \frac{P(7)-1}{P(6)-1}=\frac{a \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{a \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=6$