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Q.
A cubic polynomial $P ( x )$ is such that $P (1)=1, P (2)=2, P (3)=3$ and $P (4)=5$, then $P (6)$ is equal to
Complex Numbers and Quadratic Equations
Solution:
Let $ P ( x )= A ( x -1)( x -2)( x -3)+ x$
Now, $P(4)=5$
$\Rightarrow 5= A (3)(2)(1)+4 $
$\Rightarrow A =\frac{1}{6}$
So, $P(x)=\frac{1}{6}(x-1)(x-2)(x-3)+x$
$\therefore P (6)=16$