Question

Let a polynomial $P(x)$ when divided by $x-1,x-2,x-3$ leaves the remainder 4, 5, 6 respectively. If $P(x)$ is divided by $(x-1)(x-2)(x-3)$ and remainder is $R(x)$, then find the value of R(100).

Answer 1

Answer: 103

Let $P(x)=(x-1)(x-2)(x-3)Q(x)+R(x)$, where $R(x)=\left(a{x}^2+bx+c\right)$
Now,$P(1)=R(1)=4\Rightarrow a+b+c=4$
$P(2)=R(2)=5\Rightarrow 4a+2b+c=5$ ....(i)
$P(3)=R(3)=6\Rightarrow 9a+3b+c=6$....(ii)
$\therefore$ On solving, we get....(iii)
$a=0,b=1,c=3\text{. }$
Hence, $R(x)=x+3\Rightarrow R(100)=103$.