Let $P\left(x\right)$  be a polynomial such that  $x.P(x-1)=(x-4)P\left(x\right)\forall x\in R.$
 $P(-1)=24$
Match each entry in List – I to the correct entries in List – II.

	List – I		List – II
P)	$P\left(4\right)=$	1)	0
Q)	$P\left(5\right)=$	2)	12
R)	$\frac{P(-2)}{2}=$	3)	60
S)	$P\left(3\right)=$	4)	120
		5)	24

The correct option is

 Let  $P\left(x\right)$ be a polynomial such that  $x.P(x-1)=(x-4)P\left(x\right)\forall x\in R.$
Find all such  $P\left(x\right).$
Put  $x=0,0-4P\left(0\right)$
 $\Rightarrow P\left(0\right)=0$
Put  $x=1,1.P\left(0\right)=-3P\left(1\right)$
 $\Rightarrow P\left(1\right)=0$
Put  $x=2,2.P\left(1\right)=-2P\left(2\right)$
 $\Rightarrow P\left(2\right)=0$
Put $x=3,3.P\left(2\right)=-P\left(3\right)$
 $\Rightarrow P\left(3\right)=0$
Let us assume  $P\left(x\right)=x(x-1)(x-2)(x-3)Q\left(x\right),$ where  $Q\left(x\right)$  is some polynomial.
Now using given relation we have
 $x(x-1)(x-2)(x-3)Q(x-1)=x(x-1)(x-2)(x-3)Q\left(x\right)$
 $\Rightarrow Q(x-1)=Q\left(x\right)\forall x\in R-\{0,1,2,3,4\}$
 $\Rightarrow Q(x-1)=Q\left(x\right)\forall x\in R\left(\text{From identity theorem}\right)$
 $\Rightarrow Q\left(x\right)\text{\hspace{0.17em}is periodic}$
 $\Rightarrow Q\left(x\right)=c$
 $\Rightarrow P\left(x\right)=cx(x-1)(x-2)(x-3)$