Question

Let $P\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$ be a polynomial such that $P\left(1\right)=1,P\left(2\right)=8,P\left(3\right)=27,P\left(4\right)=64$, then find the remainder when $P\left(5\right)$ is divided by $5$.

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

$P\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$
Given $P\left(1\right)=1$
$\Rightarrow a+b+c+d=0$ .....(1)
Also, given $P\left(2\right)=8$
$\Rightarrow 8a+4b+2c+d=-8$ .....(2)
Also, given $P\left(3\right)=27$
$\Rightarrow 27a+9b+3c+d=-54$ ....(3)
Also given $P\left(4\right)=64$
$\Rightarrow 64a+16b+4c+d=-192$ .....(4)
Solving these equations, we get
$a=-9,b=35,c=-50,d=24$
So, $P\left(x\right)=x^4-9x^3+35x^2-50x+24$
Now, $P\left(5\right)=625-1125+875-250+24=149$

Dividing $P\left(5\right)$ by $5$ will give the remainder as $4$.