Question

Let $P(x)=x^4+a{x}^3+b{x}^2+cx+d$, where $a,b,c,d$ are real constants. If $P(1)=10,P(2)=20$ and $P(3)=30$ then the value of $\frac{1}{10}(P(12)+P(-8))$ is equal to

• A. 1984
• B. 2016
• C. 2000
• D. 2017

Answer 1

Answer: (A)

$P(x)-10x=0$ have roots $x=1,2,3$
$\Rightarrow P(x)=(x-1)(x-2)(x-3)(x-\alpha )+10x\text{ where }x=\alpha \text{ is fourth root }$
$\therefore P(12)=120+11\cdot 10\cdot 9(12-\alpha )$
$P(-8)=-80+9\cdot 10\cdot 11(8+\alpha )$
___________________________
$\therefore P(12)+P(-8)=40+9\cdot 10\cdot 11\cdot 20$
$=\frac{P(12)+P(-8)}{10}=4+1980=1984$