Question

Let $P(x)=x^4+a x^3+b x^2+c x+d$, 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)

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