If $f(x)=x^4+ax^3+bx^2+cx+d$ is a polynomial such that $f(1)=10, f(2)=20, f(3)=30$, find the value of $\frac{f(12)+f(-8)}{10}$

Question

If $f(x)=x^4+ax^3+bx^2+cx+d$ is a polynomial such that $f(1)=10, f(2)=20, f(3)=30$, find the value of :- $\frac{f(12)+f(-8)}{10}$

I tried to find the values of $a,b,c,d$ individually (which was a very long process).

Here's what I got

$a=\frac{-59}{12}$, $b=-54$.

The equation got too complicated before I could find the values of $c$ and $d$.

The answer is 1984.

This question is printed in an authorized book of Pearson Publication and was also asked in CMO 1984.

Answer 1

Since $f(x)-10x$ have $1,2,3$ as its roots and have leading coefficient $1$, $f(x) = 10x + (x-1)(x-2)(x-3)(x-t)$.

$$ \begin{aligned} f(12) + f(-8) &= 10(12-8) + 11\cdot 10 \cdot 9\cdot(12-t) + (-11)(-10)(-9)(-8-t)\\ &= 40+11\cdot10\cdot9\cdot 20. \end{aligned} $$

Now you should be able to finish.

Answer 2

You have to solve the system $$a+b+c+d+1=10$$ $$8a+4b+2c+d+16=20$$ $$27a+9b+3c+d+81=30$$