value of $f(2008)$ in $4$ th degree polynomial

Question

If $f(x)$ is a $4$ th degree polynomual such that

$f(2003)=24, f(2004)=-6, f(2005)=4,f(2006)=-6,f(2007)=24$

Then value of $f(2008)$ is

what i try

assuming $f(x)=ax^4+bx^3+cx^2+dx+e\cdots \cdots (1)$.

putting $x=2003,2004,2005,2006,2007$ in equation $(1)$ and solving for $a,b,c,d,e$

How to solve it using some easy way help me please

Answer 1

You can get rid of the big numbers by defining $y=x-2005$. I chose $2005$ because of the symmetry in the values. We now have $g(\pm 2)=24,g(\pm 1)=-6,g(0)=4$

We can see that $g$ is even and the constant term is $4$, so we let $g(y)=ay^4+by^2+4$. Only two equations in two unknowns, then $f(2008)=g(3)$

Answer 2

Just use that taking differences of successive terms lowers the degree of your polynomial function, so that after four applications you get a constant sequence. Then work back up from there.

$$ \matrix{24& &-6& &4 & &-6& &24 & & \color{red}{274}\\ & -30 & & 10& &-10 & & 30 & & \color{red}{250}\\ & & 40 & & -20 & & 40 & &\color{red}{220}\\ &&& -60& & 60 & & \color{red}{180}\\ & & & & 120 & & \color{red}{120} } $$ So $f(2008)=274$ is your answer.