If $p(x)$ is a polynomial of degree 3 such that $p(i) = {1\over1+i}$ for all $a=\{1,2,3,4\}$. Then find $p(5)$.

If $p(x)$ is a polynomial of degree $3$ such that $p(i)$ = $\frac{1}{1+i}$ for all $a=\{1,2,3,4\}$. Then find $p(5)$.

My attempt : (1) First obviously I thought of solving the four equations which can be generated by assuming polynomial as $ax^{3}+bx^{2}+cx+d$. But that would be a very lengthy solution.

(2) In other attempt, I let

$g(x)=p(x) - \frac1{1+x}$ believing that some calculus concepts might be applicable since $g(x)=0$ at $1,2,3,4$ but I couldn't think of anything that works out.

Also, we "cannot" make $g(x) =k(x-1)(x-2)(x-3)(x-4)$ since we can only write that for polynomials since we can comment on their maximum number of real roots by looking at the degree.

That rules out another possible method.

So, how can this question be solved?

• – Shaun Jun 19 '17 at 13:45
• You've been here long enough to use MathJax. – Shaun Jun 19 '17 at 13:45
• Why is (1) so bad? That's just 4 unknowns, so a $4x4$ matrix with integer entries that needs to be reduced, that doesn't look that hard to me. Of course there might be a really deep, nice concept here to make it easier, but such a concept has to be found and proven first, which will most likely take more time than just row elimination on a $4 x 4$ matrix. – Dirk Jun 19 '17 at 13:49
• Consider $h(x) = (x+1)p(x) - 1$ rather than $g$. – Daniel Fischer Jun 19 '17 at 13:52
• I'm not quite sure whether to call this a duplicate of this question. – Daniel Fischer Jun 19 '17 at 13:57

I am writing this answer based on the hints provided by Daniel Fischer( in the comments section) so that others can also benefit from this answer.

Consider a polynomial function $h(x)=p(x)(x+1)-1$

$x=1,2,3,4$ are clearly the roots of this polynomial therefore $h(x)$ can also be written as

$h(x)=k(x-1)(x-2)(x-3)(x-4)$

So, $h(x)=p(x)(x+1)-1=k(x-1)(x-2)(x-3)(x-4)$----(1)

To find the value of k, we put $x=-1$ in (1)

( why? Because it would make p(x) vanish, so we wouldn't need more values of p(x) to find k)

Putting $x=-1$, $0-1=k(-2)(-3)(-4)(-5)$

This gives $k=\frac{-1}{120}$

Now to find $p(5)$,

we put $x=5$ in eq(1), this gives the value of $p(5)=\frac{2}{15}$