Problematic Polynomial

Calculus Level 3

Let $P(x)$ be a polynomial of degree $n$ . If

$P(x) - P'(x) =x^{n}$ , then what is $P(0)$ ?

Note: $P'(x)$ denotes the first derivative of $P(x)$ .

n^{n}

2n+1

n

n^{2}

0

n!

Cannot be determined

1 solution

Dhanvanth Balakrishnan
Nov 1, 2016

$P(x) - P'(x) = x^{n}$

On differentiating the equation w.r.t $x$

$P'(x) - P''(x) = nx^{n-1}$

On differentiating again,

$P''(x) - P'''(x) = n(n-1)x^{n-2}$

So , we continue differentiating $n$ times and we get,

$n^{th} derivative$ $of$ $x$ $-0$ $=$ $n(n-1)(n-2)...(1)$

So on adding all the obtained equations, we get,

$P(x) = x^{n} + nx^{n-1} + n(n-1)x^{n-2} + ... + n(n-1)(n-2)...(1)$

$\boxed{P(0) = n!}$