Quartic Polynomials

Steven, that is a nice question! But the answer is not $715$ . It's $495$ . Why?

A quartic polynomial is a polynomial of the form $ax^4+bx^3+cx^2+dx+e$ where $a$ does not equal $0$ . This is the very important condition you forgot.

I'm adding a short solution for this.

Let $P(x)$ be equal to $ax^4+bx^3+cx^2+dx+e$ where $a$ , $b$ , $c$ , $d$ and $e$ are non-negative integers and $a\neq 0$ .

We have $P(1)=9$

So, $a\cdot1^4+b\cdot1^3+c\cdot1^2+d\cdot1+e=9$

In other words $a+b+c+d+e=9$

So, the problem is reduced to finding the number of ordered sets of non-negative integers that satisfies the above equation and where $a\neq 0$ .

The easiest way to do this forget about the restraint on $a$ and find all ordered sets of non-negative integers where $a+b+c+d+e=9$ and then subtract the cases we don't want to include [the sets where $a=0$ ].

The first part can be calculated as ${9+5-1}\choose {5-1}$ $=715$ [the number you put as the answer].

And we don't want to include the cases where $a=0$ . So we're going to subtract ${9+4-1}\choose{4-1}$ $=220$ cases from our total.

So the correct answer is $715-220=495$ .