Find the Number of Polynomials with Coefficients in {0-9} That Satisfy P(-1)=-9

[anemone]

Here is this week's POTW:

-----

Consider polynomials $P(x)$ of degree at most $3$, each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. How many such polynomials satisfy $P(-1) = -9$?

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Ackbach
2. kaliprasad
3. castor28
4. lfdahl

Solution from castor28:

Spoiler

If we write $P(x)=ax^3+bx^2+cx+d$, we must have $-P(-1)=a-b+c-d=9$. If $k=a-b$ and $m = c-d$, then, as $k,m\le9$ and $k+m=9$, we have $k,m\ge0$; this means that each of $k,m$ can take the $10$ values from $0$ to $9$.

A fixed value of $k=a-b$ can be achieved in $10-k$ ways, the corresponding value of $m=9-k$ can be achieved in $10-(9-k)=k+1$ ways, giving a total of $(k+1)(10-k)$ ways for each possible pair $(k,m)$. The number of polynomials is therefore:
$$
\sum_{k=0}^9{(k+1)(10-k)} = 1\times10+2\times9+\cdots+10\times1 = 220
$$
We can get a closed form by writing $n=k+1$ and evaluating the expression
$$S(N) =\sum_{n=1}^N{n(11-n)}$$
for $N=10$. Using the identities:
\begin{align*}
\sum_{n=1}^N{n} &= \frac{N(N+1)}{2}\\
\sum_{n=1}^N{n^2} &= \frac{N(N+1)(2N+1)}{6}
\end{align*}
we obtain:
$$
S(N)=\frac{-N^3+15N^2+16N}{3}
$$
and $S(10)=220$.