Polynomials with length and degree 4

Let the polynomial be $ax^4 + bx^3 + cx^2 + dx + e$ , since the length is 4. Then

$|a| + |b| + |c| + |d| + |e| = 4$ , with integers $a,b,c,d,e$ , and $a \ne 0$

I will denote $ABCDE$ as the solution of $|a|, |b|, |c|, |d|, |e|$ where $A = |a|$ , $B, C, D, E$ are equals to $|b|, |c|, |d|, |e|$ in a certain order

For $A = 1$ , $13000, 12100, 11110$ are possible solution

• If $13000$ , there's a total possibilty of $2^2$ change in sign: $(1,3), (1,-3), (-1,3), (-1,-3)$ , with $A = \pm 1$ , so the total arrangement is $2^2 \cdot \frac {4!}{3!} = 16$

• If $12100$ , using the same analogy gives a total arrangement of $2^3 \cdot \frac {4!}{2!} = 96$

• If $11110$ , total arrangement is $2^4 \cdot \frac {4!}{3!} = 64$

For $A = 2$ , $22000, 21100$ are possible solution

• If $22000$ , then $2^2 \cdot \frac {4!}{3!} = 16$

• If $21100$ , then $2^3 \cdot \frac {4!}{2! 2!} = 48$

For $A = 3 \Rightarrow 31000$ , then $2^2 \cdot \frac {4!}{3!} = 16$

For $A = 4 \Rightarrow 40000$ , then $2^1 \cdot \frac {4!}{4!} = 2$

Add all of these up gives $\boxed{258}$

The general form of the polynomial of degree 4 is $ax^4 + bx^3 + cx^2 + dx + e$ . We have to find the number of ordered sets $\{ a, b, c, d, e \}$ s.t. $a \neq 0$ and $\|a\| + \|b\| + \|c\| + \|d\| + \|e\| = 4$ .

Lets first assume $a, b, c, d, e \geq 0$ . Then there are five following cases.

I. Coefficients are 1, 1, 1, 1, 0 in some order. The number of possible orderings of $\{1, 1, 1, 1, 0\}$ is $5! / 4! = 5$ . Considering that 0 cannot be in the the first, we are left with $\mathbf{4}$ orderings.

II. Coefficients are 2, 1, 1, 0, 0 in some order. The number of possible orderings of $\{2, 1, 1, 0, 0\}$ is $5! / ( 2! \cdot 2! ) = 30$ . Considering that 0 cannot be in the the first (and it occurs in the first place in $2/5$ cases out of 30), we are left with $\mathbf{18}$ orderings.

III. Coefficients are 2, 2, 0, 0, 0 in some order. The number of possible orderings of $\{2, 2, 0, 0, 0\}$ is $5! / ( 3! \cdot 2! ) = 10$ . Considering that 0 cannot be in the the first (and it occurs in the first place in $3/5$ cases out of 10), we are left with $\mathbf{4}$ orderings.

IV. Coefficients are 3, 1, 0, 0, 0 in some order. The number of possible orderings of $\{3, 1, 0, 0, 0\}$ is $5! / 3! = 20$ . Considering that 0 cannot be in the the first (and it occurs in the first place in $3/5$ cases out of 20), we are left with $\mathbf{8}$ orderings.

V. Coefficients are 4, 0, 0, 0, 0 in some order. The number of possible orderings of $\{4, 0, 0, 0, 0\}$ is $5! / 4! = 5$ . Considering that 0 cannot be in the the first (and it occurs in the first place in $4/5$ cases out of 5), we are left with $\mathbf{1}$ ordering.

Now we consider that non-zero coefficients can be either positive or negative. Multiplying the number of orderings in each case by 2^(number of non-zero coefficients in this case) and summing them all up, we get that the answer is $$4\cdot2^4 + 18\cdot2^3 + 4\cdot2^2 + 8\cdot2^2 + 1\cdot2^1 = \boxed{258}$$

One way to do this problem would be to separate the possible polynomials into cases regarding how many terms they have. That is, since the coefficients of the polynomial must be integers, the least absolute value each of the five terms could have (and still be considered a term) is $1$ .

Case 1: The polynomial has $4$ terms. All of the coefficients of the terms must have absolute value of $1$ . We must also choose $1$ out of $4$ terms to be $0$ (not $5$ terms because the $x^4$ coefficient can't be zero). Also, for each term, we can choose whether its coefficient is negative or positive. The number of possibilities for this case is therefore $2^4 \cdot 4 = 2^6$ .

Case 2: The polynomial has $3$ terms. Exactly one of the terms' coefficients must have absolute value $2$ , so we choose one of the $3$ terms to have coefficient absolute value of $2$ and $2$ (of $4$ ) terms to have coefficient $0$ . And also, as in the last case, we can choose for all $3$ terms to have positive or negative coefficient, so the number of possibilities for this case is $\binom{4}{2} \cdot 3 \cdot 2^3 = 2^4 \cdot 3^2$ .

Case 3: The polynomial has $2$ terms. This case yields two subcases: first, we can have absolute value of coefficients equal to $\{1, 3\}$ , and second, we have $\{2, 2\}$ . Collectively, this yields $3$ possibilities for the absolute values of the coefficients. So, following the reasoning of the previous cases, the number of possibilities in this case is equal to $\binom{4}{3} \cdot (2 + 1) \cdot 2^2 = 2^4 \cdot 3$ .

Case 4: The polynomial has $1$ term. It is not hard to see that this case only has $2$ solutions.

So, summing our cases, we see that the number of polynomials is equal to $2^6 + 2^4 \cdot 3^2 + 2^4 \cdot 3 + 2 = \boxed{258}$ .