Question

Let a, b, c, d be four integers (not necessarily distinct) in the set {1, 2, 3, 4, 5}. The number of polynomials x4 + ax3 + bx2 + cx + d which is divisible by x + 1 is(
a. between 55 and 65. (
b. between 66 and 85.(
c. between 86 and 105. (
d. more than 105.

Answer 1

By the polynomial remainder theorem, $x+1$ will be a factor of $f(x)=x^4+ax^3+bx^2+cx+d$ if the remainder upon division is 0, and this remainder is given by $f(-1)$:

$f(-1)=(-1)^4+a(-1)^3+b(-1)^2+c(-1)+d$
$0=1-a+b-c+d$
$a+c=1+b+d$

Since $a,c\in\{1,\ldots,5\}$, it follows that $a+c\in\{2,\ldots,10\}$. But notice that if $a+c=2$, then we have

$2=1+b+d\implies 1=b+d$

and since $b,d\in\{1,\ldots,5\}$, the equation above requires that either $b=0$ or $d=0$, which is impossible. So $a+c\in\{3,\ldots,10\}$.

So we have 8 cases to check:

(1) Notice that if $a+c=10$, we have $b+d=9$. This is only possible for $(b,d)\in\{(4,5),(5,4)\}$.

(2) If $a+c=9$, then $b+d=8$, and so we can have $(b,d)\in\{(3,5),(4,4),(5,3)\}$.

(3) If $a+c=8$, then $b+d=7$, and so $(b,d)\in\{(2,5),(3,4),(4,3),(5,2)\}$.

(4) If $a+c=7$, then $(b,d)\in\{(1,5),(2,4),(3,3),(4,2),(5,1)\}$.

(5) If $a+c=6$, then $(b,d)\in\{(1,4),(2,3),(3,2),(4,1)\}$.

(6) If $a+c=5$, then $(b,d)\in\{(1,3),(2,2),(3,1)\}$.

(7) If $a+c=4$, then $(b,d)\in\{(1,2),(2,1)\}$.

(8) If $a+c=3$, then $(b,d)\in\{(1,1)\}$.

At the same time, we have 8 cases to consider to find how many options there are for $(a,c)$.

(1) $a+c=10$. We have only one choice of $(a,c)=(5,5)$.

(2) $a+c=9$. This is the same as when $b+d=9$, which we found to be 2 choices.

(3) Same as $b+d=8$; 3 choices.

(4) Same as $b+d=7$; 4 choices.

(5) 5.

(6) 4.

(7) 3.

(8) 2.

In total, there are

$2\times1+3\times2+4\times3+5\times4+4\times5+3\times4+2\times3+1\times2$
$=2(2\times1+3\times2+4\times3+5\times4)$
$=2\displaystyle\sum_{n=1}^4n(n+1)$
$=80$

ways to choose $a,b,c,d$ such that $x+1$ is a factor of $x^4+ax^3+bx^2+cx+d$, so the answer is B.

Note the symmetry of the sum above. You can easily give a slightly briefer combinatorial argument for this answer, but I figured a more brute-force approach would be easier to follow.