By the polynomial remainder theorem, [tex]x+1[/tex] will be a factor of [tex]f(x)=x^4+ax^3+bx^2+cx+d[/tex] if the remainder upon division is 0, and this remainder is given by [tex]f(-1)[/tex]:
[tex]f(-1)=(-1)^4+a(-1)^3+b(-1)^2+c(-1)+d[/tex]
[tex]0=1-a+b-c+d[/tex]
[tex]a+c=1+b+d[/tex]
Since [tex]a,c\in\{1,\ldots,5\}[/tex], it follows that [tex]a+c\in\{2,\ldots,10\}[/tex]. But notice that if [tex]a+c=2[/tex], then we have
[tex]2=1+b+d\implies 1=b+d[/tex]
and since [tex]b,d\in\{1,\ldots,5\}[/tex], the equation above requires that either [tex]b=0[/tex] or [tex]d=0[/tex], which is impossible. So [tex]a+c\in\{3,\ldots,10\}[/tex].
So we have 8 cases to check:
(1) Notice that if [tex]a+c=10[/tex], we have [tex]b+d=9[/tex]. This is only possible for [tex](b,d)\in\{(4,5),(5,4)\}[/tex].
(2) If [tex]a+c=9[/tex], then [tex]b+d=8[/tex], and so we can have [tex](b,d)\in\{(3,5),(4,4),(5,3)\}[/tex].
(3) If [tex]a+c=8[/tex], then [tex]b+d=7[/tex], and so [tex](b,d)\in\{(2,5),(3,4),(4,3),(5,2)\}[/tex].
(4) If [tex]a+c=7[/tex], then [tex](b,d)\in\{(1,5),(2,4),(3,3),(4,2),(5,1)\}[/tex].
(5) If [tex]a+c=6[/tex], then [tex](b,d)\in\{(1,4),(2,3),(3,2),(4,1)\}[/tex].
(6) If [tex]a+c=5[/tex], then [tex](b,d)\in\{(1,3),(2,2),(3,1)\}[/tex].
(7) If [tex]a+c=4[/tex], then [tex](b,d)\in\{(1,2),(2,1)\}[/tex].
(8) If [tex]a+c=3[/tex], then [tex](b,d)\in\{(1,1)\}[/tex].
At the same time, we have 8 cases to consider to find how many options there are for [tex](a,c)[/tex].
(1) [tex]a+c=10[/tex]. We have only one choice of [tex](a,c)=(5,5)[/tex].
(2) [tex]a+c=9[/tex]. This is the same as when [tex]b+d=9[/tex], which we found to be 2 choices.
(3) Same as [tex]b+d=8[/tex]; 3 choices.
(4) Same as [tex]b+d=7[/tex]; 4 choices.
(5) 5.
(6) 4.
(7) 3.
(8) 2.
In total, there are
[tex]2\times1+3\times2+4\times3+5\times4+4\times5+3\times4+2\times3+1\times2[/tex]
[tex]=2(2\times1+3\times2+4\times3+5\times4)[/tex]
[tex]=2\displaystyle\sum_{n=1}^4n(n+1)[/tex]
[tex]=80[/tex]
ways to choose [tex]a,b,c,d[/tex] such that [tex]x+1[/tex] is a factor of [tex]x^4+ax^3+bx^2+cx+d[/tex], 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.
