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loris [4]
3 years ago
6

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.
Mathematics
1 answer:
fgiga [73]3 years ago
4 0
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.
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