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 - QAmmunity.org

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

asked May 26, 2018 11.5k views

1 Answer

By the polynomial remainder theorem,
x+1

x+1

will be a factor of
f(x)=x^4+ax^3+bx^2+cx+d

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)

:

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

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

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

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

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

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

a+c=7

, then
$(b,d)\in\{(1,5),(2,4),(3,3),(4,2),(5,1)\}$ .

(5) If
a+c=6

a+c=6

, then
$(b,d)\in\{(1,4),(2,3),(3,2),(4,1)\}$ .

(6) If
a+c=5

a+c=5

, then
$(b,d)\in\{(1,3),(2,2),(3,1)\}$ .

(7) If
a+c=4

a+c=4

, then
$(b,d)\in\{(1,2),(2,1)\}$ .

(8) If
a+c=3

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)

(a,c)

.

(1)

a+c=10

. We have only one choice of
(a,c)=(5,5)

(a,c)=(5,5)

.

(2)

a+c=9

. This is the same as when
b+d=9

b+d=9

, which we found to be 2 choices.

(3) Same as
b+d=8

b+d=8

; 3 choices.

(4) Same as
b+d=7

b+d=7

; 4 choices.

(5) 5.

(6) 4.

(7) 3.

(8) 2.

In total, there are

2*1+3*2+4*3+5*4+4*5+3*4+2*3+1*2

2*1+3*2+4*3+5*4+4*5+3*4+2*3+1*2

=2(2*1+3*2+4*3+5*4)

$=2\displaystyle\sum_(n=1)^4n(n+1)$

=80

=80

ways to choose
a,b,c,d

a,b,c,d

such that

x+1

is a factor of
x^4+ax^3+bx^2+cx+d

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.

IanNorton answered May 30, 2018

6.3k points

Search QAmmunity.org