Asked by KK
A polynomial ax^3+bx^2+cx+d leaves a remainder 10x+11 upon division by x^2-1 and also leaves a remainder -6 upon division by x^2+x+1. What are the values of a,b,c,d?
Answers
Answered by
Count Iblis
If you compute Modulo (x^2-1), then you have:
x^2 = 1
x^3 = x
So, we have:
ax^3+bx^2+cx+d =
(a + c) x + b + d
Equating this to 10x+11 for all x, gives:
a + c = 10
b + d = 11
Modulo (x^2+x+1), we have:
x^2 = -(x+1)
x^3 = -x(x+1) = -x^2 - x = 1
So, we have:
ax^3+bx^2+cx+d =
(c-b) x + a - b + d
Equating this to -6 for all x gives:
c-b = 0
a - b + d = -6
Solving the 4 equations for the 4 unknowns gives:
a = 1
b = 9
c = 9
d = 2
x^2 = 1
x^3 = x
So, we have:
ax^3+bx^2+cx+d =
(a + c) x + b + d
Equating this to 10x+11 for all x, gives:
a + c = 10
b + d = 11
Modulo (x^2+x+1), we have:
x^2 = -(x+1)
x^3 = -x(x+1) = -x^2 - x = 1
So, we have:
ax^3+bx^2+cx+d =
(c-b) x + a - b + d
Equating this to -6 for all x gives:
c-b = 0
a - b + d = -6
Solving the 4 equations for the 4 unknowns gives:
a = 1
b = 9
c = 9
d = 2
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