Posted by betty on Tuesday, March 31, 2009 at 6:27pm.
You first compute the remainder of the division of ax^6+bx^5+1 by x^2-x-1. This can be conveniently done by doing computations Modulo x^2-x-1.
Modulo x^2 - x - 1 we have:
x^2 = x + 1 (1)
Squaring both sides gives:
x^4 = x^2 + 2 x + 1
Applying (1) on the right hand side gives:
x^4 = 3 x + 2
Multiplying both sides by x gives:
x^5 = 3 x^2 + 2 x
Applying (1) to the right hand side gives:
x^5 = 5 x + 3 (2)
Multiplying by x gives:
x^6 = 5 x^2 + 3 x
Applying (1) gives:
x^6 = 8 x + 5 (3)
From (2) and (3) it follows that:
ax^6+bx^5+1 = (8a+5b) x + 5a + 3b + 1
Since the remainder is zero, this means that
(8a+5b) x + 5a + 3b + 1 = 0
for all x.
You can equate the coefficient of x and the term 5a + 3b + 1 to zero to obtain two equations and solve those for a and b.
thank you soooo much!!! =)
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