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March 4, 2015

March 4, 2015

Posted by **betty** on Tuesday, March 31, 2009 at 6:27pm.

Please, please help me out!! Thanks alot. =)

- algebra 2 -
**Count Iblis**, Tuesday, March 31, 2009 at 7:16pmYou 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.

- algebra 2 -
**betty**, Tuesday, March 31, 2009 at 7:21pmthank you soooo much!!! =)

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