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

If x^2-x-1 divides ax^6+bx^5+1 evenly, find the sum of a and b. Show all steps and exlain your answer.

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

  • algebra 2 - , Tuesday, March 31, 2009 at 7:16pm

    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.

  • algebra 2 - , Tuesday, March 31, 2009 at 7:21pm

    thank you soooo much!!! =)

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