Asked by betty
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. =)
Please, please help me out!! Thanks alot. =)
Answers
Answered by
Count Iblis
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.
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.
Answered by
betty
thank you soooo much!!! =)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.