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. =)
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!!! =)
To find the sum of a and b, we need to find the polynomial expression that results when we divide ax^6 + bx^5 + 1 by x^2 - x - 1.
To divide polynomials, we can use long division or synthetic division. Let's use long division in this case:
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x^2 - x - 1 | ax^6 + bx^5 + 1
We start by dividing the highest degree term of the dividend (ax^6) by the highest degree term of the divisor (x^2). This gives us ax^4 as the first term of the quotient. We then multiply the entire divisor (x^2 - x - 1) by ax^4 to get ax^6 - ax^5 - ax^4 as the product:
ax^4
________________________
x^2 - x - 1 | ax^6 + bx^5 + 1
- (ax^6 - ax^5 - ax^4)
Next, we subtract this product from the dividend to get the new dividend:
ax^4
________________________
x^2 - x - 1 | ax^6 + bx^5 + 1
- (ax^6 - ax^5 - ax^4)
________________________
bx^5 + ax^4 + 1
Now, we repeat the process. We divide the highest degree term of the new dividend (bx^5) by the highest degree term of the divisor (x^2), which gives us bx^3 as the next term of the quotient. We multiply the entire divisor (x^2 - x - 1) by bx^3 to get bx^5 - b*x^4 - bx^3 as the product:
ax^4 + bx^3
________________________
x^2 - x - 1 | ax^6 + bx^5 + 1
- (ax^6 - ax^5 - ax^4)
________________________
bx^5 + ax^4 + 1
- (bx^5 - bx^4 - bx^3)
Continuing this process, we eventually get:
ax^4 + bx^3 + (a - b)x^2 - (a + b)x - (a + b)
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x^2 - x - 1 | ax^6 + bx^5 + 1
- (ax^6 - ax^5 - ax^4)
- (bx^5 - bx^4 - bx^3)
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(a - b)x^4 + (a - b - b)x^3 + (a - b - a + b)x^2 - (a - b + a + b)x - (a - b + a + b - 1)
Now, for the polynomial to be divisible by x^2 - x - 1, the remainder must be zero. This means that the coefficient of each term in the remainder polynomial must be zero:
(a - b)x^4 + (a - b - b)x^3 + (a - b - a + b)x^2 - (a - b + a + b)x - (a - b + a + b - 1) = 0
Simplifying...
(a - b)x^4 + (a - 2b)x^3 + (-b)x^2 - (2b)x - (2a - 2b - 1) = 0
To satisfy this equation, each coefficient must be zero:
a - b = 0 --> a = b
a - 2b = 0
-b = 0 --> b = 0
-2b = 0
From the equations above, we can see that b = 0.
Now, since a = b, a must also be 0.
Therefore, the sum of a and b is a + b = 0 + 0 = 0.