if ax^17+bx^16+1 is divisble by x^2-x-1,thwn find the integral value of a
Answers
If ax^17 + bx^16 + 1 is divisble by x^2 - x - 1 then there must be a quotient ( result).
let, the quotient be d.
so, the expression becomes,
(ax^17 + bx^16 + 1) / (x^2 - x - 1) = d ……………….. (equation no. 1)
Now, if you deduce the value of “a” from the equation no. 1
it will be,
a = {d(x^2 + x + 1) - bx^16 -1} / x^17 ………………….(equation no. 2)
Now, if you put the value of “d” from equation no. 1 to the equation no. 2
it becomes,
a = [{(ax^17 + bx^16 + 1) / (x^2 - x - 1) }( x^2 + x + 1) - bx^16 - 1] / x^17 …………………..(equation no. 3)
and if you solve the equation no. 3
You’ll get
a = - (bx^16 + 1) / { x^15 ( x + 1 ) }
Step-by-step explanation:
How do you find [a] such that ax^17+bx^16+1 is divisible by x^2-x-1?
I started thinking about long division but that seems tedious. How about the theorem that x−a divides polynomial f(x) precisely when f(a)=0
f(x)=ax17+bx16+1=x16(ax+b)+1
So we need to solve x2−x−1=0. That’s the Golden Ratio equation, isn’t it? The quadratic formula gives:
x=12(1±5–√)
Let’s call the roots r=12(1+5–√) and s=12(1−5–√).
From the roots or the quadratic equation we know r+s=1 , rs=−1
0=f(r)=r16(ar+b)+1
−r−16=ar+b
Similarly
−s−16=as+b
That’s a nice 2x2 linear system, in matrices:
(−r−16−s−16)=(rs11)(ab)
so solution
(ab)=1r−s(1−s−1r)(−r−16−s−16)
a=s−16−r−16r−s
b=sr−16−rs−16r−s
The question becomes can we calculate these? The denominator is easy,
r−s=5–√
It’s probably best just to generate the powers of r , otherwise b is going to be complicated.
r=−s−1=12(1+5–√)
r2=s−2=14(6+25–√)=12(3+5–√)
r4=14(14+65–√)=12(7+35–√)
r8=12(47+215–√)
r16=s−16=12(2207+9875–√)
r−16=1/r16=22207+9875–√⋅2207−9875–√2207−9875–√=12(2207−9875–√)
a=s−16−r−16r−s=12(2207+9875–√)−12(2207−9875–√)5–√=987
b=sr−16−rs−16r−s
sr−16=12(1−5–√)(12(2207−9875–√))=12(3571−15975–√)
rs−16=12(1+5–√)(12(2207+9875–√))=12(3571+15975–√)
b=12(3571−15975–√)−12(3571+15975–√)5–√=−1597
f(x)=987x17−1597x16+1
That’s the answer. Let’s check it by asking to divide,
(987x17−1597x16+1)/(x2−x−1)
Math works! ✓
Look at the Fibonacci numbers in the factor and the product; that’s probably a hint at a faster way to do it.
How do you find [a] such that ax^17+bx^16+1 is divisible by x^2-x-1?
. ……………………………………………………………………………………….
I had set the problem aside for sometime and when i got back to it and solved it, I found already Roberto had published the same solution. Not ready to give up I cooked up another solution.
ax17+bx16+1=0(1)
x2−x−1=0(2)
ax+b=−1x16
Substitutex=1y,Then
a+by=−y17(3)
y2+y−1=0(4)
y2=1−y
Squaring
y4=1–2y+y2=2–3y
y8=4–12y+9y2=13–21y
y16=169–546y+441y2=610–987y
So
a+by=−y(610–987y)=−610y+987y2
=−610y+987(1−y)=987–1597y
Comparing both sides
a=987
b=−1597
It’s hard to write mathematical expressions in quora editor.
If ax^17 + bx^16 + 1 is divisble by x^2 - x - 1 then there must be a quotient ( result).
let, the quotient be d.
so, the expression becomes,
(ax^17 + bx^16 + 1) / (x^2 - x - 1) = d ……………….. (equation no. 1)
Now, if you deduce the value of “a” from the equation no. 1
it will be,
a = {d(x^2 + x + 1) - bx^16 -1} / x^17 ………………….(equation no. 2)
Now, if you put the value of “d” from equation no. 1 to the equation no. 2
it becomes,
a = [{(ax^17 + bx^16 + 1) / (x^2 - x - 1) }( x^2 + x + 1) - bx^16 - 1] / x^17 …………………..(equation no. 3)
and if you solve the equation no. 3
You’ll get
a = - (bx^16 + 1) / { x^15 ( x + 1 ) }
and that’s the answer, you are looking for.