Math, asked by sspp9100, 2 months ago

if ax^17+bx^16+1 is divisble by x^2-x-1,thwn find the integral value of a

Answers

Answered by Anonymous

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 ) }

Answered by gyanjyotidas74

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

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.