Evaluating a Quintic Polynomial from a Quadratic

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[kaarthikananthanarayanan]

There's an easier way to present the second method. You can perform a polynomial division and say x^5 -5x = (x^2 - x -1)(x^3 + x^2 +2x +3) + 3(the remainder!). Now the first term is zero as x^2-x-1= 0 is given. So remainder=3 is the answer!

[GirishManjunathMusic]

Given:
x² = x + 1
To find:
x⁵ – 5x

x² = x + 1
Multiplying both sides by x (as x ≠ 0):
x³ = x² + x = 2x + 1

Multiplying both sides by x²:
x⁵ = (2x + 1)(x²)
= (2x + 1)(x + 1)
= 2x² + 3x + 1
= 2(x + 1) + 3x + 1
= 2x + 2 + 3x + 1
= 5x + 3

Now,
x⁵ – 5x = 5x + 3 – 5x
= 3.

[kurtlichtenstein]

Simple rule for powers of phi:
Phi^n=F(n)times phi plus F(n-1), it works for all positive and negative integers of n. F(x) is the Fibbonaci sequence, F(x)=F(x-1)+F(x-2), where F(0)=0 and F(1)=1

[sacoolman]

Im glad I dont have to do these eq anymore.Well done to the solution solver...

[jmart]

This time I solved it using both of the two methods. Very nice problem.
Thank you and take care.

[riccardofroz]

x^2=1+x
square and multiply by x both sides
x^5=x+2x^2+x^3
replace x^2 with 1+x and x^3 with x+x^2
x^5=x+2+2x+x+x^2
replace x^2 with 1+x resulting in:
x^5=5x+3
therefore
x^5-5x=3

Really fun problem

[stevenlitvintchouk]

I had a "golden" opportunity to solve this problem easily. The first equation is the equation of the golden ratio, so x = phi. A property of the golden ratio phi, is that for some power n, phi^n = phi^(n-1) + phi^(n-2). So let's keep applying these recurrences: x^5 = x^4 + x^3 = 2*x^3 + x^2 = 3*x^2 + 2*x. The first equation tells us that x^2 = x+1. Plug that in and we get x^5 = 5*x + 3. Therefore, x^5 - 5*x = 3.

[tbg-brawlstars]

This is the first time on your channel that I have no Idea where to start solving this....

[tapassikder]

Last method was just a magical one. Thanks

[seegeeaye]

Square the given equation and then multiply x to both sides, we have x^5 – 2x^4 + x^3 – x = 0, then x^5 – 5x -2x^2(x^2-x) - x^3 + 4x = 0, then x^5 – 5x – x(x^2 – x) – 3x^2 + 4x = 0. then x^5 – 5x – 3(x^2 – x) = 0, then we get the result 3.

[MichaelRothwell]

Nice video & solutions as always.

As others have pointed out, in the 2nd method we can use the formula φⁿ = Fₙφ+Fₙ₋₁ to simplify φ⁵ (where Fₙ is the nth term of the Fibonacci sequence, starting F₀=0, F₁=1; the formula is easily proved by induction).
So φ⁵ = F₅φ+F₄= 5φ+3, φ⁵-5φ=3.

Note that the formula for φⁿ, and hence this last equation, both remain true if we replace φ by the other root of x²-x-1=0, namely -φ⁻¹, since in the proof by induction, we only use that φ satisfies that quadratic equation.

Regarding the 1st method, note that φⁿ can be written in radical form as
φⁿ = (Fₙ₊₁+Fₙ₋₁ + Fₙ√5)/2 (easy to check from the above formula for φⁿ by substituting φ=(1+√5)/2) and using Fₙ₊₁=Fₙ₋₁ + Fₙ).
So φ⁵ = (8+3 + 5√5)/2=(11+5√5)/2, which nicely explains where the coefficients come from.

The corresponding result for the other root of x²-x-1=0, namely -φ⁻¹, is (-φ⁻¹)⁵ = (8+3 - 5√5)/2=(11-5√5)/2, and can be justified by noting that the radical expressions in calculations involving this other root are the conjugate surds of those for φ (i.e. replace √5 by -√5).

As variant to the 1st method, the binomial theorem could, of course, have been used to evaluate [(1+√5)/2]⁵.

[adandap]

Thus got me wondering... if given the quintic x^5 - 5x - 3 =0, is it possible to somehow reverse this to get the roots phi, -1/phi?

[liberalaccidental]

You can easily prove that if x^2 = x + 1, then x^n = f(n)*x + f(n-1) where f(n) is the nth Fibonacci number, given by f(0) = 0, f(1)=1, and f(n) = f(n-1)+f(n-2) for n>=2. Since f(5)=5 and f(5-1)=3, you get the result

[angelmendez-rivera]

If x^2 – x – 1 = 0, then x^n = F(n)·x + F(n – 1), where F(n) is the nth Fibonacci number. Hence, x^5 = F(5)·x + F(4), so x^5 – 5·x = (F(5) – 5)·x + F(4). Notice that F(5) = 5, so x^5 – 5·x = F(4). F(4) = 3.

[damirdukic]

It's not necessary to evaluate x^4. Just calculate x^5 as x^3 * x^2, i.e. (2x + 1)(x + 1).

[christianthomas]

let us remember that phi is the solution of the golden equation and that (phi)^2 = Phi + 1
Let us go ahead and substitute x by phi in te equation to resolve x^5 - 5x;
we have: phi^5 - 5phi = phi(phi^4-5) = phi((phi+1)^2 - 5) = phi(phi^2 + 2phi + 1- 5) = phi(phi + 1 + 2phi + 1 - 5) = phi( 3phi -3) = 3phi^2 - 3phi = 3(phi + 1) - 3phi, which gives us by identification x = 3 for the solution.

[scottleung]

Took me 3 tries to get 3 (using the polynomial method) - oh well, like they always say, third time's the charm!

[lioneldarles]

SyberMath, c'est le Dieu des Maths...

[Dmh]

Your voice isn’t fine, i think you are sick. So i hope you fast healing

[ricardoquaresma]

Can you add Turkish subtitles to your videos, Turks are watching you too