Rational Roots Theorem

Rational Roots Theorem

In this video, I give you a cool theorem that helps us factor out polynomials, provided that they have a rational root, enjoy!

[sanjeevkumardhiman]

He is such a student-friendly teacher; I wish I had a teacher like him

[nourchalhoub]

Hello Professor, I am watching this in TEXAS, I am a graduate mathematics student and this theorem is helping me in my Advanced Linear Algebra course! My final examination is tomorrow, thank you for explaining this so well!!!

[eliyasne]

This theorem is surprisingly easy to prove:
Let there be that rational root described at the start of the video.
Therefore:
an(p/q)^n+a(n-1)(p/q)^(n-1)+ • • • +a0 = 0
=> an(p)^n+a(n-1)(p)^(n-1)q+ • • • +a0q^n = 0
Applying mod p destroys everything except:
a0q^n Ξ 0 mod p
Since p, q are relatively prime p||a0
On the other hand, Applying mod q destroys everything except:
anp^n Ξ 0 mod p
Since p, q are relatively prime q||an
[|||]

[NazarTimofeyev]

You may not be a medical doctor, but you ARE a lifesaver.

[robsbackyardastrophotograp]

I wish you were my math teacher. Your enthusiasm is contagious!

[MathemaTeach]

This is one of the topics i gave my students last week while we are on online teaching. 👍👍👍

[AnkitSharma-efmd]

Dr. Peyam, I wish I had a teacher like you . I chose mathematics because I want to be a teacher like you. It's an honor to watch you teaching us. Thank you so much for another powerful methodology. Live long sir...

[lightning]

Thank you so much, my online teachers dont teach us in a way i understand. Your a lifesaver

[bishakhasarmah]

Sir you are so quite that we fall in love with your smile

[nathanisbored]

i learned it as c/d rather than p/q. c stands for 'Constant term', d stands for 'leaDing coefficient'. that's how i remember it

[thomasborgsmidt]

Hmm.... can that theorem be extended?
I.e.

If we have a polynomial of degree n does it have a polynomial of degree m. where m<n that is divisible in the polynomial - even though there are no rational roots, so the polynomial can be split up into two (or more) polynomials? These splinters of the polynomial have all rational coefficients, but the polynomial as such has no rational roots.

I.e. Can we isolate f.i. x^2 + 1 = 0 which has no rational roots out of a larger polynomial (or polynomial of degree - say 27 - which also has no rational roots????

[Evan-nebu]

I really like this theorem! In Italy we call it Ruffini's theorem

[Apollorion]

-8, -4, -2, -1, 1, 2, 4, 8 <= Are that *16* different numbers, Dr Peyam?

[nahweh]

What a wonderful opportunity to bring in synthetic division. You could determine whether a potential root was indeed a root but also be left with the coefficients of the resulting quadratic in the event a root was found.

[harrywotton]

Interesting. I didn't know this theorem before watching. Is it a famous theorem? Anyway I'm lucky to know it today. Thanks!

[thomasborgsmidt]

Now that is a highly interesting theorem.
The general problem I am struggling with is finding the roots in a polynomial of degree n, where n is fairly large, say 100 in order to find the real roots of a polynomial.
If you quickly can suck out the rational roots, then the problem reduces to a polynomial of degree n-1.

If you then note, that the number of swiches between + and - is equal to the number of real roots in the polynomial. The task becomes somewhat easier - if you quickly can find the rational ones.

This should mean, that after long division the number of switches between + and - should also have been reduced by 1. This further simplifies the question.

I would like to see a proof of the Rational Roots Theorem.
Then I would like to see a proof of the Switch Theorem.

Because then the problem of finding roots - when you have found both the rational and irrational roots the rest must be irrational or complex - which we know occur in pairs. (I would like to see a proof of that too).

Now bankers claim that a property value is whatevet they want it to be - which might be true or not. The problem is here to establish under what conditions the claim is true.
If you find the rational and real (but not rational) roots then the imaginary roots can be discarded as gibberish. In effect you need to find a very limited number of root in a polynomial of a high degree.

[CliffStamp]

I have to say, I wish I had the enthusiasm for Math as you display teaching it.

[plaustrarius]

Im interested to write down the proof for this, it intuitive in my head but of course the detail is in the steps.

The only possible rational roots, after you plug in must cancel out with the constant.

With A equals 1 (leading coefficient), the coefficients being integers, and all natural number exponents, that limits your choices to only the factors (plus or minus) of the constant.

If A isn't 1, we divide by by the factors of A, heres where i feel like i would mess up in the proof but that makes intuitive sense.

Remainder theorem coming next? Or possible cardano's formula??

[andreapaps]

Thought of an easier way to remember if you arrange the polynomial with decending powers.... (p points right so it divides the constant term.... q points left so it divides the highest power of x term) :D

[paulkohl]

Are p and q in lowest form? Reminds me of Eisenstein's Irreducibility Criteria... Love your videos!