Factoring a Quintic | Challenging Algebra

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[nguyenminhquang]

Fun fact: Vietnamese 8th graders will found these types of algebra problems to be extremely easy as they study this for their exams.

Also here is the general method to solving this problem and similar problems of this type (which Vietnamese 8th graders learn in school):
x^5+x^4+1
(adding in the missing degree)

=(x^2+x+1)(x^3-x+1)

[cameronspalding]

@11:30 I was worried we wouldn’t be able to find a solution that involved radicals since it was a quintic polynomial

[timothybolshaw]

A nice clear teaching style, very appropriate to someone like myself who is not an idiot, but was never exposed much to this particular area of algebra.

[davidblauyoutube]

Another good video! Here's some food for thought:
Each factor of f(x) reduces to an equivalent factor modulo each prime p, so you might consider factoring f(x) mod p.
The first try gives f(x) = 1 (mod 2) which has no polynomial factors, so that's a bust.
However, f(x) = x^5 + x^4 +1 (mod 3) clearly has x=1 as a root since 1+1+1=3=0 (mod 3).
* Factoring, f(x)=(x-1)(x^4-x^3-x^2-x-1) (mod 3), and the latter factor also has x=1 as a root.
* Factoring again, f(x)=(x-1)^2 (x^3-x+1) (mod 3) and this is as far as we can go. Testing these factors, x-1 does not divide f(x) over the integers but x^3-x+1 does, with quotient x^2+x+1. A quick check (using the rational root theorem) shows that neither factor itself has a further linear factor, so we are done!

[gopalakrishnachalla]

Very nice and brief explanation. Very good.

[adandap]

I wrote the original expression as (x^5 +x^4+... +1) - (x^3+x^2+x) = (x^6-1)/(x-1) - x (x^3-1)/(x-1) Then X^6 - 1 = (x^3-1) (X^3+1) etc. I think that's a bit more intuitive - at least it is to me!

[wesleydeng]

(x^5+x^4+1)(x-1) = x^6-x^4+x-1 = (x^3+1)(x^3-1)-x(x^3-1) = (x^3-1)(x^3-x+1). So, x^5+x^4+1 = (x^2+x+1)(x^3-x+1). In this kind of questions, x^2+x+1 is always a factor.

[djvalentedochp]

I got the same result by adding and subtracting x^8, nice video!

[aa]

I solved this problem using the properties of cube roots of 1.

[cameronspalding]

@ 12:39 the polynomial is quintic, thus has odd degree, thus has at least one real root

[robyzr]

hello teacher . Maybe it's stupid my question but I d like to know how understand/realize when is not possible keep factoring. Just for knowing cause i thought was possible factoring even this part, but i couldn t ... (X^3-x+1)

[kinshuksinghania]

If I assume x³ = 1. Then it turns into x² + x + 1 = 0 which it is actually true when x³ = 1, so that's our solution.

[kienthanhle]

so in a nutshell, this problem can be generalized as x^a+x^b+x^c =(x^2+x+1).P(x)

[Alians]

Hello again, x^2+x+1. I just divided by it and got the answer :(

[leecherlarry]

Not too challenging for my compi hehe:
*x^5 + x^4 + 1 // Factor*

[shunsukeotsuki]

I searched up "Rick roll" and found this.

[shm]

इसका आंसर जिनको पता है कमेंट करे
👉 if x+1/x= 2 find x^99+1/x^99 -2 =? 👇


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