Factoring a septic polynomial. A challenge in algebra

This video is about factoring a polynomial. A challenge in algebra
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SIMILAR VIDEOS:

[kurtlichtenstein]

At 7:48 when you redistribute the x^4, you should get x^5 -x^4, not x^5-x.

[bart]

An intuitive approach to the factorization is that x^3-1 has 3 roots spread evenly on the unit circle in the complex plane. One is of course 1, so the other 2 are e^(2*π"i/3) and its conjugate e^(-2"π*i/3), which are the roots of x^2+x+1.
If you square these roots you get them again but swapped, so they are also roots of x^4+x^2+1 which is the same polynomial but with x replaced with x^2.

[dryplace]

Actually just try 10000101/111 in the calculator and it gives 90091.So, at once we can get the answer (x^2+x+1)(x^5-x^4+x^2-x+1). Solve it in 10 seconds.

[songguo]

Instead of x^4, you may choose to add and subtract x,

[JackPowerX]

If it's septic, shouldn't we sterilize it before factoring?

[KRYPTOS_K]

7:21 how could you identity that two trinomials in the right?

[BCS-IshtiyakAhmadKhan]

Just use the fact that ω, ω^2 are the roots of this equation so x^2+x+1 must be a factor of it.now solve by taking x^2+x+1 common.

[subhashkumarsinha]

Instead of adding and subtracting x2, we can try by adding and subtracting x and that will be more simpler. I'll try then confirm

[anastasissfyrides]

Claiming the "difference of 3 in the powers" trick will almost always do the job, is consequentially silently claiming that all of these weird polynomials will almost always have x²+x+1 as a factor, which i highly doubt

[MS-cjuw]

Of course I like the vedio you are marvelous

[robyzr]

Hello teacher, maybe it could be faster adding and subtract x. U find soon (x^2 +x +1) + (x ^7 - x), then u factoring last one in x (x^3 - 1)(x^3 +1)....etc until u find (x^2 +x +1)[x(x - 1)(x +1)(x^2 - x +1) +1]...

[djvalentedochp]

we can make a hard looking problem then: prove that if x^3 - x is greater than or equal to zero, then x^7 + x^2 + 1 is always greater than zero.

[ramaprasadghosh]

if one write omega as a complex cubic root of unity,
(x^7 + x^2 +1)= 0 at x= omega
This means x*x+x+1 is a divisor of x^7 + x^2 +1 .
Hereby
x^7 + x^2 +1 .
= (x^2+x+1)"x^5 - x^4*(x^2+x+1)
+ x^2*(x^2+x+1) - x* (x^2+x+1)
+ 1*(x^2+x+1)
= (x^2+x+1)(x^5 -x^4 + x^2 -x +1)

Wow! That's something new and cool! 👌

[cameronspalding]

If you assume that x=exp(2*i*pi/3), then x^2+x+1=0 and x^7=x,
You can use this to prove that x^7+x^2+1=0

[sssh]

just add x is fine too. x^7-x+x^2+x+1=x(x^6-1)+x^2+x+1, x^6-1 has one factor x^2+x+1

[coolmangame]

factor the quintic too lol i gotta find all solutions

[safwanmfarij]

You made a mistake by multiply x^4(x-1) =x^5-$^4

[razor-xnve]

Alternate approach:
X^7-x+x^2+x+1
X(X^6-1) +x^2+x+1
Difference of two squares, difference of two cubes .
Game over

[leecherlarry]

Not too hard for my compi hehe:
*x^7 + x^2 + 1 // Factor*