Factor x^7-1

[Anonymous No.10233031]

reading a book reviewing my algebra when i come in to contact with:

>Factor: x^7-1

I left as much info from the preceding question to give a hint as to what is going on, so the question wouldn't be too far out of context.

I'll probably just use a online tool to factor this and get the answer, but I'm very curious to know if /sci/ is able to do this equation easily.

personally, i can't. the most math i've completed is calc 1 atm,

[Anonymous No.10233034]

x-1 times something, work it from there

[Anonymous No.10233037]

>>10233031
by inspection, a-1 divides a^7-1.
Divide it out and keep going until you're done.

[Anonymous No.10233041]

>>10233034
very nice anon, I just used wolfram alpha and that's what is the factor

(x-1) (x^6+x^5+x^4+x^3+x^2+x^1+1)

much cleaner than i expected

i guess i'm a brainlet heheh

[Anonymous No.10233131]

the x spits out the x^7 to x
the -1 then cancels x^6 to x
leaving only the highest and the lowest

[Anonymous No.10233183]

>>10233031
>protip 1 = 1^7

Also, cs majors belong on >>>/g/

[Anonymous No.10233209]

>>10233031
you should have learned about the finite geometric series in calc 1

[Anonymous No.10233210]

>>10233209
it was right side up, i fucking double checked it before i posted to god dammit

[Anonymous No.10233257]

>>10233031
x-1 will obviously be a factor, since 1 is a root.

Recreate annihilated terms like so:
x^7 - x^6 +x^6-x^5+x^5-... Etc and factor out the x-1.
I'm actually also going through this book.

[Anonymous No.10233469]

>>10233031
1 x (x^7 - 1)

[Anonymous No.10233474]

>>10233210
You have to rotate the pixels, not just change the exif tag

[sage goes in all fields No.10233645]

Jeez OP, didn't you ever learn polynomial division? It's high school/freshman level stuff.

If you replace x=1, then you get 1^7-1=0. This means that 1 is a root of (x^7-1). Then, (x-1) divides (x^7-1): https://proofwiki.org/wiki/Polynomial_Factor_Theorem

You can then use either of those to calculate (x^7-1)/(x-1), giving a null remainder and thus a quotient that also divides the original polynomial:
http://mathworld.wolfram.com/RuffinisRule.html
https://en.wikipedia.org/wiki/Polynomial_long_division

[Anonymous No.10234174]

>>10233645
you don't need to do polynomial division, as the book he's going through doesn't cover it until the next section

[Anonymous No.10234185]

(a-1)(1+a+a^2+...+a^6)
how fucking hard is it to use the difference rule for polynomials or whatever it's called.

[Anonymous No.10234230]

>>10233031
In the complex plane, the roots of the polynomial will be 7 points distributed evenly (i.e. with identical angular separations) on a circle of radius 1. We know from inspection one root is x=1=exp(i*0). Thus, the nth root will be given as [math] x = e^{i* 2\pi n / 7} [/math] where n = 0,1,2,3,4,5,6. Factorizing thus becomes trivial since the polynomial can be written in the form of the products of (x-r_i), where r_i is the ith root

[Anonymous No.10234556]

>>10233031
1 is just 1^7

[Anonymous No.10235498]

>>10234185
Hey fellas look, I factored this polynomial real good!!

[Anonymous No.10235509]

[math]a^{3}^{4} + (-1^{3}^{4})[/math] then I think you can factor it just like the example above

[Anonymous No.10237276]

>>10233031
(a^3.5 - 1)(a^3.5 + 1)

[Anonymous No.10238655]

>>10237276
What is the picture?

[Anonymous No.10238657]

>>10238655
What?

[Anonymous No.10238665]

>>10238655

Ugandan knuckles newfag

[Anonymous No.10238687]

>>10237276
We must go further!
[math](a^{3.5}+1)(a^{1.75}+1)(a^{0.875}+1)(a^{0.4375}+1)(a^{0.21875}+1) ...[/math]

[Anonymous No.10238914]

why do [math](x-1)(x^6+x^5+x^4+x^3+x^2+x^1+1)[/math] when you can do the obvious [math](a^3.5 - 1)(a^3.5 + 1)[/math]?

[Anonymous No.10238917]

>>10238914
second one wrongly formatted, [math](a^(3.5) - 1)(a^(3.5) + 1)[/math]

[Anonymous No.10238943]

>>10238917
Try the squiggly brackets my dude. "^{stuff}"

[Anonymous No.10238980]

>>10238917
[math]
\left ( a^{3.5} \right -1)\left ( a^{3.5} \right +1)
[/math]
Optimized.

[Anonymous No.10238984]

>>10238917
[math]
\left ( a^{3.5} -1 \right ) \left ( a^{3.5} +1 \right )
[/math]
Optimized.

[Anonymous No.10239193]

>>10238984
[eqn](a^{3.5}-1)+(a^{1.75}-1)+(a^{1.75}+1)[/eqn]

[Anonymous No.10239203]

>>10239193
You added the factors, retard.

[Anonymous No.10240051]

>>10233041
Now generalize this result

[Anonymous No.10240695]

>>10233209
you literally don't even need that. you should have learned polynomial division in fucking middle school/high school algebra.

[Anonymous No.10240728]

>>10240695
then, a variant of the eisenstein criterion shows the other factor is irreducible.