One satisfying factoring problem x^5+x^4+1

Long version: x⁵ + x⁴ +1 = x⁵+x⁴+x³+x²+x+1 – (x³+x²+x) = (x⁶ -1)/(x-1) – x(x²+x+1) = ((x⁶ -1) – x(x-1)(x²+x+1)) /(x-1) = ((x⁵+x⁴+x³+x²+x+1)/(x²+x+1) – x)(x²+x+1) (x⁵+x⁴+x³+x²+x+1)/(x²+x+1) = Ax³+Bx²+Cx+1 => 1= A, 1=A+B, 1=B+A+C, 1= B+C+1, 1= C+1 => B = 0, C= 0 => (x⁵+x⁴+x³+x²+x+1)= (x³+1)(x²+x+1) ✓ Therefore: x⁵ + x⁴ +1 = ((x³+1) – x)(x²+x+1)✓ Short version: x⁵ + x⁴ +1 = (x⁵+x⁴+x³)+(x²+x+1) – (x³+x²+x) = (x²+x+1)(x³+1 –x) So, x⁵ + x⁴ +1 = 0 if either: (x³+1) – x= 0 or x²+x+1 =0.

Nice!

Muito engenhoso!

ahh yes, another wild sighting of adding zero to an equation

x⁵+ x⁴+x³ -x³+1 x³(x²+x+1) -(x³-1) x³(x²+x+1) -(x-1)(x²+x+1) (x²+x+1)(x³-x+1)

Notice that 5, 4 and 0 (from 1=x⁰) form a complete residue class mod 3, so it's divisible by x²+x+1

I think you can also do euclidian division for polynomials

Just factor it in Z mod 3 😂

Why no use Horner scheme. Dead simple instead of adding and substracting.

Very clever

It's not an equation but a polynomial. For the rest: excellent work.

I just came.

Where did x^3, x^2, and x^1 come from? They came out of nowhere and aren’t in the original question

If you put omega (w) - the cube root of unity, it is zero. So 1 + x + x^2 is a factor. Do long division

What facturiel 1/2

The marker switch is so smooth

Can someone explain why we’re able to write x^5 + x^4 + 1 as x^5 + x^4 + x^3 + x^2 + x + 1? Aren’t they two different equations? Unless x^3 through x are just place holders, but wouldn’t we write them as 0x^3, etc?

Stupid beautiful 💪 ♥️ 👍

Very nice

Factor theorem my beloved