Actually I tried both methods, too. But we can also factorize (1+x⁴) = (x² + √2x +1)(x² - √2x + 1) and of course (1+x⁸) too. Latter left as an exercise to the viewer 😁)
@@Straight_Talk Interesting (but wrong) answer as I already showed in my post how to factorize (x⁴ + 1). Even if we avoid complex numbers we can still split 1 + x + ... + x^15 into one liniear factor (x+1) and seven quadratic factors with three of them being (x² + 1) and the two I posted abpve. The other four can be derived from factorizing (x⁸ + 1).
You are a genius! This one had me bamboozled! It was intimidating but I had not the knowledge to do it! Thanks for doing such an amazing factoring job 🙏🙏🙏🙏🙏👍🏼❤️😎 problem Can you factor ? 1+x+x²+x³+x⁴+x⁵+x⁶+x⁷+x⁸+x⁹+x¹⁰+x¹¹+x¹²+x¹³+x¹⁴ +x¹⁵ The sum of the evens is 8 and the sum of the odds is 8. Therefore x= -1 is a root and (x+1) a factor. 1+x+x²+x³+x⁴+x⁵+x⁶+x⁷+x⁸+x⁹+x¹⁰+x¹¹+x¹²+x¹³+x¹⁴ +x¹⁵ = (x+1)(1+x²+x⁴+x⁶+x⁸+x¹⁰+x¹²+x¹⁴) = (x+1)(1+x²+x⁴(1+x²)+x⁸(1+x²)+x¹² (1+x²)) = (x+1)(1+x²)(1+x⁴+x⁸+x¹²) = (x+1)(1+x²)(1+x⁴+x⁸ (1+ x⁴)) = (x+1)(1+x²)(1+x⁴)(1+x⁸) These factors found by factoring. answer (x+1)(1+x²)(1+x⁴)(1+x⁸)
Why not group first half and second half then factor out x^8 from second half and repeat the setps? My second method was to multiply bu x-1 and observe we get x^16-1=0 this is the 16th roots of unity but we introduced x=1 as an extra solution. So to factor into quadratic polynomials we just have to find sin and cos of 45/2 degrees and we are set but will take a bit of work to sort everything into irreducible quadratic polynomials x=45/2 degrees From half angle formula for sin √2/2=2sin(x)cos(x) From half angle formula for cos √2/2=cos^2(x)-sin^2(x) √2/2=2cos^2(x)-1 cos(x)=√(2+√2)/4 then sin(x)=√2/√(2(√2)=1/2*√2/2*√(2-√2)/√(4-2)=√(2-√2)/2 now we will have complex conjugated roots +-45/2 degrees √(2+√2)/2+√(2-√2)i and √(2+√2)/2-√(2-√2)i And √2/2+√2/2i and √2/2-√2/2i for +-45 degrees Also √(2-√2)/2+√(2+√2)/2i and √(2+√2)/2-√(2-√2)/2i for +-135/2 degrees + I and - i for +-90 degrees And so on With the conjugate roots we form the polynomials x^2-Sx+P and this is it. I feel the problem was left half finished the way you did it. After all Euler proved all polynomials can be decomposed as products of quadratic polynomials. We cannot decompose more only of we set the conditions of having integer or rational coefficients
That was fun. I didn't have any paper, so I had to do this mentally. My method was to first recognize the form, then try a smaller example, then see the pattern, then verify the pattern. First the form. I knew it would be something like (x^n+1) times other expressions of the same form with different n. For the small example, I tried n values of 1 and 2 and got a result of x^3+x^2+x+1. Seeing the pattern, I realized that keeping every coefficient as 1 meant that every sum of combinations of the different n values had to be distinct and also cover every integer from 0 to 15. And hey, that's binary. 15 is 2^4-1, so this works perfectly with n values of 8, 4, 2, and 1. It also generalizes to any expression like this where the largest exponent is a power of 2, minus 1. So for x^31+...+x+1 you only need to multiply by another x^16+1.
Okay, I finished watching, and my method was definitely more hand-wavey than either of the ones shown. I could call my method "solving by constraints". That's how I narrowed down the form the answer would need to take, and it's also how I determined what property the exponents needed to have. I was simply lucky that these constraints were enough to narrow the possible solution space down to a single easy-to-check possibility.
Just think of what happens if x = 2. You end up with a binary number: 111...1. Add 1 to that number and you end up with 1000...00 = 2^16 - 1. So right there, we know that the binary number = (x^16 - 1) / (x - 1). Then of course, (x^16 - 1) = (x^8 + 1) * (x^8 - 1) and so on.
Both methods are equally elegant!
Thanks for sharing.
Both methods are beautiful and cunning!!!!!
The second method was awesome!
This is yet another Highly Satisfying Problem 😊
Yay, thank you!
This polynomial is the result of factoring x-1 from x^16-1.
(1+x)(1+x^2+x^4+x^6+x^8+x^10+x^12+x^14);
(1+x)(1+x^2)(1+x^4+x^8+x^12);
(1+x)(1+x^2)(1+x^4)(1+x^8).
Actually I tried both methods, too. But we can also factorize (1+x⁴) = (x² + √2x +1)(x² - √2x + 1) and of course (1+x⁸) too. Latter left as an exercise to the viewer 😁)
That's good fun 😊
1 + x^4 can’t be factorised.
@@Straight_Talk Sophie Germain's formula .
@@Straight_Talk Interesting (but wrong) answer as I already showed in my post how to factorize (x⁴ + 1). Even if we avoid complex numbers we can still split 1 + x + ... + x^15 into one liniear factor (x+1) and seven quadratic factors with three of them being (x² + 1) and the two I posted abpve. The other four can be derived from factorizing (x⁸ + 1).
@@Straight_Talk Did you mean √2 x? There was no restriction given to use integer coefficients only.
You are a genius! This one had me bamboozled! It was intimidating but I had not the knowledge to do it! Thanks for doing such an amazing factoring job 🙏🙏🙏🙏🙏👍🏼❤️😎
problem
Can you factor ?
1+x+x²+x³+x⁴+x⁵+x⁶+x⁷+x⁸+x⁹+x¹⁰+x¹¹+x¹²+x¹³+x¹⁴ +x¹⁵
The sum of the evens is 8 and the sum of the odds is 8. Therefore x= -1 is a root and (x+1) a factor.
1+x+x²+x³+x⁴+x⁵+x⁶+x⁷+x⁸+x⁹+x¹⁰+x¹¹+x¹²+x¹³+x¹⁴ +x¹⁵ =
(x+1)(1+x²+x⁴+x⁶+x⁸+x¹⁰+x¹²+x¹⁴)
= (x+1)(1+x²+x⁴(1+x²)+x⁸(1+x²)+x¹² (1+x²))
= (x+1)(1+x²)(1+x⁴+x⁸+x¹²)
= (x+1)(1+x²)(1+x⁴+x⁸ (1+ x⁴))
= (x+1)(1+x²)(1+x⁴)(1+x⁸)
These factors found by factoring.
answer
(x+1)(1+x²)(1+x⁴)(1+x⁸)
I did your method 2, but method 1 has its own elegance. Not brutal. Both work only because it's 16 terms in the expression, being a power of 2.
I saw that -1 is a root, so I long divided by x+1. The result suggested long dividing by x^2+1, and so forth.
The given expression is f(x) = (x^16-1)/(x-1) = (x^8+1)(x^4+1)(x^2+1)(x+1).
Why not group first half and second half then factor out x^8 from second half and repeat the setps?
My second method was to multiply bu x-1 and observe we get x^16-1=0 this is the 16th roots of unity but we introduced x=1 as an extra solution. So to factor into quadratic polynomials we just have to find sin and cos of 45/2 degrees and we are set but will take a bit of work to sort everything into irreducible quadratic polynomials
x=45/2 degrees
From half angle formula for sin √2/2=2sin(x)cos(x)
From half angle formula for cos √2/2=cos^2(x)-sin^2(x)
√2/2=2cos^2(x)-1
cos(x)=√(2+√2)/4
then sin(x)=√2/√(2(√2)=1/2*√2/2*√(2-√2)/√(4-2)=√(2-√2)/2
now we will have complex conjugated roots +-45/2 degrees
√(2+√2)/2+√(2-√2)i and √(2+√2)/2-√(2-√2)i
And √2/2+√2/2i and √2/2-√2/2i for +-45 degrees
Also
√(2-√2)/2+√(2+√2)/2i and √(2+√2)/2-√(2-√2)/2i for +-135/2 degrees
+ I and - i for +-90 degrees
And so on
With the conjugate roots we form the polynomials x^2-Sx+P and this is it.
I feel the problem was left half finished the way you did it. After all Euler proved all polynomials can be decomposed as products of quadratic polynomials.
We cannot decompose more only of we set the conditions of having integer or rational coefficients
That was fun. I didn't have any paper, so I had to do this mentally. My method was to first recognize the form, then try a smaller example, then see the pattern, then verify the pattern.
First the form. I knew it would be something like (x^n+1) times other expressions of the same form with different n.
For the small example, I tried n values of 1 and 2 and got a result of x^3+x^2+x+1.
Seeing the pattern, I realized that keeping every coefficient as 1 meant that every sum of combinations of the different n values had to be distinct and also cover every integer from 0 to 15. And hey, that's binary.
15 is 2^4-1, so this works perfectly with n values of 8, 4, 2, and 1.
It also generalizes to any expression like this where the largest exponent is a power of 2, minus 1. So for x^31+...+x+1 you only need to multiply by another x^16+1.
You could generalize further to get other factorings using other bases, but I'll save that for morning. 😅
Okay, I finished watching, and my method was definitely more hand-wavey than either of the ones shown.
I could call my method "solving by constraints". That's how I narrowed down the form the answer would need to take, and it's also how I determined what property the exponents needed to have. I was simply lucky that these constraints were enough to narrow the possible solution space down to a single easy-to-check possibility.
Just think of what happens if x = 2. You end up with a binary number: 111...1.
Add 1 to that number and you end up with 1000...00 = 2^16 - 1.
So right there, we know that the binary number = (x^16 - 1) / (x - 1).
Then of course, (x^16 - 1) = (x^8 + 1) * (x^8 - 1) and so on.
（x-1)(x^15+^4+....+1)/(x-1)=(x^16-1)/(x-1)=(x^8+1)(x^8-1)/(x-1)=(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)/(x-1)=(x^8+1)(x^4+1)(x^2+1)(x+1)
Factors of 1+x+'''''''''''+x^14=?
(1-x^16)/(1-x)=(1+x^8)(1-x^8)/(1-x)=(1+x^8)(1+x^4)(1-x^4)/(1-x)=(1+x^8)(1+x^4)(1+x^2)(1+x)
Thumbs-down. Why? You did not show enough steps. You have to be consistent.
1 + x + x^2 + ... + x^15 =
(1 - x^16)/(1 - x) =
(1 + x^8)(1 - x^8)/(1 - x) =
(1 + x^8)(1 + x^4)(1 - x^4)/(1 - x) =
(1 + x^8)(1 + x^4)(1 + x^2)(1 - x^2)/(1 - x) =
(1 + x^8)(1 + x^4)(1 + x^2)(1 + x)(1 - x)/(1 - x) =
(1 + x^8)(1 + x^4)(1 + x^2)(1 + x)
Халява же. (1+x⁸)(1+x⁴)(1+x²)(1+x)