From about 2:28, it's actually pretty easy to factor to get (x^4-1)(x^3+1) = 0, so either: • x^4 = 1 --> x=±1 or ±i, or • x^3 = -1 --> 3 evenly-spaced solutions around the unit circle in the complex plane --> x = -1 or 0.5±0.5i*sqrt3 So, there are 4 non-integral roots, all unique.
Sir, as a matter of fact. By further examination by Descartes rule of signs and Rational + Integer root theorem, we find out that the 4 non-integer solutions are actually all complex solutions. Thus we can say that the equation has 3 Real (3 Integer + 0 Non-Integer) & 4 Complex Solutions. 😊
First: such a beautiful exposition! Prime Newton's love for his subject shines through his videos. Second: look at the factorisation of the expression! It is (x-1) . (x+1)^2 . (x*2+1) . (x^2-x+1), The last factor shouts TRIANGLE if you recognise a cyclotomic polynomial. If you now take one of the factors (x+1) and multiply it into the last factor and also multiply the remaining factors, ypu get a beautiful and revealing form of the polynomial! (x^4 - 1) . (x^3 + 1). The roots of the first factor are obviously +/- 1 and +/1 i, the x (or y!) coordinates of the vertices of a square. Similarly the second factor for an equilateral trangle. The solution now stares you in the face: exactly three integer roots. I have not done the work but I am pretty sure that the polynomial can be written as a trigonimetric identity and with help of Euler and de Moivre as some beautiful function of a complex variable. Were the framers of this problem inspired by geometry? It certainly looks like it... All this is not to suggest a better solution but merely to have a closer look at the beauty of the problem. This is just one of the ways in which the universe sings.
The only thing I did slightly different was the polynominal/syntetic division ... I divided the original equation in one go by x^2 -1 (which is (x+1)(x-1) ), and therefore 'saved' one division step. (the remaining division attempts were the same as shown in the video).
It is so easy to simplify this equation into the basic factors that give us an overview of all the solutions: x4(x3+1)-(x3+1) = (x3+1)(x4-1) = (x3+1)(x2+1)(x2-1) = (x2-x+1)(x2+1)(x-1)(x+1)(x+1) So we have 3 integer solutions (one is double) from the last three factors, and 4 real solutions from the first two factors
That's how I did it. But the first two factors don't give real solutions. The first gives two complex solutions and the second gives two imaginary ones. I started with synthetic division but life is too short the way I do it.
Ans. There are 4 non-integer roots of this polynomial. It can be factored as: (x - 1)(x + 1)^2(x^4 - x^3 + 2x^2 - x + 1) ◼ Solution obtained via synthetic division.
NOTE: That right hand side you can already tell has no integer solutions by rational root theorem but you can still factor it into (x^2 - x + 1)(x^2 + 1) if you so wish
@@nanamacapagal8342 Nice catch. I worked to solve the problem as written (before watching the video). It didn't occur to me to clean up the solution any further. Of course I'm of the opinion that once you've found what you're looking for, you should stop looking.
Decartes' Rule of Signs tells us that there is one positive real root and zero or two negative real roots. The Rational Roots Theorem finds that 1 and -1 are rational roots. As we found one negative real root, then by Descartes' Rule of Signs, there is a second real root. Unfortunately, it doesn't tell us that the other negative root is rational. However, it does tell us that the other four roots are complex.
x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1) (x-1)(x^6+x^5+x^4+2x^3+x^2+x+1)=0 x=1 integer, rejected x=-1 is also a solution by observation. It's also an integer, but i can divide by x+1 x^5+x^3+x^2+1=0 🚨quintic alert🚨 luckily, -1 is a solution to this as well so we can divide again. x^4-x^3+2x^2-x+1=0 I see an overlap (x^2-x+1)(x^2+1)=0 do complex numbers with nonzero integer imaginary parts count as integers? x^2-x+1=0 x=(1+-sqrt(-3))/2 x^2+1=0 x=+-i there are either 2 or 4 non-integer solutions depending how integers are defined.
This one is interesting... Usually we are asked about integer solutions, but this question asks about NOT integer solutions... Though we aren't calculating what are those for non integer solutions are, i still think this one is different.
x^7+x^4-x^3-1=0 or x^7-x^3+x^4-1 =x^3(x^4-1)+(x^4-1)=0 divide all sides by x^4-1 and we get: x^3+1=0 =>x^3=-1 we know (-1)^3=-1, factor out the root and you get: x^2-x+1 [1 quadratic formula later] x=(1+/-sqrt(-3))/2 (both not integers so we’ll ignore them) next, x^4-1=0 it’s obvious that x is 1, -1, and x=i is also a solution and x=-i is a solution too, since every negative version of a solution counts as well so, recap: x=-1 x=(1+isqrt3))/2 x=(1-isqrt3))/2 x=1 x=-1 x=i x=-i so 3 solutions are integers (unless you consider i an integer, in which case 5 solutions are integers)
At 2:38 by looking at the equation it has one real positive root and 2 or zero negative roots. Only possible rational roots are +- 1, so only need to check for double root on the -1 solution Edit: It would have saved you from having to test the value of 1. Edit: You need to be careful when you say integers. Because there is a definition of integers in complex we can not test without actually factoring.
@@PrimeNewtons Because Gaussian integers Z[i] are integers and have their own subset of primes, which some are the same as the regular primes (if p = 3 mod 4) otherwise you can factor.
Since any number is divisible by 1, then they're also divisible by -1. The first one wouldn't change the number, while the other one would change its sign. We can make these divisions as many times as we want. Synthetic division is therefore not dividing by 1, or -1, since that would not give us any reduced polynomial (i.e, of lower degree), therefore not being really any synthetic division, but by (x - 1), and [x - (-1)] = (x + 1). By the way, when have you solved a cubic equation in this video? I've only seen that you found the number of non-integer solutions (though for me are sets of non-integer solutions, since I consider that two complex numbers with same value, but different arguments are different because they live in different complex planes. Since the number of arguments are infinite, the number of complex solutions are always infinite, though the number of sets of complex solutions are finite).
Other name for that, which I learned is Gorner scheme ( or method), I used to learn it in slightly different format, where we omit writing that intermediate line with products, rather doing two actions(multiplication and then addition) mentally writing the third line at once,without the second line. In fact it's a simplified version of long division for polynomials, without writing all those x's with exponents and repeating the same terms at least twice. But it only works if you divide by (x-a) otherwise it won't give you correct results, than you put a in that left corner ( in the video there were 1 and - 1 that means it can be divided by (x-1)(x+1)²and the figures in the last line are coefficients of the quotient)
Turns out factoring this is rather simple. x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1). (this is true for all x^n-1, if n is rational, since all the terms cancel out and you'll be left with x^n-1) x^6+x^5+x^4+x^3+x^2+x+1+x^3=(x^3+x^2+x+1)(x^3+1)=(x+1)(x^2+1)(x^3+1)=(x+1)(x^2+1)(x+1)(X^2-x+1) you end up getting (x+1)^2(x-1)(x^2+1)(x^2-x+1)=0, solve the 2 quadratics to get 4 complex solutions that aren't integers.
Take x cube as factor from x power seven and x power four -1 as factor from -x cube and -1 Finally take x cube +1 Factor You will get x power 4 -1 and x cube +1 =0 Further you get x square =-1 or + 1 and x cube =-1 from x square =. -10you get tow non integer roots From x cube = -1 you get two non integer roots Finally four non integer roots and x=-1,-1,1 are integer roots Answer is four non integer roots
How do we know the 4 noninteger solutions are unique? If two of them are say i, wouldnt you call that one noninteger solution with multiplicity 2? Also this doesnt seem like a cubic equation since leading term is x^7. Otherwise, great explanations of rational root theorem, remainder theorem, and synthetic divison :)
The rational root theorem tells us that the only possible rational roots of this polynomial are 1 and -1. So, there must indeed be some multiplicity among 3 integer roots.
It allows you to count real solutions by counting the signs of the polynomial (and the polynomial after substuting -x in for x) x is positive signs from largest to smallest power of x are: + + - -; Which has one change of sign, which means there is ONLY one positive solution (so the x=1 is not a double root) x is negative signs from largest to smallest power of x are (flip any sign that is on an odd power) : - + + -; Which has 2 changes of signs which means there are either 2 or 0 solutions. And he later shows that -1 is a double root and counts for both Rule of signs only tells you the MAX possible solutions, and that number can always be reduced by 2n since if you have real coefficients, they come in pairs, which was why we could not reduce the value of 1 since -1 solutions does not make sense.
The way i was watching this channel since its begging seeing it grow makes me so happy fr
Thank you
From about 2:28, it's actually pretty easy to factor to get (x^4-1)(x^3+1) = 0, so either:
• x^4 = 1 --> x=±1 or ±i, or
• x^3 = -1 --> 3 evenly-spaced solutions around the unit circle in the complex plane --> x = -1 or 0.5±0.5i*sqrt3
So, there are 4 non-integral roots, all unique.
non-integer*
Yeah I got that straight away it’s actually very straight forward
Sir, as a matter of fact. By further examination by Descartes rule of signs and Rational + Integer root theorem, we find out that the 4 non-integer solutions are actually all complex solutions.
Thus we can say that the equation has 3 Real (3 Integer + 0 Non-Integer) & 4 Complex Solutions. 😊
x=-1 is a repeat
First: such a beautiful exposition! Prime Newton's love for his subject shines through his videos.
Second: look at the factorisation of the expression! It is (x-1) . (x+1)^2 . (x*2+1) . (x^2-x+1), The last factor shouts TRIANGLE if you recognise a cyclotomic polynomial. If you now take one of the factors (x+1) and multiply it into the last factor and also multiply the remaining factors, ypu get a beautiful and revealing form of the polynomial!
(x^4 - 1) . (x^3 + 1). The roots of the first factor are obviously +/- 1 and +/1 i, the x (or y!) coordinates of the vertices of a square. Similarly the second factor for an equilateral trangle. The solution now stares you in the face: exactly three integer roots.
I have not done the work but I am pretty sure that the polynomial can be written as a trigonimetric identity and with help of Euler and de Moivre as some beautiful function of a complex variable.
Were the framers of this problem inspired by geometry? It certainly looks like it...
All this is not to suggest a better solution but merely to have a closer look at the beauty of the problem. This is just one of the ways in which the universe sings.
Well done. I also found the factorization, but not that most elegant form.
Alternatively we can factor x⁷+x⁴-x³-1 as (x³+1)(x⁴-1)=(x+1)(x²-x+1)(x-1)(x+1)(x²+1) where quadratics clearly all have imaginary roots - so 1,-1,-1
those are integer solutions
Teacher, you always explain super great. You are a charismatic teacher. Thank you very much!!
The only thing I did slightly different was the polynominal/syntetic division ... I divided the original equation in one go by x^2 -1 (which is (x+1)(x-1) ), and therefore 'saved' one division step. (the remaining division attempts were the same as shown in the video).
It is so easy to simplify this equation into the basic factors that give us an overview of all the solutions: x4(x3+1)-(x3+1) =
(x3+1)(x4-1) =
(x3+1)(x2+1)(x2-1) =
(x2-x+1)(x2+1)(x-1)(x+1)(x+1)
So we have 3 integer solutions (one is double) from the last three factors, and 4 real solutions from the first two factors
That's how I did it. But the first two factors don't give real solutions. The first gives two complex solutions and the second gives two imaginary ones. I started with synthetic division but life is too short the way I do it.
Simplify!
I used this method too : )
@@XiOjala yes true, the others are complex solutions because both factors have Delta
@@azzteke sorry 😃
Congratulations 🎉on 100k+ subscribers.Great teachers like you always teach not to stop learning, cuz those who stop learning, stop living 😊😊
Your presentation is didactic.
Verry useful.
Thank you
Ans. There are 4 non-integer roots of this polynomial. It can be factored as:
(x - 1)(x + 1)^2(x^4 - x^3 + 2x^2 - x + 1) ◼
Solution obtained via synthetic division.
NOTE: That right hand side you can already tell has no integer solutions by rational root theorem but you can still factor it into (x^2 - x + 1)(x^2 + 1) if you so wish
@@nanamacapagal8342 Nice catch. I worked to solve the problem as written (before watching the video). It didn't occur to me to clean up the solution any further. Of course I'm of the opinion that once you've found what you're looking for, you should stop looking.
Prime Newtons does it again! 🎉😊
Decartes' Rule of Signs tells us that there is one positive real root and zero or two negative real roots.
The Rational Roots Theorem finds that 1 and -1 are rational roots. As we found one negative real root, then by Descartes' Rule of Signs, there is a second real root. Unfortunately, it doesn't tell us that the other negative root is rational. However, it does tell us that the other four roots are complex.
x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)
(x-1)(x^6+x^5+x^4+2x^3+x^2+x+1)=0
x=1 integer, rejected
x=-1 is also a solution by observation. It's also an integer, but i can divide by x+1
x^5+x^3+x^2+1=0
🚨quintic alert🚨
luckily, -1 is a solution to this as well so we can divide again.
x^4-x^3+2x^2-x+1=0
I see an overlap
(x^2-x+1)(x^2+1)=0
do complex numbers with nonzero integer imaginary parts count as integers?
x^2-x+1=0
x=(1+-sqrt(-3))/2
x^2+1=0
x=+-i
there are either 2 or 4 non-integer solutions depending how integers are defined.
If you factor the equation you get:
(x^4-1)(x^3+1) = 0
Then
(x^2-1)(x^2+1)(x+1)(x^2-x+1)=0
Integer roots:
x^2-1 = 0 -> 2 int roots
x+1 =0 -> 1 int root
Non Integer roots
x^2+1 =0 -> 2 complex roots
x^2-x+1 =0 -> 2 complex roots
This one is interesting... Usually we are asked about integer solutions, but this question asks about NOT integer solutions...
Though we aren't calculating what are those for non integer solutions are, i still think this one is different.
x^7+x^4-x^3-1=0
or x^7-x^3+x^4-1
=x^3(x^4-1)+(x^4-1)=0
divide all sides by x^4-1 and we get:
x^3+1=0
=>x^3=-1
we know (-1)^3=-1, factor out the root and you get:
x^2-x+1
[1 quadratic formula later]
x=(1+/-sqrt(-3))/2 (both not integers so we’ll ignore them)
next, x^4-1=0
it’s obvious that x is 1, -1, and x=i is also a solution
and x=-i is a solution too, since every negative version of a solution counts as well
so, recap:
x=-1
x=(1+isqrt3))/2
x=(1-isqrt3))/2
x=1
x=-1
x=i
x=-i
so 3 solutions are integers (unless you consider i an integer, in which case 5 solutions are integers)
At 2:38 by looking at the equation it has one real positive root and 2 or zero negative roots. Only possible rational roots are +- 1, so only need to check for double root on the -1 solution
Edit: It would have saved you from having to test the value of 1.
Edit: You need to be careful when you say integers. Because there is a definition of integers in complex we can not test without actually factoring.
Every time someone says 'complex', I quake. I will watch out from now on.
@@PrimeNewtons Because Gaussian integers Z[i] are integers and have their own subset of primes, which some are the same as the regular primes (if p = 3 mod 4) otherwise you can factor.
Beautiful!
Thank you! Cheers!
thank You very much, I really enjoy Your videos!
why do all of this instead of easily factoring the polynomyal as (x^2 + 1)(x - 1)(x + 1)(x^2 - x + 1)(x+1)
Since any number is divisible by 1, then they're also divisible by -1. The first one wouldn't change the number, while the other one would change its sign. We can make these divisions as many times as we want. Synthetic division is therefore not dividing by 1, or -1, since that would not give us any reduced polynomial (i.e, of lower degree), therefore not being really any synthetic division, but by (x - 1), and [x - (-1)] = (x + 1).
By the way, when have you solved a cubic equation in this video? I've only seen that you found the number of non-integer solutions (though for me are sets of non-integer solutions, since I consider that two complex numbers with same value, but different arguments are different because they live in different complex planes. Since the number of arguments are infinite, the number of complex solutions are always infinite, though the number of sets of complex solutions are finite).
Do you have a video on Descartes rule of signs?, we haven't learnt one at school, where I live
Synthetic division: completely new to me. Please give a separate video about this subject. It was dazzling me.
Other name for that, which I learned is Gorner scheme ( or method), I used to learn it in slightly different format, where we omit writing that intermediate line with products, rather doing two actions(multiplication and then addition) mentally writing the third line at once,without the second line. In fact it's a simplified version of long division for polynomials, without writing all those x's with exponents and repeating the same terms at least twice. But it only works if you divide by (x-a) otherwise it won't give you correct results, than you put a in that left corner ( in the video there were 1 and - 1 that means it can be divided by (x-1)(x+1)²and the figures in the last line are coefficients of the quotient)
what do you mean you don't have access to the blackboard?
Turns out factoring this is rather simple. x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1). (this is true for all x^n-1, if n is rational, since all the terms cancel out and you'll be left with x^n-1)
x^6+x^5+x^4+x^3+x^2+x+1+x^3=(x^3+x^2+x+1)(x^3+1)=(x+1)(x^2+1)(x^3+1)=(x+1)(x^2+1)(x+1)(X^2-x+1)
you end up getting (x+1)^2(x-1)(x^2+1)(x^2-x+1)=0, solve the 2 quadratics to get 4 complex solutions that aren't integers.
Solving for x would yield x=1, x=-1 ,x=-1 and two complex numbers x=i and x= -i only
Take x cube as factor from x power seven and x power four
-1 as factor from -x cube and -1
Finally take x cube +1
Factor
You will get x power 4 -1 and x cube +1 =0
Further you get x square =-1 or + 1 and x cube =-1 from x square =. -10you get tow non integer roots
From x cube = -1 you get two non integer roots
Finally four non integer roots and x=-1,-1,1 are integer roots
Answer is four non integer roots
Never knew about this, thanks.
thank you. well done ! 💥
What if you have a whiteboard and a black marker in every video?
Great viedo but what happened to the background
What background? You mean blackboard?
Yes@@PrimeNewtons
@@itachu. Went on a trip.
How do we know the 4 noninteger solutions are unique? If two of them are say i, wouldnt you call that one noninteger solution with multiplicity 2? Also this doesnt seem like a cubic equation since leading term is x^7. Otherwise, great explanations of rational root theorem, remainder theorem, and synthetic divison :)
The rational root theorem tells us that the only possible rational roots of this polynomial are 1 and -1. So, there must indeed be some multiplicity among 3 integer roots.
sir,can you please make a video on pentation.
whats decots rules of sines???
Descartes rule of signs
@@PrimeNewtons ohh thanks
It allows you to count real solutions by counting the signs of the polynomial (and the polynomial after substuting -x in for x)
x is positive signs from largest to smallest power of x are: + + - -; Which has one change of sign, which means there is ONLY one positive solution (so the x=1 is not a double root)
x is negative signs from largest to smallest power of x are (flip any sign that is on an odd power) : - + + -; Which has 2 changes of signs which means there are either 2 or 0 solutions. And he later shows that -1 is a double root and counts for both
Rule of signs only tells you the MAX possible solutions, and that number can always be reduced by 2n since if you have real coefficients, they come in pairs, which was why we could not reduce the value of 1 since -1 solutions does not make sense.
I am not sure what make you switch background but I hope it is not a problem that would affect your life too much, be safe :)
Great episode. Loved the efficient synthetic division and the neat handwriting
I LOVE synthetic division
x^4(x^3+1)-1(x^3+1)=0
X=1
bring back chalk board