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.

The way i was watching this channel since its begging seeing it grow makes me so happy fr

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. 😊

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

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

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).

Teacher, you always explain super great. You are a charismatic teacher. Thank you very much!!

Congratulations 🎉on 100k+ subscribers.Great teachers like you always teach not to stop learning, cuz those who stop learning, stop living 😊😊

thank You very much, I really enjoy Your videos!

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.

Your presentation is didactic. Verry useful. Thank you

Prime Newtons does it again! 🎉😊

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.

thank you. well done ! 💥

Beautiful!

Never knew about this, thanks.

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)

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

Synthetic division: completely new to me. Please give a separate video about this subject. It was dazzling me.

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)

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.

Do you have a video on Descartes rule of signs?, we haven't learnt one at school, where I live

Solving for x would yield x=1, x=-1 ,x=-1 and two complex numbers x=i and x= -i only

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).

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.

Great episode. Loved the efficient synthetic division and the neat handwriting

sir,can you please make a video on pentation.

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

What if you have a whiteboard and a black marker in every video?

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 :)

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 :)

whats decots rules of sines???

I LOVE synthetic division

Great viedo but what happened to the background

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.

x^4(x^3+1)-1(x^3+1)=0

X=1

bring back chalk board