I absolutely appreciated the humor in this video! 🤣 On a side note, and while continuing to acknowledge how good these videos are, I'd highly suggest double-checking the examples. Unfortunately I think the errors can get to the point where they create genuine confusion.
Hi Matthew, You have an error in the polynomial you present at around the 2 minute mark. (t+1)(t-2) is the factorisation of t^2 - t - 2, not +2. Thank you for these videos. I am currently studying group theory in my undergraduate course, and your explanations are much better than the texts I've been using. Keep up the good work, Paul
For quadratics, you can always fall back to the quadratic formula. Then once you have the roots, the original polynomial is just the product (x - root1) * (x - root2).
That is easy. (x-alfa)(x-bèta)(x-gamma)(x-delta) = 0 then multipy everything out. You get (x-a)(x-b)(x-c)(x-d) = 0 x^4+(-a-b-c-d)x^3+(ab+ac+ad+bc+bd+cd)x^2+(-bcd-acd-abd-abc)x +abcd = 0then the sum of the roots are for example 0 if there is no X^3 termif you have x^4+(-4)x^3+(8)x^2+(-12)x +18 = 0then -a-b-c-d = -4 ab+ac+ad+bc+bd+cd = 8 -bcd-acd-abd-abc = =12 and abcd = 18
I absolutely appreciated the humor in this video! 🤣
On a side note, and while continuing to acknowledge how good these videos are, I'd highly suggest double-checking the examples. Unfortunately I think the errors can get to the point where they create genuine confusion.
Hi Matthew, You have an error in the polynomial you present at around the 2 minute mark. (t+1)(t-2) is the factorisation of t^2 - t - 2, not +2.
Thank you for these videos. I am currently studying group theory in my undergraduate course, and your explanations are much better than the texts I've been using.
Keep up the good work,
Paul
(t+1)(t-2) = t^2 - t - 2
t = 1/2 + sqrt(7)i/2 and 1/2 - sqrt(7)i/2 are the roots to t^2 -t +2 = 0
I am enjoying this series on Galois theory.
how you factor t^2-t+2?
Complete the square. Roots are complex.
For quadratics, you can always fall back to the quadratic formula. Then once you have the roots, the original polynomial is just the product (x - root1) * (x - root2).
You can also factor x^5 + x+1 = 0 (x^2 + x+1)(x^3 + -x^2 + 1) = 0 and is thus solvable. Some Quintics are. But how can you spot that?
That is easy. (x-alfa)(x-bèta)(x-gamma)(x-delta) = 0 then multipy everything out.
You get (x-a)(x-b)(x-c)(x-d) = 0
x^4+(-a-b-c-d)x^3+(ab+ac+ad+bc+bd+cd)x^2+(-bcd-acd-abd-abc)x +abcd = 0then the sum of the roots are for example 0 if there is no X^3 termif you have x^4+(-4)x^3+(8)x^2+(-12)x +18 = 0then
-a-b-c-d = -4 ab+ac+ad+bc+bd+cd = 8
-bcd-acd-abd-abc = =12 and abcd = 18