I was illustrating that multiple polynomials could satisfy the definition of GCD. One could decide to go after a monic polynomial always or show that the two GCDs are associates by showing that they are associates of the same monic polynomial, but there wasn't really any great reason to do so.
Thank you!
Thank u, bro!!
what happens if I want to find GCD of polynomial over Q[x]?
I really hated this unit in abstract algebra.
Where is the 2X+4 coming from,, explain further please.
Dry English Salim Working modulo 5, so -3 is the same as 2 and -1 is the same as 4.
thanks
Why don't you divide 3x+1 (or 2x+4) by the leading coefficient to get 2+x?
I was illustrating that multiple polynomials could satisfy the definition of GCD. One could decide to go after a monic polynomial always or show that the two GCDs are associates by showing that they are associates of the same monic polynomial, but there wasn't really any great reason to do so.
why did you multiply by 4? to find the other gcd?
Basically. The idea is that each GCD is a (constant) multiple of the other.
Mitch Keller still i didn't get that.....so if it so,can we use any constant 4 this....won't it affect the answer
I dont understand 2x+4, how? Where have this teorem?
Hopefully you found the reason in the past 3 years but its because -3 mod 5 is 2 and -1 mod 5 is 4
boss, i don't understand how (-x+3)-(9x+3)=0.
because its mod 5.
+Michael Merkle
Hi
in mod 5: (-x+3) -(9x+3) = (-x+3) -(4x+3) = -5x +0 = 0
Lol MATH161.