- #1
Kys91
- 12
- 0
I can solve:
(2x+3y-4z)(2x-3y+4z) = [2x +(3y-4z)] [2x -(3y-4z)] = (2x)^2 -(3y-4z)^2 = (2x)^2 - (9y^2 - 24yz - 16z) = 4x^2 - 9y^2 + 24yz + 16z which is fine, but if I try to solve:
(m^2-m-1)(m^2+m-1) = [m^2 -(m+1)][m^2+(m-1) = (m^2)^ 2 - (m-1)^2 = (m^2)^2 - (m^2 -2m +1) = m^4 - m^2 +2m -1 which is not, and I also tried doing:
[(m^2 - m)-1] [(m^2+m)-1] = (m^4 - m^2) + 1 = this latest one is near, but anyway it should be m^4-3m^2 + 1
Doing normal multiplication I get: m^4 +m^3 -m^2 -m^3 -m^2+ m-m^2-m+ 1 = m^4-3m^2 + 1
So why is it that I have -m^2 when is it really -3m^2?
Thanks
(2x+3y-4z)(2x-3y+4z) = [2x +(3y-4z)] [2x -(3y-4z)] = (2x)^2 -(3y-4z)^2 = (2x)^2 - (9y^2 - 24yz - 16z) = 4x^2 - 9y^2 + 24yz + 16z which is fine, but if I try to solve:
(m^2-m-1)(m^2+m-1) = [m^2 -(m+1)][m^2+(m-1) = (m^2)^ 2 - (m-1)^2 = (m^2)^2 - (m^2 -2m +1) = m^4 - m^2 +2m -1 which is not, and I also tried doing:
[(m^2 - m)-1] [(m^2+m)-1] = (m^4 - m^2) + 1 = this latest one is near, but anyway it should be m^4-3m^2 + 1
Doing normal multiplication I get: m^4 +m^3 -m^2 -m^3 -m^2+ m-m^2-m+ 1 = m^4-3m^2 + 1
So why is it that I have -m^2 when is it really -3m^2?
Thanks