how do u find the root of
m^3+3m^2+3m+1 = 0
please help me find what are the value of what m is equal to
2 answers
assistance needed
I would let f(m) = m^3+3m^2+3m+1
on second try f(-1) = 0, so m+1 is a factor
using sysnthetic division, or long algebraic division, the other factor is (m^2 + 2m + 1)
which is (m+1)(m+1)
so m^3+3m^2+3m+1 = 0
(m+1)^3 = 0
m = -1
(I should have recognized 1 3 3 1 as the third row of Pascal's triangle)
on second try f(-1) = 0, so m+1 is a factor
using sysnthetic division, or long algebraic division, the other factor is (m^2 + 2m + 1)
which is (m+1)(m+1)
so m^3+3m^2+3m+1 = 0
(m+1)^3 = 0
m = -1
(I should have recognized 1 3 3 1 as the third row of Pascal's triangle)
10 answers