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
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)
To find the roots of a given equation, you can utilize the concept of factoring and the properties of polynomial equations. In this case, the equation you provided is a cubic equation of the form m^3 + 3m^2 + 3m + 1 = 0.
Step 1: Factor the cubic equation:
To factor this equation, start by checking if any integers are solutions. By trial and error, you can find that when m is equal to -1, the equation becomes zero. So, m = -1 is one of the roots.
Using polynomial division or synthetic division, you can divide the given equation by (m + 1). This will yield a quadratic equation.
(m^3 + 3m^2 + 3m + 1) ÷ (m + 1) = m^2 + 2m + 1
So, the factored form becomes (m + 1)(m^2 + 2m + 1) = 0.
Step 2: Solve the quadratic equation:
Now, you need to solve the quadratic equation (m^2 + 2m + 1) = 0. This quadratic equation can also be factored as (m + 1)(m + 1) = 0.
So, the factored form now becomes (m + 1)^2 = 0.
Step 3: Find the roots:
From the factored form, you can see that (m + 1)^2 = 0. To find the roots, set (m + 1) equal to zero and solve for m.
m + 1 = 0
m = -1
So, the equation m^3 + 3m^2 + 3m + 1 = 0 has a single real root, which is m = -1.
In summary, the root of the cubic equation m^3 + 3m^2 + 3m + 1 = 0 is m = -1.