let f(m)=m^3 + 3m^2 + 3m + 1
notice f(-1) = 0, so m+1 is a factor.
Using division, the answer is m^2 + 2m + 1 which factors once more to (m+1)(m+1)
so you have (m+1)^3 = 0
and then m = -1
Are you familiar with Pascal's triangle?
Did you notice the pattern in the coefficients 1 3 3 1 and descending powers of m?
good work Reiny.
can anybody help with factoring this problem
m^3 + 3m^2 + 3m + 1 = 0
Does it help to check to see if -1 is a root? If so, x-1 can be divided into the polynomial.
To factor the equation m^3 + 3m^2 + 3m + 1 = 0, we can follow the steps you provided.
Step 1: Checking if -1 is a root:
We can substitute -1 for m, and if the equation evaluates to zero, it means that -1 is a root. Let's do that:
(-1)^3 + 3(-1)^2 + 3(-1) + 1 = 0
-1 + 3 + (-3) + 1 = 0
0 = 0
Since the equation evaluates to zero, we can conclude that -1 is indeed a root.
Step 2: Dividing the polynomial:
Since we know that (m+1) is a factor of the polynomial, we can divide the polynomial by (m+1) using long division or synthetic division. Let's use long division:
m^2 + 2m + 1
m + 1 | m^3 + 3m^2 + 3m + 1
- (m^3 + m^2 )
-----------------
2m^2 + 3m + 1
- (2m^2 + 2m)
-----------------
m + 1
The result of the division is m^2 + 2m + 1, which can be further factored as (m+1)(m+1) or (m+1)^2.
Step 3: Factoring the polynomial:
Now we have the factored form: (m+1)(m+1)(m+1) = 0. Since (m+1) repeats 3 times, we can write it as (m+1)^3 = 0.
Step 4: Solving for m:
To find the value of m, we set the factored form equal to zero and solve for m:
(m+1)^3 = 0
By taking the cube root of both sides:
m + 1 = 0
m = -1
So the solution to the equation m^3 + 3m^2 + 3m + 1 = 0 is m = -1.