# Factor. `m^3 + 3m^2 + 3m + 1`

lemjay | Certified Educator

`m^3+3m^2+3m+1`

To factor the given polynomial, we may apply the Rational Zeros Theorem.

Since the coefficient of the leading term is 1 and the constant is 1, the possible zeros are `+-1/1 = -1` and `1` .

Now that we know the possible zeros, use synthetic division to verify.

`-1` `|` `1`     `3`     `3`       `1`
`-1`  `-2`    `-1`
`-----------`
`1`     `2`      `1`       `0`

Since the last number at the third row is 0, it means that -1 is one of the zeros.

So the factor of the given polynomial is,

`m^3+3m^2+3m+1 = (m+1)(m^2+2m+1)`

To factor further, apply synthetic division again to m^2+2m+1.

`-1` `|`  `1`     `2`       `1`
`-1`    `-1`
`--------`
`1`       `1`      `0`

Since the last number at the third row is 0, then we have:

`m^3+3m^2+3m+1`

`= (m+1)(m^2+2m+1)`

`= (m + 1)(m+1)(m+1)`

`=(m+1)^3`

Hence, `m^3+3m^2+3m+1= (m+1)^3` .

oldnick | Student

`m^3+3m^2+3m+1=m^3+1 +3m(m+1)=`

`=(m+1)(m^2-m+1) +3m(m+1)=`

`=(m+1)(m^2-m+1+3m)=(m+1)(m^2+2m+1)= `

`=(m+1)(m^2+m+m+1)=(m+1)[m(m+1)+(m+1)]=`

`=(m+1)(m+1)(m+1)=(m+1)^3`