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

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### 2 Answers

`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` .**

`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`