How to factor third degree polynomial like this.

Question

I have problem with factoring 3rd degree polynomial. $$m^3 - 2m + 1$$ I know answer is $(m-1)(m^2+m-1)$ but I'd like to know the way to do it. Thanks for help.

Answer 1

Because $1^3-2\cdot1+1=0$, which by the Bézout's theorem(https://en.wikipedia.org/wiki/Polynomial_remainder_theorem) gives the factor $m-1$ and $$m^3-2m+1=m^3-m^2+m^2-m-m+1=$$ $$=m^2(m-1)+m(m-1)-(m-1)=(m-1)(m^2+m-1).$$

Answer 2

Using the rational root theorem, you can determine that $1$ is a root of the polynomial. In other words, $m-1\mid m^3-2m+1$. After that, you can use long polynomimal division (or Ruffini's rule) to determine the quotient.

Answer 3

Note that

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

Answer 4

There was a trick that I once learned (Don't remember what it is called) for a cubic equation that stated that if the sum of the coefficients of the first and third terms is the opposite of the sum of the coefficients of the second and fourth terms, then one of the roots is $1$. If the sums are the same, then one of the roots is $-1$. Since in this case $1 + 0$ is the opposite of $-2 + 1$, one of the roots is $1$. Using synthetic division, we get:

  ________________
1 |  1   0   -2  1
  |      1    1 -1
     1   1   -1| 0

$(m-1)(m^2+m-1)$

The trinomial cannot be factored further, so we are done.

Answer 5

completing the quadratic: $$m^3-2m+1=m^3-m^2+m^2-2m+1=m^2(m-1)+(m-1)^2=(m-1)(m^2+m-1)$$