I know this may seem trivial, but I, for the life of me, I cannot figure out how to factor this polynomial, I know that the root is $$(x-1)^3=0$$ because of wolframalpha, but I don't know how to get there. any help would be greatly appreciated. and also if you have any recommended web sites that help with higher order polynomial factoring that would be extremely helpful.

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    $\begingroup$ It's kind of a good idea to familiarize yourself with the first few entries of Pascal's triangle... $\endgroup$ – J. M. is a poor mathematician Nov 28 '11 at 12:28
  • $\begingroup$ I remember leaning about that in one of my calculus classes, but how does that apply to this? Can you use that to factor a third degree polynomial? $\endgroup$ – AlexW.H.B. Nov 28 '11 at 12:31
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    $\begingroup$ The $1,3,3,1$ pattern ought to have "screamed" at you... :) $\endgroup$ – J. M. is a poor mathematician Nov 28 '11 at 12:36
  • $\begingroup$ Oh i now see what your getting at. thanks for the tip. $\endgroup$ – AlexW.H.B. Nov 28 '11 at 12:39

First, guess a root (this is the hard part). The so called "rational roots test" will be helpful here.

Eventually, you'll discover that $x=1$ is a root. This will imply that your polynomial has the form $$ \tag{1}(x-1)(ax^2+bx+c), $$ for some constants $a, b, c$.

To find those constants, you could do one of two things (and maybe more)

  1. perform the division $ x^3-3x^2+3x+1\over x-1$.
  2. expand (1) and set it equal to the original polynomial. Setting the coefficients of the two sides of this equation equal to each other will give you a system of equations that are solvable for $a$, $b$, and $c$.

Once you've figured out what $a,b$, and $c$ are, factor the quadratic.

  • $\begingroup$ Thank you so much for you thorough answer and quick response. this is very helpful. :) $\endgroup$ – AlexW.H.B. Nov 28 '11 at 12:21
  • $\begingroup$ You're welcome. Glad to help. $\endgroup$ – David Mitra Nov 28 '11 at 12:52

$$ \begin{align*} x^3-3x^2+3x-1 &=x^3-x^2-2x^2+3x-1 \\ &=x^2(x-1)-2x^2+2x+x-1 \\ &=x^2(x-1)-2x(x-1)+1(x-1) \\ &=(x-1)(x^2-2x+1) \\ &=(x-1)(x-1)^2 \\ &=(x-1)^3 \end{align*}$$

  • $\begingroup$ Thank you very much... this helped me a lot. I do have one question though. does this method of splitting up terms work in most or all cases of third degree polynomial factoring? $\endgroup$ – AlexW.H.B. Nov 29 '11 at 6:14


$=x(x^2-2x+1)-(x^2-2x+1)$ ....etc.

  • $\begingroup$ so you are saying you can factor this as two trinomial degree two polynomials just by grouping? very interesting. $\endgroup$ – AlexW.H.B. Nov 28 '11 at 12:24


If you know a solution for the equation (in this case $x=1$) then you can use the Briot-Ruffini method to reduce this 3-degree polinomyal to a 2-degree.

Just find the other roots applying Baskara.

If the roots are $x_1$, $x_2$, $x_3$ you can factore your polinomyal as:

$a(x-x_1)(x-x_2)(x-x_3)$, where $a$ is the coefficient of the highest degree.

Other hint:

A notable product is:


See this is quite similar to your polynomial.

  • $\begingroup$ hey thank you very much for the tips. I will keep them in mind. $\endgroup$ – AlexW.H.B. Nov 28 '11 at 12:29
  • $\begingroup$ @AlexW.H.B. You're welcome .-. $\endgroup$ – GarouDan Nov 28 '11 at 12:34

Another way: $\rm\ f\:' = 3\ (x-1)^2\ $ and $\rm\ gcd(f,f\:') = (x-1)^2\ $ by Euclid's algorithm (or by inspection).

In fact most polynomial factorization algorithms start by reducing to the squarefree case by factoring out $\rm\ gcd(f,f')\:.$


From binomial formula $$a^3-3a^2b+3ab^2-b^3=(a-b)^3$$for$a=x,b=1$ we get that

$$x^3-3x^2+3x-1 =x^3-3x^2\times1+3x\times1^2-1^3=(x-1)^3$$

  • $\begingroup$ I was about to say the same thing, observing that my answer would not properly be the same as the other answers. And then I come across the very last answer...haha $(+1)$ $\endgroup$ – Feeds Oct 16 '17 at 23:40

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