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$$x^3-3x^2+3x-1?$$

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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6 Answers

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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.

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

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$x^3-3x^2+3x-1=x^3-2x^2-x^2+x+2x-1=x^3-2x^2+x-x^2+2x-1=$

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

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Hint

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:

$(x-y)^3=x^3-3x^2y+3xy-y^3$

See this is quite similar to your polynomial.

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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 $\rm\ gcd(f,f')\:.$

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

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

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