Finding Roots of Multiplicity in a Polynomial

Solution 1:

The quoted passage shows how to get $X_1, X_2, \dots$ starting from a generic $f(x)$ polynomial.

Underlying idea is that $x=a$ is a root of multiplicity $m$ if $f(a)=f'(a)=\cdots=f^{(m-1)}(a)=0$.

For a minimalistic step by step example (as asked), let:

$$ f(x)=x^3-5 x^2+8x-4 $$

Find all the greatest common measure of $f(x)\text{ and }f'(x)=F_1(x)$

$$F_1(x) = \gcd(\,x^3-5 x^2+8x-4\,, \;3x^2-10 x+8\,) = x-2$$

$\text{,,}\qquad F_1(x)\text{ and } F_1'(x)=F_2(x),$

$$F_2(x)=\gcd(\,x-2,\, 1 \,) = 1$$

Lastly, the greatest common measure of $F_{m-1}(x)$ and $F'_{m-1}(x)=F_m(x)=1$

Therefore $m=2$ since $F_2(x)=1$.

Next, perform the divisions $f(x)\div F_1(x)=\phi_1(x)$

$$\phi_1(x)=f(x)\div F_1(x)=x^2-3x+2$$

$F_1(x)\div F_2(x)=\phi_2(x)$

$$\phi_2(x) = F_1(x)\div F_2(x) = x-2$$

And finally $\phi_1(x)\div\phi_2(x)=X_1$

$$X_1 = \phi_1(x)\div\phi_2(x) = x-1$$

$F_{m-1}(x)=\phi_m(x)=X_m$

$$X_2= \phi_2(x)=x-2$$

Then, since it was established earlier that $m=2\,$:

$$f(x)=X_1\,X_2^2=(x-1)(x-2)^2$$