Q.Find the multiple roots of $x^4-6x^3+13x^2-12x+4=0$ by the H.C.F. method.

Step 1. $f(x)=x^4-6x^3+13x^2-12x+4$, so $f'(x)=4x^3-18x^2+26x-12=2\big(2x^3-9x^2+13x-6\big)$. Write $g(x)=2x^3-9x^2+13x-6$ (a constant multiple does not affect the H.C.F.).

Step 2. Divide $f$ by $g$:

So the first remainder is $r_1(x)=-\dfrac14(x^2-3x+2)$.

Step 3. Divide $g(x)$ by $x^2-3x+2$ (the constant factor $-1/4$ is dropped, as it does not affect the H.C.F.):

The division is exact -- the remainder is $0$.

Step 4. Since $x^2-3x+2$ divides $g(x)$ exactly, it is the H.C.F. of $f$ and $f'$:

Step 5. By the H.C.F. method, every multiple root of $f(x)=0$ is a root of this H.C.F., appearing there with multiplicity one less than in $f$. So $x=1$ and $x=2$ are each multiple roots of $f$, each of multiplicity $2$ (i.e. multiplicity $1+1$, since they appear to the first power in the H.C.F.).

Step 6 (check). Indeed $(x-1)^2(x-2)^2=(x^2-3x+2)^2=x^4-6x^3+13x^2-12x+4=f(x)$, confirming the factorisation exactly.

H.C.F. method for multiple roots: compute $\mathrm{HCF}(f,f')$ by the Euclidean algorithm on polynomials.

• Forgetting that a constant factor may be dropped at each stage of the polynomial Euclidean algorithm (it does not change the H.C.F.), which otherwise makes the arithmetic unmanageable.
• Misreading the multiplicity: a root appearing to the first power in $\mathrm{HCF}(f,f')$ has multiplicity two, not one, in $f$.