How to tell if a quartic equation has a multiple root.

Solution 1:

The quartic $x^4 + a x^3 + b x^2 + c x + d = 0$ has a multiple root if it has a common root with $4x^3 + 3a x^2 + 2b x +c = 0$.

Any common factor can be determined using Euclid's algorithm. The "worst case" is that there are two double roots and the common factor is a quadratic - but that's easy to factor. A triple root will give a quadratic common factor too, but recognisable as having two equal roots.

This is not a simple criterion in terms of the coefficients, but it is a practical way of computing multiple roots when they exist and if the factorisation is not obvious.

The question is tagged pre-calculus - the easiest way to prove that this works is with some simple calculus (the second equation is obtained by differentiating the first). But the criterion can be applied without calculus, and is valid in many situations where limits do not make sense - and this motivates generalisations and extensions of the notion of derivative in algebra.

Solution 2:

To answer your question: yes.

As Neil mentioned in the comments, a quartic can have between 0 and 4 different roots. Other than the obvious way of checking the number of roots by actually solving the equation, the following also applies, as listed in J.M.'s link to this monthly article:

For an equation of the type $x^4+qx^2+rx+s=0$

If you were looking for a quick way, easy to memorize, I'm afraid this won't be of much help after all. Also, notice the absence of a $x^3$-term.

Solution 3:

Maybe I misunderstood the question, but I think this might be useful for you.

Let $f(x)\in F[x]$ be a polynomial and $f'(x)$ be the formal derivative of $f(x)$ where $F$ is any field.

Then $f(x)$ has multiple roots if and only if $\gcd(f(x),f'(x))\ne 1$.

If $F$ is a field of characteristic zero then you know more: If you denote $g(x)=\gcd(f(x),f'(x))$ and divide $f(x)$ by $g(x)$, i.e. you find the polynomial $h(x)$ such that $f(x)=g(x)\cdot h(x)$, then the polynomials $f(x)$ and $h(x)$ have the same roots, but the multiplicity of each root of $h(x)$ is one.

You are asking about polynomial over $\mathbb C$; for this field the above results are true. Moreover, $\mathbb C$ is algebraically closed, hence every quartic polynomial has 4 roots in $\mathbb C$, if we count them with multiplicities.

So in your case, you can compute $g(x)=\gcd(f(x),f'(x))$, e.g. using Euclidean algorithm. If $g(x)=1$, then all (complex) roots of $f(x)$ have multiplicity one. If $g(x)=f'(x)$ then $f(x)$ has only one root of multiplicity 4. In the remaining two cases there is a root of multiplicity 2 or 3.

(However, if you're only interested in multiplicity of real roots, the situation is slightly more complicated.)