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Given a polynomial of degree four: $ax^4+bx^3+cx^2+dx+e$, with $a,b,c,d,e$ real and $a\neq 0$, how do I derive the condition for there to be exactly distinct 3 real roots (i.e., one root is repeated)? I know that the discriminant is zero when there is a double root. But how do I enforce the condition that there be only one double root?

If this is known, a link to the resource would be appreciated. If not, helpful guidance in how to proceed will be nice. If you already know the answer, that would be great!

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up vote 2 down vote accepted

If $\alpha$ is a root of order $m>1$, then $(x-\alpha)^{m-1}$ must be a common factor of your polynomial $p(x)$ and its derivative $p'(x)$. So you run Euclid's algorithm to find $d(x)=gcd(p(x),p'(x))$. It must be linear in order to $p(x)$ have exactly one (necessarily) double root. Check that the roots of $p(x)/d(x)^2$ are real to make sure that there are 3 real roots.

I don't know, if this is the kind of answer that you wanted, though?

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Thanks for the response. So I tried reducing $p(x)$ by $(x-\alpha)^2$ and $p'(x)$ by $(x-\alpha)$. From each of these, I get quotients $q_1(x)$, $q_2(x)$ and remainders $r_1(x)$, $r_2(x)$ respectively. Would I be correct in saying that if $r_1$ and $r_2$ are both zero, then it is satisfied? – user7815 Jul 9 '11 at 20:26
@doob: You still might have to worry that the roots of $q_1(x)$ might not be real. Also, the point of calculating the gcd of $p(x)$ and $p'(x)$ was that you don't need to guess what $\alpha$ (=the double root) is. You are familiar with Euclid's algorithm, aren't you? Also, if $p(x)$ is divisible by $(x-\alpha)^2$, then there is no need to check the divisibility of $p'(x)$ it is automatic. I brought the derivative into the picture for the sole purpose of computing the gcd, and not having to guess the value of $\alpha$. – Jyrki Lahtonen Jul 9 '11 at 20:45
Running Euclid's algorithm for the general quartic above leaves me with $d(x)$ that is independent of $x$, i.e., involving only the terms $a,b,c,d,e$. I'm guessing this term must be $0$ so that the previous step (linear term) in the algorithm will be the GCD. Is that correct? – user7815 Jul 9 '11 at 23:09
Yes. A general quartic will not have a double root, so a constant gcd is to be expected. You get a condition for the the existence of a double root by setting that quantity to zero. If the gcd were of degree $>1$ (you don't want that to happen), then the polynomial of the previous step would also be zero. – Jyrki Lahtonen Jul 10 '11 at 5:01

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