Let $a,b$ and $c$ be real numbers. Then prove that the fourth degree polynomial in $x$
$acx^4+b(a+c)x^3+(a^2+b^2+c^2)x^2+b(a+c)x+ac$ has either 4 real roots or 4 complex roots.
I have never solved a fourth degree polynomial and don't know the conditions for it to have real/complex roots. How do we approach this problem?

  • $\begingroup$ You know that if a polynomial with real coefficients has a root $z = a + bi$ that's not real, then $a - bi$ is also a root. So there are only three possibilities: 4 real, 2 real + 2 complex, 4 complex. Your job is to figure out why the middle condition cannot occur. In general, it can occur, as in $(x^2 - 1) (x^2 + 1)$. But presumably the arrangement of coefficients in your problem places some constraint on the possible roots preventing this from happening. $\endgroup$ – John Hughes Mar 28 '14 at 14:54
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    $\begingroup$ One can notice that the coefficients are palindromic. That is, if $\alpha$ is a root then $1/\alpha$ is a root. So one has to eliminate the roots being of the following type: $\alpha,\alpha^{-1}$ real, $\beta, \bar \beta$ complex of unit modulus. $\endgroup$ – P Vanchinathan Mar 28 '14 at 15:34
  • $\begingroup$ @PVanchinathan Palindromic coefficients imply roots that are reciprocal of each other? $\endgroup$ – idpd15 Mar 30 '14 at 8:42
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    $\begingroup$ That's right, the polynomial obtained by changing $x\mapsto \frac1x$ and multiplying by a power of $x$ would be the original polyomial. $\endgroup$ – P Vanchinathan Mar 30 '14 at 13:27

Notice that the given expression,

$$acx^4+b(a+c)x^3+(a^2+b^2+c^2)x^2+b(a+c)x+ac = 0$$

can be factorized into (see below for derivation):

$$(ax^2 + bx + c)(cx^2 + bx + a) = 0$$

The discriminant for each is the same, $b^2 - 4ac$.

If this common discriminant is zero or more, then the roots for both $ax^2 + bx + c = 0$ and $cx^2 + bx + a = 0$ are all real (possible with multiplicity $2$). If not, they are all imaginary. Hence the roots are either all real or all imaginary.

Here's how I did the factorization:

$$acx^4+b(a+c)x^3+(a^2+b^2+c^2)x^2+b(a+c)x+ac$$ $$=acx^4 + abx^3 + bcx^3 + a^2x^2 + b^2x^2 + c^2x^2 + abx + bcx + ac$$ $$=(acx^4 + abx^3 + a^2x^2) + (bcx^3 + b^2x^2 + abx) + (c^2x^2 + bcx + ac)$$ $$=ax^2(cx^2 + bx + a) + bx(cx^2 + bx + a) + c(cx^2 + bx + a)$$ $$=(ax^2 + bx + c)(cx^2 + bx + a)$$


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