What can be said about the roots of $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2 )x^2 + b(a + c)x + ac$?

Question

Let a, b and c be real numbers. Then the fourth degree polynomial in $x$, $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2 )x^2 + b(a + c)x + ac$

(a) Has four complex (non-real) roots

(b) Has either four real roots or four complex roots

(c)Has two real roots and two complex roots

(d) Has four real roots

This question is from a book called 'Test of Mathematics at the $10+2$ Level' published by the Indian Statistical Institute

We can see that the expression factorizes to $(ax^2+bx+c)(a+bx+cx^2)$. If $\alpha,\beta$ are the roots of the first factor then $1/\alpha , 1/\beta$ are the roots of the second expression. And we know that complex roots occur in conjugation. So, we can easily understand that option (b) is a correct.

But today, I want to solve this problem in some different way in which it does not require us to factorize the expression in a hit and trial way as I did. Even if we have to factorize, we must use a proper way to find the factors. Please help.

Answer 1

Hint: It is $$ac\left(x^2+\frac{1}{x^2}\right)+b(a+c)\left(x+\frac{1}{x}\right)+a^2+b^2+c^2=0$$ and now substitute $$t=x+\frac{1}{x}$$

Answer 2

Presumably "complex roots" in (b) and (c) should be "nonreal complex roots" as in (a), lest some of the answers (a)-(d) overlap.

Hint Write the given quartic polynomial as $$f(x) = \sum_{k = 0}^4 d_n x^n ,$$ so that $$d_4 = d_0 = ac, \qquad d_3 = d_1 = b (a + c), \qquad \textrm{and} \qquad d_2 = a^2 + b^2 + c^2 .$$

If $0$ were a root of $f$, then $d_4 = d_0 = f(0) = 0$, which contradicts the hypothesis that $\deg f = 4$, so $f(0) \neq 0$. Now, since $f$ is palindromic, we have that $$f(x) = x^4 p\left(\frac{1}{x}\right) .$$ In particular, if $r$ is a root of $p$, so is $\frac{1}{r}$.

You've already pointed out that the set of roots is closed under complex conjugation, so if $r$ is a nonreal complex root it has four nonreal complex roots: $r, 1 / r, \bar r, 1 / \bar r$, counting multiplicity. Hence if $f$ has a real root then all four of its roots (again counting multiplicity) are real.