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

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.

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 $$t=x+\frac{1}{x}$$
Hint Glancing at the given quartic polynomial, which we write as $$f(x) = \sum_{k = 0}^4 d_n x^n$$, has palindromic coefficients: $$d_4 = d_0 = ac$$, $$d_3 = d_1 = b (a + c)$$. So, for $$x \neq 0$$, we have that $$f(x) = x^4 p\left(\frac{1}{x}\right) .$$ Thus, if $$r$$ is a nonzero root of $$p$$, so is $$\frac{1}{r}$$.
Additional hint You've already pointed out that the set of roots is closed under complex conjugation, so if $$r$$ is a nonreal complex root, so are $$1 / r, \bar r, 1 / \bar r$$.