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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.

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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 $$t=x+\frac{1}{x}$$

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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 $1 / r, \bar r, 1 / \bar r$.

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