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It seems that a quartic polynomial (degree $4$) either can have $0$ real, $1$ real, $2$ real, or $4$ real roots, and the rest is complex roots. Why can't it have $3$ real roots and $1$ complex?

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    $\begingroup$ If it has complex coefficients, then it can have just 1 real root. If not, the note that if $\alpha$ is a complex root, then so is the complex conjugate of $\alpha$. $\endgroup$ – almagest Jun 21 '16 at 15:21
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    $\begingroup$ Consider $p(x)=(x-1)(x-2)(x-3)(x-i)$ as a counterexample. (Its coefficients are not all real.) $\endgroup$ – Wouter Jun 21 '16 at 15:36
  • $\begingroup$ Not only must there be an even number of complex roots, that must remain true if you count their multiplicities! $\endgroup$ – Hurkyl Jun 21 '16 at 17:21
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    $\begingroup$ For a polynomial with real coefficients: The case with three real roots is just like the case with one real root, so be aware that the premise of the question is not correct. If the number of roots is counted with multiplicity, the number of real roots will be even for an even-degree polynomial, and odd for an odd-degree polynomial. However if you count distinct roots only, i.e. roots equal to other roots are not counted again, you can get any positive number of real distinct root not exceeding the degree. For example a polynomial with roots 7, 7, 9 and 13 has three distinct roots. $\endgroup$ – Jeppe Stig Nielsen Jun 22 '16 at 6:37
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If $z$ is a root of a real polynomial, say $p(z) =\sum_{j=0}^n r_jz^j = 0$, then $\overline{z}$ is also a root of $p$ as $p(\overline{z}) =\sum_{j=0}^n r_j\overline{z}^j = \sum_{j=0}^n r_j\overline{z^j}= \sum_{j=0}^n \overline{r_jz^j} = \overline{p(z)} = 0$. Thus, non-real roots of real polynomials always come in pairs and their number is thus even.

There is no restriction (but the degree) on the number of real roots, though; it is possible that the polynomial of degree $4$ has $3$ real roots too, like $x^2 (x-1)(x-2)$.

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    $\begingroup$ That polynomial actually has 4 real roots, two of them are equal. $\endgroup$ – Somnium Jun 21 '16 at 18:11
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    $\begingroup$ This depends on which convention one uses. Note that OP says there are polynomials of degree 4 with 1 real root. They thus use the convention not to consider the multiplicity when they talk about the number of roots, which I thus did, too. $\endgroup$ – quid Jun 21 '16 at 18:36
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Who told a quartic polynomial can't have 3 real roots.? It can have provided the coefficients are complex... If coefficients have to be real then you must see that to cancel the $i$ of one root there must be one more $i$ in another root.!

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if so, then by the complex conjugate root theorem, there is a conjugate root thus raising the degree from $4$ to $5$ which is not the case.

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  • $\begingroup$ could you expand on why there has to be a conjugate root? $\endgroup$ – costrom Jun 21 '16 at 15:20
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    $\begingroup$ @DeepSea No, it's called the complex conjugate root theorem. $\endgroup$ – Noble Mushtak Jun 21 '16 at 15:21
  • $\begingroup$ @costrom Just replace $i$ by $-i$. The same arithmetic that shows $\alpha$ is a root will show that the complex conjugate is too. $\endgroup$ – almagest Jun 21 '16 at 15:22
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let real roots be $x_1,x_2 , x_3$ and complex be $C$ then polynomial becomes $$p(x)=(x-x_1)(x-x_2)(x-x_3)(x-C)$$ this makes polynomial have complex coefficients as only complex part of equation is $C$ , thus quartic cannot have three real and one complex root

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