Why can a quartic polynomial never have three real and one complex root?
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?
$\endgroup$ 44 Answers
$\begingroup$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 $x^2 (x-1)(x-2)$.
$\endgroup$ 2 $\begingroup$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.!
$\endgroup$ $\begingroup$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.
$\endgroup$ 3 $\begingroup$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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