Zero this, part 4

Let $f(x) = a_nx^n + a_{n - 1}x^{n - 1} + ... + a_1x + a_0$ be a polynomial function. Suppose $f(z) = 0,$ for some complex number $z.$ Denote $\overline{a}$ as the conjugate of $a.$ We have

$\begin{aligned} \overline{f(z)} &= \overline{a_nz^n + a_{n - 1}z^{n - 1} + ... + a_1z + a_0} \\ &= \overline{a_nz^n} + \overline{a_{n - 1}z^{n - 1}} + ... + \overline{a_1z} + \bar{a_0} \\ &= a_n\overline{z^n} + a_{n - 1}\overline{z^{n - 1}} + ... + a_1\overline{z} + a_0 \\ &= a_n\overline{z}^n + a_{n - 1}\overline{z}^{n - 1} + ... + a_1\overline{z} + a_0 \\ &= f(\overline{z}) \\ &= 0. \end{aligned}$

Thus, if $f(z) = 0,$ then $f(\overline{z}) = 0.$ This means that complex solutions to a polynomial must come in pairs. The given polynomial has $5$ total roots, so there are at most $2$ pairs of complex solutions, and at least $\boxed{1}$ real solution.

By-end-pieces the function acts like $x^5$ . So for a sufficiently large negative value of $x$ it is always negative at or below that value, and for a sufficiently large positive value of $x$ it is always positive at or above that value. Since the function is everywhere continuous (as are all polynomial functions), and has both negative and positive values, by Bolzano's Theorem it has at least one zero value.

By considering the function $x(x^2+1)(x^2+2)$ , where the two irreducible quadratics have no zeroes, we show that an example can be achieved with ONLY one zero.