# Polynomial with one rational root or one imaginary root

In my textbook there is an example where we have to find all the roots of $2x^3-5x^2+4x-1$. After applying the Rational Root Theorem we can conclude that $1$ and $1/2$ are two solutions to this equation. Now we have to find the third root.

It says, that we can exclude the irrational or imaginary numbers as the third root since a polynomial can not just have one irrational or one imaginary root.

But why is it so?

(It turns out that $x = 1$ is a double root.)

• – lab bhattacharjee Sep 7 '14 at 10:41
• Since $1$ and $\frac 1 2$ are roots of $2x^3-5x^2+4x-1$, you know that $2x^3-5x^2+4x-1=(x-1)\left(x-\frac 1 2\right)(x-\alpha)$, for a certain $\alpha\in \mathbb C$. Plus, by equating coefficients you must have $1\cdot \frac 1 2 \cdot \alpha=-1$. – Git Gud Sep 7 '14 at 10:44
• I meant $(-1)^3\cdot1\cdot \frac 1 2 \cdot \alpha=-1$. – Git Gud Sep 7 '14 at 10:56
• Yes, that is the neat solution, unlike those below! – almagest Sep 7 '14 at 10:57
• if x=1, polynomial = 0 so (x-1) is a factor of the polynomial. If you use long division $({2x^3-5x^2+4x-1}) \div ({x-1})$ gives $2x^2-3x+1$ so $2x^3-5x^2+4x-1 = (x-1)(2x^2-3x+1) = (x-1)(2x-1)(x-1)$ which makes me think that only $1$ and $\frac{1}{2}$ are roots. With that double $x=1$. Aren't all the roots of this real anyway? – StephanCasey Sep 7 '14 at 11:43

If the roots of the order-$n$ polynomial $p(x)$ are (including multiplicity) $\lambda_1, \ldots, \lambda_n$, then, the polynomial can be written as $$\alpha(x - \lambda_1) \cdots (x - \lambda_n),$$ where $\alpha$ is the leading coefficient of $p$. Now, expanding gives that the constant term is $$p_0 = (-1)^n \alpha \lambda_1 \cdots \lambda_n.$$
If $p$ is rational (so that $\alpha$ is rational) and precisely one root $\lambda_a$ is not rational, then $p_0$ is not rational; contrapositively, if $p_0$ is ration, $p$ cannot have precisely one nonreal root.
Similarly, if $p$ is real and precisely one root $\lambda_a$ is not real, then $p_0$ is not real, and so, if $p_0$ is real, it cannot have a single nonreal root. In fact, we can show that nonreal roots of real polynomials always come in complex conjugate pairs: If $z_0$ is a root of the real polynomial $p$, so that $p(z_0) = 0$, then applying conjugation to both sides gives that $$0 = \overline{p(z_0)} = \bar{p}(\overline{z_0}) = p(\overline{z_0}),$$ and so $\overline{z_0}$ is also a root of $p$.