# A real polynomial P(x) is the product of real linear and real quadratic factors. Can you that if a real quadratic factor has one zero, a+bi, then the other zero must be the complex conjugate, a-bi?

## Hint: Let the quadratic be $\alpha {x}^{2} + \beta x + \gamma$ and use the properties of complex conjugates

Feb 23, 2017

See below.

#### Explanation:

$P \left(a + i b\right) = \text{Re"(P(a+ib))+i "Im} \left(P \left(a + i b\right)\right)$

where $\text{Re} \left(\cdot\right)$is the real component of $P \left(a + i b\right)$ and
$\text{Im} \left(\cdot\right)$ is the real part of the imaginary component. Now

$P \left(a + i b\right) = 0 \to \text{Re} \left(P \left(a + i b\right)\right) = 0$ and $\text{Im} \left(P \left(a + i b\right)\right) = 0$

but

$\text{Re} \left(z\right) = \frac{z + \overline{z}}{2}$ where $\overline{z}$ is the conjugate of $z$ and
$\text{Im} \left(z\right) = \frac{z - \overline{z}}{2 i}$ so now if $P$ has real coefficients, $P \left(x\right) = \overline{P} \left(x\right)$ and $\overline{P \left(x\right)} = P \left(\overline{x}\right)$ and also

$\text{Re} \left(P \left(a + i b\right)\right) = \frac{P \left(a + i b\right) + P \left(a - i b\right)}{2}$

and

$\text{Im} \left(P \left(a + i b\right)\right) = \frac{P \left(a + i b\right) - P \left(a - i b\right)}{2 i}$

So if $P \left(a + i b\right) = 0$ then

$\left\{\begin{matrix}P \left(a + i b\right) + P \left(a - i b\right) = 0 \\ P \left(a + i b\right) - P \left(a - i b\right) = 0\end{matrix}\right.$

concluding $P \left(a - i b\right) = 0$

Another version.

If $P \left(a + i b\right) = 0$ then $P \left(x\right) = {Q}_{1} \left(x\right) \left(x - \left(a + i b\right)\right)$ but $P \left(x\right)$ is a real polynomial. That means all its coefficients real. So $P \left(x\right)$ must have another root ${x}_{0}$ such that $\left(x - {x}_{0}\right) \left(x - \left(a + i b\right)\right) = {x}^{2} + \alpha x + \beta$ with $\alpha , \beta$ real. Or

${x}^{2} - \left({x}_{0} + a + i b\right) x + {x}_{0} \left(a + i b\right) = {x}^{2} + \alpha x + \beta$

or

$\left\{\begin{matrix}- \left({x}_{0} + a + i b\right) = \alpha \\ {x}_{0} \left(a + i b\right) = \beta\end{matrix}\right.$

The feasible solution is

${x}_{0} = a - i b$

because

$- \left(a - i b + a + i b\right) = - 2 a = \alpha$ and also

$\left(a - i b\right) \left(a + i b\right) = {a}^{2} + {b}^{2} = \beta$

with both $\alpha , \beta$ real.

So if $P \left(x\right)$ is a real polynomial and $P \left(a + i b\right) = 0$ then necessarily $P \left(a - i b\right) = 0$ and $P \left(x\right) = {Q}_{2} \left(x\right) \left({x}^{2} + \alpha x + \beta\right)$ with $\alpha = - 2 a$ and $\beta = {a}^{2} + {b}^{2}$

Feb 24, 2017

#### Explanation:

If one of the real quadratic factors is:

$\alpha {x}^{2} + \beta x + \gamma$

then using the quadratic formula, it has zeros:

$\frac{- \beta \pm \sqrt{{\beta}^{2} - 4 \alpha \gamma}}{2 \alpha}$

If ${\beta}^{2} - 4 \alpha \gamma \ge 0$ then the square root is real and the zeros are real, resulting in real linear factors.

Otherwise, ${\beta}^{2} - 4 \alpha \gamma < 0$ so:

$\sqrt{{\beta}^{2} - 4 \alpha \gamma} = \left(\sqrt{4 \alpha \gamma - {\beta}^{2}}\right) i$

resulting in a complex conjugate pair of non-real zeros.