Question #a65c9

Mar 24, 2016

$k = \frac{2}{3}$

$P \left(x\right) = 3 \left(x + \frac{2}{3}\right) \left(x - \sqrt{15} i\right) \left(x + \sqrt{15} i\right)$

Explanation:

Since 1 zero is purely imaginary, we denote it as $i y$, where $y$ is the imaginary part of the zero.

The conjugate of $i y$ is $- i y$. We know from the complex conjugate root theorem that roots of a polynomial with real coefficients exists as conjugate pairs. Therefore, $- i y$ is also a root of the cubic equation.

We let the last root be $z$. $z$ is purely real.

We can then rewrite $P \left(x\right)$ in product form. We know that the leading coefficient is 3.

$P \left(x\right) = 3 \left(x - z\right) \left(x - i y\right) \left(x + i y\right)$

$= 3 \left(x - z\right) \left({x}^{2} + {y}^{2}\right)$

Upon expanding, we get

$P \left(x\right) = 3 {x}^{3} + \left(- z\right) {x}^{2} + \left({y}^{2}\right) x + \left(- z {y}^{2}\right)$

Comparing the coefficients with the original $P \left(x\right)$

• Coefficient of ${x}^{0}$ term

$- z {y}^{2} = 10$

• Coefficient of ${x}^{1}$ term

${y}^{2} = 15$

• Coefficient of ${x}^{2}$ term

$- z = k$

Solving the 3 equations, we have

$y = \pm \sqrt{15}$
$z = - \frac{2}{3}$
$k = \frac{2}{3}$

Hence, the roots of P(x) are $- \frac{2}{3}$, $\sqrt{15} i$, $- \sqrt{15} i$.

$P \left(x\right)$ in linear product form

$P \left(x\right) = 3 \left(x + \frac{2}{3}\right) \left(x - \sqrt{15} i\right) \left(x + \sqrt{15} i\right)$

$= \left(3 x + 2\right) \left({x}^{2} + 15\right)$