Question #a65c9

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Question #a65c9

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Answer 1 · 2016-03-24T07:26:05

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

$P (x) = 3 (x + \frac{2}{3}) (x - \sqrt{15} i) (x + \sqrt{15} i)$ $P(x) = 3(x + 2/3)(x - sqrt15 i)(x + sqrt15 i)$

Explanation:

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

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

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

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

$P (x) = 3 (x - z) (x - i y) (x + i y)$ $P(x) = 3(x - z)(x - iy)(x + iy)$

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

Upon expanding, we get

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

Comparing the coefficients with the original $P (x)$ $P(x)$

Coefficient of $x^{0}$ $x^0$ term

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

Coefficient of $x^{1}$ $x^1$ term

$y^{2} = 15$ $y^2 = 15$

Coefficient of $x^{2}$ $x^2$ term

$- z = k$ $-z = k$

Solving the 3 equations, we have

$y = \pm \sqrt{15}$ $y = +-sqrt15$
$z = - \frac{2}{3}$ $z = -2/3$
$k = \frac{2}{3}$ $k = 2/3$

Hence, the roots of P(x) are $- \frac{2}{3}$ $-2/3$ , $\sqrt{15} i$ $sqrt15 i$ , $- \sqrt{15} i$ $-sqrt15 i$ .

$P (x)$ $P(x)$ in linear product form

$P (x) = 3 (x + \frac{2}{3}) (x - \sqrt{15} i) (x + \sqrt{15} i)$ $P(x) = 3(x + 2/3)(x - sqrt15 i)(x + sqrt15 i)$

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

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Question #a65c9

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