If 2-3i is a root of z^3-7z^2+25z-39=0, find the other two roots?

Jun 6, 2017

2+3i, and 3.

Explanation:

I polynomial equations of order 3 (cubics like this) written in the form $a {x}^{3} + b {x}^{2} + c x + d = 0$, with three roots, $\alpha$, $\beta$, and $\gamma$, it is a rule that $\alpha + \beta + \gamma = - b \div a$, also $\alpha \cdot \beta + \beta \cdot \gamma + \gamma \cdot \alpha = c \div a$, and that $\alpha \cdot \beta \cdot \gamma = - d \div a$.
Now because $d$ is real and not imaginary, that means $\alpha \cdot \beta \cdot \gamma$ is real. You should know that to get a real number by multiplying imiginary numbers, you times it by it's conjugate, which for $2 - 3 i$ is $2 + 3 i$. So now we know that $\alpha$ and $\beta$ are $2 - 3 i$ and $2 + 3 i$.
We know $\alpha \cdot \beta \cdot \gamma = - d \div a$, so $\left(2 - 3 i\right) \left(2 + 3 i\right) \gamma = - \left(- 39\right) \div 1$. Expanding $\left(2 - 3 i\right) \left(2 + 3 i\right)$ we get $13$, so $13 \gamma = 39$, thus $\gamma = 39 \div 13 = 3$, therefore giving us the other two roots: $2 + 3 i$ and $3$.

Jun 6, 2017

$2 + 3 i \text{ and } 3$

Explanation:

$\text{the complex roots of polynomial equations always }$
$\text{occur in "color(blue)"conjugate pairs}$

$2 - 3 i \text{ is a root "rArr2+3i" is also a root}$

$\text{the quadratic factor formed by these roots is }$

$\left(z - \left(2 - 3 i\right)\right) \left(z - \left(2 + 3 i\right)\right)$

$= \left(\left(z - 2\right) + 3 i\right) \left(\left(z - 2\right) - 3 i\right)$

$= {\left(z - 2\right)}^{2} - 9 {i}^{2}$

$= {z}^{2} - 4 z + 4 + 9$

$= {z}^{2} - 4 z + 13$

$\Rightarrow {z}^{3} - 7 {z}^{2} + 25 z - 39$

$= \textcolor{red}{z} \left({z}^{2} - 4 z + 13\right) \textcolor{m a \ge n t a}{+ 4 {z}^{2}} - 7 {z}^{2} \textcolor{m a \ge n t a}{- 13 z} + 25 z - 39$

$= \textcolor{red}{z} \left({z}^{2} - 4 z + 13\right) \textcolor{red}{- 3} \left({z}^{2} - 4 z + 13\right) \textcolor{m a \ge n t a}{- 12 z} + 12 z \textcolor{m a \ge n t a}{+ 39}$
$\textcolor{w h i t e}{=} - 39$

$= \textcolor{red}{z} \left({z}^{2} - 4 z + 13\right) \textcolor{red}{- 3} \left({z}^{2} - 4 z + 13\right) + 0$

$\Rightarrow \left(z - 3\right) \text{ is a root}$

$\Rightarrow \left(z - 3\right) \left(z - \left(2 - 3 i\right)\right) \left(z - \left(2 + 3 i\right)\right) = 0$

$\Rightarrow \text{roots are " z=3" and } z = 2 \pm 3 i$