Question

Solve the equation $x^5-x^4+8x^2-9x-15=0$, one root being $\sqrt{3}$ and another $1-2\sqrt{-1}$.

Answer 1

$p\left(x\right)=x^5-x^4+8x^2-9x-15=0$

If $\sqrt{3}$ is one root then $-\sqrt{3}$ is also a root

$\Rightarrow (x-\sqrt{3})(x+\sqrt{3})$ is a factor of $p\left(x\right)$

$\Rightarrow x^2-3$ is a factor of $p\left(x\right)$

Dividing $p\left(x\right)$ by $\Rightarrow x^2-3$

$p\left(x\right)=(x^3-x^2+3x+5)(x^2-3)$

Let $q\left(x\right)=x^3-x^2+3x+5$

Now $(-1-2i)$ is a factor of $q\left(x\right)$ then $(-1+2i)$ is also a factor

$\Rightarrow \{x-(-1-2i\left)\right\}\{x-(-1+2i\left)\right\}$ is a factor of $q\left(x\right)$

$\Rightarrow {(x+1)}^2-{\left(2i\right)}^2$ is a factor of $q\left(x\right)$

$\Rightarrow x^2+2x+5$ is a factor of $q\left(x\right)$

Dividing $q\left(x\right)$ by $x^2+2x+5$

$\Rightarrow q\left(x\right)=(x+1)(x^2+2x+5)\Rightarrow (x+1)(x^2+2x+5)=0\Rightarrow x+=0\Rightarrow x=-1$

So the remaining root is $-1$