help asap!!! polynomial question. need help with 8 b-d

I actually really need help with e

I think it might help if you watch (and study) this video.     

https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials/polynomial-remainder-theorem/v/polynomial-remainder-theorem

e

Given that $(\sqrt{2}x-i\sqrt{3})$ divides P(x) find all roots of P(x)

$\begin{array}{|rcll|} \hline P(x) &=& x^6-6x^5+\frac{17}{2}x^4-7x^3+\frac{21}{2}x^2+3x \\ P(x) &=& x\cdot \left( x^5-6x^4+\frac{17}{2}x^3-7x^2+\frac{21}{2}x+3 \right)\\ \hline \end{array}$

If $(\sqrt{2}x-i\sqrt{3})$ divides P(x) then $(\sqrt{2}x+i\sqrt{3})$ divides P(x)

$\begin{array}{|rcll|} \hline (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) &=& (\sqrt{2}x)^2 -(i\sqrt{3})^2 \\ &=& 2x^2 -(i^2\cdot 3) \quad &| \quad i^2 = -1 \\ &=& 2x^2 -[(-1)\cdot 3] \\ &=& 2x^2 +3\\ \hline \end{array}$

{nl} $\begin{array}{rcll} \hline \left( x^5-6x^4+\frac{17}{2}x^3-7x^2+\frac{21}{2}x+3 \right) : 2x^2 + 3 &=& \frac12x^3-3x^2+\frac72x+1\\ \hline \end{array}$

$\begin{array}{rcll} \Rightarrow P(x) &=& x\cdot (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) \cdot \left( \frac12x^3-3x^2+\frac72x+1 \right)\\ \end{array}$

test x=2: $\Rightarrow \frac12x^3-3x^2+\frac72x+1 = 0$

$\begin{array}{rcll} \hline \left( \frac12x^3-3x^2+\frac72x+1 \right) : x-2 &=& \frac12x^2-2x-\frac12\\ \hline \end{array}$

$\begin{array}{rcll} \Rightarrow P(x) &=& x\cdot (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) \cdot (x-2) \cdot \left( \frac12x^2-2x-\frac12 \right)\\ P(x) &=& x\cdot (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) \cdot (x-2) \cdot \frac12 \cdot \left( x^2-4x-1 \right)\\ \end{array}$

Roots:

$\small{ \begin{array}{|lrccccccccc|} \hline P(x)=& 0 &=& x &\cdot& (\sqrt{2}x-i\sqrt{3}) &\cdot& (\sqrt{2}x+i\sqrt{3}) &\cdot& (x-2) \cdot \frac12 \cdot &\left( x^2-4x-1 \right)&\\ \text{1. Root} & && \mathbf{x_1 = 0} \\ \text{2. Root} & & & & & \sqrt{2}x-i\sqrt{3} = 0 \\ & & & & & \sqrt{2}x=i\sqrt{3} \\ & & & & & \mathbf{x_2=i\frac{\sqrt{3} } {\sqrt{2}}} \\ \text{3. Root} & & & & & & & \sqrt{2}x+i\sqrt{3} = 0 \\ & & & & & & & \sqrt{2}x=-i\sqrt{3} \\ & & & & & & & \mathbf{x_3=-i\frac{\sqrt{3} } {\sqrt{2}}} \\ \text{4. Root} & & & & & & & & & x-2=0 \\ & & & & & & & & & \mathbf{x_4=2} \\ \text{5.6. Root} & & & & & & & & & & x^2-4x-1 = 0\\ & & & & & & & & & & x = \frac{4\pm\sqrt{4^2-4\cdot(-1)}}{2} \\ & & & & & & & & & & x = \frac{4\pm\sqrt{4\cdot 5}}{2} \\ & & & & & & & & & & x = \frac{4\pm 2 \sqrt{ 5}}{2} \\ & & & & & & & & & & x = 2\pm \sqrt{ 5} \\ & & & & & & & & & & \mathbf{x_5= 2+ \sqrt{ 5}} \\ & & & & & & & & & & \mathbf{x_6 = 2- \sqrt{ 5}} \\ \hline \end{array} }$