This is my question :
Given that $2 - 3i$ is a root of the equation :
$x^4 - 10x^3 + 48 x^2 -122x + 143 =0$
Solve the equation , giving the answer in exact form.
I am going to devide the LHS for the given root , $2-3i$ . Yet , it is difficult to find the quotient .
On
Note that the coefficients of the polynomial are all real. So complex roots must appear in conjugate pairs. Hence if $2-3i$ is a root, then $2+3i$ should also be a root of the polynomial. Now $(x-(2-3i))(x-(2+3i))=x^2-4x+13$. Note that the given polynomial is $$x^2(x^2-4x+13)-6x(x^2-4x+13)+11(x^2-4x+13)=(x^2-4x+13)(x^2-6x+11)$$
On
Complex roots occurs in conjugate pairs
$$x=2-3i \text{ is a root }\implies x=2+3i\text{ is a root}$$ $$\begin{align}x-2&=3i&x-2=-3i\\\text{Multiplying both }&\text{gives...}\\(x-2)^2&=-9\\x^2-4x+13&=0\end{align}\tag*{}$$
Now we can either factor the original polynomial or perform long division. I'll factor the entire polynomial. $$\begin{align}x^4-10x^3+48x^2-122x+143&=0\\x^4-4x^3+13x^2-6x^3+35x^2-122x+143&=0\\x^2(x^2-4x+13)-6x^3+24x^2-78x+11x^2-44x+143&=0\\x^2(x^2-4x+13)-6x(x^2-4x+13)+11(x^2-4x+13)&=0\\(x^2-4x+13)(x^2-6x+11)&=0\\\hline\text{Now, we solve the equations}\\x^2-4x+13&=0\\x^2-4x+4&=-9\\(x-2)^2&=-9\\x&=2\pm3i\\\hline x^2-6x+11&=0\\x^2-6x+9&=-2\\(x-3)^2&=-2\\x&=3\pm\sqrt 2i\end{align}$$
Since the coefficients of your polynomial are real, the complex roots will appear in conjugates. Therefore you have 2+3i and 2-3i as roots of your polynomial. Upon multiplying ( x-2-i) and (x-2+3i) we get x^2 -4x + 13 which is a quadratic factor of your original polynomial. Now divide the original polynomial by x^2 -4x + 13 to get x^2 -6x +11. This new polynomial has 3+(2^(1/2))i and 3-(2^(1/2))i as its roots.