So after calculating and factorising a little it seems that $x^4 - 2x^3 + 13x^2 - 18x - 36 = 0$ is the solution to the problem , with $k = 13$ and $p = - 18$ for the coefficients , and the roots of the polynomial being equal to $(x + 3i)(x - 3i)(x - 1 + \sqrt{3}i)(x - 1 - \sqrt{3}i)$
Also , this is a great way to solve it and I did not even think that would so simple.It seems like a valid way to get the coefficients,and yet I've never seen someone do it in that way?
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Just updated the original post thank you,the question does state that it is indeed a...
So I have tried using long division and synthetic division,but since we have to either divide using a complex number (which is not too much of a problem) or we use the known factor of $x^2 + 9$ ,but the unknown coefficients are still both present in the new polynomial created, namely being $x^2...
Firstly,thanks for welcoming me :) .Secondly , the question states that we have to prove all the factors are complex,which answers your question that the remaining $x^2+ax+b=0$ would not have any real roots as it is a factor.Thanks you for your contribution.
One of the solutions to
x4-2x3+kx2+px+36 = 0 is x = 3[FONT=times new roman]i
[FONT=arial]Prove that this polynomial has no real solutions (roots) and find the real values of k and p...