Question

Solve the equation 2z^4 -14z^3+33z^2-26z+10 = 0, zeC [where C denotes the set of complex numbers] given that it has root 3+i

Answer 1

$\displaystyle\\ 2z^4 -14z^3+33z^2-26z+10 = 0. \\ \textsf{By the fundamental theorem of}\\ \textsf{algebra, if}\, 3 + i \, \textsf{is a root, its conjugate,}\\ 3 - i \, \textsf{is also a root of the polynomial.}\\ (z - (3 + i))(z - (3 - i))\, \textsf{is a factor of the polynomial}\\ (z - (3 + i))(z - (3 - i)) \\= z^2 - (3 + i + 3 - i)z + (3 + i)(3 - i) \\=z^2 - 6z + 10 \\ \textsf{By long division}\\ z^2 - 6z + 10 | 2z^2 - 2z + 1\\ 2z^4 -14z^3+33z^2-26z+10 \\ -(2z^2 - 12z^3 + 20z^2)\\ -2z^3 + 13z^2 -26z + 10\\ -(-2z^3 + 12z^2 - 20z)\\ z^2 - 6z + 10\\ -(z^2 - 6z + 10)\\ \underline{0}\\ \therefore (z^2 - 6z + 10)(2z^2 - 2z + 1) = 0\\ 2z^2 - 2z + 1 = 0\\ \begin{aligned} z &= \frac{2 \pm \sqrt{4 - 8}}{4} \\&= \frac{2 \pm 2i}{2} = 1 \pm i \end{aligned}\\ z = 1 + i, 1 - i\\ \therefore z = 1 + i, 1 - i, 3 + i, 3 - i$