Question

The following equation is given z^3+Pz^2+6z+Q=0, P and QeR has a solution 5-i
(a)Find the other two solutions of the equation (b) Determine the value of P and Q

Answer 1

$\displaystyle z^3+Pz^2+6z+Q=0, P \, \textsf{and}\, Q\in R\, \textsf{has a solution}\, 5 - i\\ (5 - i)^3 + P(5 - i)^2 + 6(5 - i) + Q = 0\\ 125 + 3(5)^2(-i) + 3(5)(-i)^2 + (-i)^3 + P(25 - 10i - 1) + (30 - 6i) + Q = 0\\ (125 - 75i - 15 + i) + P(25 - 10i - 1) + (30 - 6i) + Q = 0\\ (110 - 74i) + P(24 - 10i) + (30 - 6i) + Q = 0 + 0i\\ \textsf{Comparing the real and imaginary parts to}\, 0\\ 110 + 24P + 30 + Q = 0\\ -80 - 10P = 0\\ 10P = -80, P = \frac{-80}{10} = -8\\ 140 + 24P + Q = 0\\ 140 + 24(-8) + Q = 0\\ 140 - 192 + Q = 0\\ Q = 52\\ \therefore P = -8, Q = 52.\\ \therefore z^3+Pz^2+6z+Q = z^3-8z^2+6z+52=0\\ \textsf{By the fundamental theorem of Algebra,}\\ \textsf{if the root of a polynomial equation}\\ \textsf{is a complex number, the polynomial}\\\textsf{has its conjugate as another root.}\\ \textsf{Therefore}\, 5 + i \, \textsf{is also a root of the equation.}\\ \therefore (z - (5 - i))(z - (5 + i)) \, \textsf{is a factor of the polynomial.}\\ \begin{aligned} (z - (5 - i))(z - (5 + i)) &= z^2 - (5 - i + 5 + i)z + (5 - i)(5 + i)\\&=z^2 - 10z + 26 \end{aligned}\\ \frac{z^3-8z^2+6z+52}{z^2 - 10z + 26} = z + 2 \\ (a)\\ z = -2, 5 + i\, \textsf{are the other two solutions}\\\textsf{to the equation}\\ (b)\\ P = -8, Q = 52$