Answer: All the four roots of the given equation are
\(x=3i,~-3i,~1+2i,~1-2i.\)
Step-by-step explanation: Given that (1 + 2i) is one of the roots of the following fourth degree polynomial equation:
\(x^4-2x^3+14x^2-18x+45=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\)
We are to find all the roots of the above equation.
Since the given equation is of degree 4, so there will be 4 roots of the equation.
We know that the complex roots of any polynomial equation occur in complex conjugate pairs.
Since (1 + 2i) is root of equation (i), so (1 - 2i) is also a root of equation (i).
FACTOR THEOREM : It states that if x = a is a root of the polynomial equation Q(x) = 0, then (x - a) is a factor of the polynomial Q(x).
That is, by factor theorem, we can say
\(\{x-(1+2i)\}~\textup{and}~\{x-(1-2i)\}\) are factors of the polynomial \(x^4-2x^3+14x^2-18x+45.\)
So, their product is also a factor of the polynomial.
We have
\(\{x-(1+2i)\}\{x-(1-2i)\}\\\\=x^2-(1+2i+1-2i)x+(1+2i)(1-2i)\\\\=x^2-2x+(1-4i^2)\\\\=x^2-2x+(1+4)\\\\=x^2-2x+5.\)
∴ (x² - 2x + 5) is a factor of te polynomial and so we have
\(x^4-2x^3+14x^2-18x+45\\\\=x^2(x^2-2x+5)+9(x^2-2x+5)\\\\=(x^2+9)(x^2-2x+5).\)
Therefore, the other two roots of equation (i) are given by
\(x^2+9=0\\\\\Rightarrow x^2=-9\\\\\Rightarrow x^2=9i^2\\\\\Rightarrow x=3i,~~-3i.\)
Thus, all the four roots of the given equation are
\(x=3i,~-3i,~1+2i,~1-2i.\)
