Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 294: 48

$3,2-i,2+i$

Consider $P(x)=x^3+7x^2+17x-15=0$ In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows: $x^3+7x^2+17x-15=(x-3)(x^2-4x+5)=0$ This gives: $(x-3)(\dfrac{4\pm\sqrt{16-20}}{2})=0$ This can be re-written as: $(x-3)(\dfrac{4\pm\sqrt{-4}}{2})=0$ $(x-3)(\dfrac{4\pm 2i}{2})=0$ Thus the zeroes of $P(x)$ are: $3,2-i,2+i$