Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 293: 31

$Q(x)=(x-i)^{2}(x+i)^{2}$ Zeros: $i $ and $-i,$ each with multiplicity 2

A square of a sum has three terms when expanded, ( first + second$)^{2}=$ (first squared) + (2$\times$first$\times$second) + (second squared), $Q(x)=(x^{2})^{2}+2(x^{2})(1)+(1)^{2}=(x^{2}+1)^{2}$ $x^{2}+1=x^{2}-(-1)=x^{2}-i^{2} \quad $...difference of squares... $=(x-i)(x+i)$ $Q(x)=[(x-i)(x+i)]^{2}=(x-i)^{2}(x+i)^{2}$ Zeros: $i $ and $-i,$ each with multiplicity 2