36 CHAPTER 1. BASIC NOTIONS

(a) x2 +2x+1+ i = 0(b) 4x2 +4ix−5 = 0(c) 4x2 +(4+4i)x+1+2i = 0

(d) x2−4ix−5 = 0

(e) 3x2 +(1− i)x+3i = 0

28. Prove the fundamental theorem of algebra for quadratic polynomials having coef-ficients in C. That is, show that an equation of the form ax2 + bx+ c = 0 wherea,b,c are complex numbers, a ̸= 0 has a complex solution. Hint: Consider the fact,noted earlier that the expressions given from the quadratic formula do in fact serveas solutions.

29. Verify DeMorgan’s laws,

(∪{A : A ∈ C })C = ∩{

AC : A ∈ C}

(∩{A : A ∈ C })C = ∪{

AC : A ∈ C}

where C consists of a set whose elements are subsets of a given set S. Hint: Thissays the complement of a union is the intersection of the complements and the com-plement of an intersection is the union of the complements. You need to show eachset on either side of the equation is a subset of the other side.

30. Find the partial fractions expansion of x6+3x4−x3−4x2−2x−4x(x4−4)

with field of scalars equal

toQ the rational numbers. Then find it for field of scalars equal toR. Note(x4−4

)=(

x2 +2)(

x2−2)

and both of these are irreducible with field of scalarsQ but the sec-

ond is not irreducible with field of scalars R because x2−2 =(

x−√

2)(

x+√

2)

.You will need to first do a division because the degree of the top is larger than thedegree of the bottom.

31. If you have any polynomial p(λ ) with coefficients from a field of scalars, showp(λ )=∏

mk=1 qk (λ )

rk where the rk are positive integers and the polynomials {qk (λ )}are irreducible, meaning they cannot be factored further. ( qk (λ ) = φ (λ )ψ (λ ) thenone of φ (λ ) or ψ (λ ) is a scalar.) Explain why any subset of {qk (λ )} having two ormore entries is relatively prime.