1.16. EXERCISES 45
28. Write x3 +27 in the form (x+3)(x2 +ax+b
)where x2 +ax+b cannot be factored
any more using only real numbers.
29. Completely factor x4 +16 as a product of linear factors.
30. Factor x4 +16 as the product of two quadratic polynomials each of which cannot befactored further without using complex numbers.
31. If z,w are complex numbers prove zw = z w and then show by induction ∏nj=1 z j =
∏nj=1 z j. Also verify that ∑
mk=1 zk = ∑
mk=1 zk. In words this says the conjugate of a
product equals the product of the conjugates and the conjugate of a sum equals thesum of the conjugates.
32. Suppose p(x) = anxn +an−1xn−1 + · · ·+a1x+a0 where all the ak are real numbers.Suppose also that p(z) = 0 for some z ∈ C. Show it follows that p(z) = 0 also.
33. Show that 1+ i,2+ i are the only two zeros to p(x) = x2−(3+2i)x+(1+3i) so thezeros do not necessarily come in conjugate pairs if the coefficients are not real.
34. I claim that 1 =−1. Here is why. −1 = i2 =√−1
√−1 =
√(−1)2 =
√1 = 1. This
is clearly a remarkable result but is there something wrong with it? If so, what iswrong?
35. Suppose p(x) = anxn + an−1xn−1 + · · ·+ a1x+ a0 is a polynomial and it has n ze-ros, z1,z2, · · · ,zn listed according to multiplicity. (z is a root of multiplicity m ifthe polynomial f (x) = (x− z)m divides p(x) but (x− z) f (x) does not.) Show thatp(x) = an (x− z1)(x− z2) · · ·(x− zn) .
36. Give the solutions to the following quadratic equations having real coefficients.
(a) x2 −2x+2 = 0(b) 3x2 + x+3 = 0(c) x2 −6x+13 = 0
(d) x2 +4x+9 = 0
(e) 4x2 +4x+5 = 0
37. Give the solutions to the following quadratic equations having complex coefficients.Note how the solutions do not come in conjugate pairs as they do when the equationhas real coefficients.
(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
38. 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 ΜΈ= 0 has a complex solution. Hint: Consider the fact,noted earlier that the expressions given from the quadratic formula do in fact serveas solutions.