Math 3118, section 4
Spring 2002
Review 2 solutions

Answers are shown in magenta.

1.  Let f(x) = x³ + 3x² - 3,  where  x  is a real variable.

1. Fill in the following table of values:
             

   x	-4	-3	-2	-1	0	1	2
   f(x)	-19	-3	1	-1	-3	1	17

   
    
2. On which of the following intervals does the Intermediate Value Theorem say that  f(x)  has a real zero?
     
   • Between  -4  and  -3      No sign change, so IVT does not predict a zero.
   • Between  -3  and  -2      Changes sign, so there is a zero.
   • Between  -2  and  -1      Changes sign, so there is a zero.
   • Between  -1  and   0      No sign change, so IVT does not predict a zero.
   • Between   0  and   1      Changes sign, so there is a zero.
   • Between   1  and   2      No sign change, so IVT does not predict a zero.
      
3. (Optional) if you have a graphing calculator, plot the graph of  f.  
     

2.   Compute the following complex numbers:

1. (3 + 4i)·(4 + i) =12 + 3i+ 16i+ 4i² = 12 + 3i+ 16i- 4 = 8 + 19i
    
2. (3 + 2i)·(2 - i) = 6 - 3i+ 4i- 2i² = 6 - 3i+ 4i+ 2 = 8 + i
    
3. the complex conjugate of 4 + 3i is   4 - 3i
    
4. the complex conjugate of - 6i is   6i
    
5. = = =
    
6. = = =
    

3.  Use the method of exercise 10.2.9 to approximate . Start with  a=3, and stop when the difference between the upper and lower estimates is less than .000001.

• First stage
• Start with  a = 3
• Divide into 10:    b = 10 ÷ a = =
• Average previous two:    c = =
• Square to check:   c² = =
• Second stage:
• a = = previous c
• Divide into 10:    b = 10 ÷ a = =
• Average previous two:    c = · ( + ) = · = = 3.162280702
• Square to check:   c² = = = 10.00001924.
• Third stage:
• a = 3.162280702 = previous c
• Divide into 10:    b = 10 ÷ a = 3.162274619
• Average previous two:    c = 3.16227766
• Square to check:   c² = 10 + e,   where   e   is approximately equal to   10^-11.
    
• So, the answer is   3.16227766.
   

4.   Find all of the zeros of the polynomial:

              x³ + 2x² - x - 2

Factor this polynomial as a product of 3 linear factors.
  

5.   Find a rational zero of each polynomial, and then factor the polynomial as a product of a linear factor and a quadratic factor.

1. 2x³ - 7x² + 5x - 1
    
   • The rational numbers which  conceiveably could be zeros of this polynomial are the (positive and negative) rational numbers [quotients of integers] whose numerators are factors of 1 and whose denominators are factors of 2, thus:     1, -1, 1/2, -1/2.
   • Substituting 1 and -1 gives odd numbers (thus not zero).
     Substituting 1/2 gives 0.
     Substituting -1/2 gives -11/2 (also not zero).
     So, 1/2 is the only rational zero.
   • Using division of polynomials, we find the following factorization:
            2x³ - 7x² + 5x - 1 = (x - ¹/₂)(2x² - 6x + 2)
      
2. x³ - 7x² +5x + 1    
• The rational numbers which  conceiveably could be zeros of this polynomial are 1 and -1.
  Substituting 1 gives 0.
  Substituting -1 gives -14 (certainly not zero).
  So, 1 is the only rational zero.
• Using division of polynomials, we find the following factorization:
         x³ - 7x² + 5x + 1 = (x - 1)(x² - 6x - 1)
   

6.   For each polynomial in the preceding problem, find all of its zeros (whether rational, real, or complex).
Suggestion: Use the quadratic formula to find the zeros of the quadratic factor.

• Solution:
• 1. 2x³ - 7x² + 5x - 1 = (x - 1/2)(2x² - 6x + 2)
     • Applying the quadratic formula to 2x² - 6x + 2, we find the roots ;
       thus .
     • So, the zeros of the third degree polynomial are  
  2. x³ - 7x² +5x - 1 = (x - 1)(x² - 6x - 1)
  • Applying the quadratic formula to  x² - 6x -1,  we find the roots  ,  
    thus   and  .
  • So, the zeros of the third degree polynomial are  1,  ,  and  .

  7.  

  1. Explain why   and    are algebraic.
      
     Because they are roots of the polynomials  x² - 2  and  x² - 5  respectively.
  
   
  2. Write  ( + )²  as  A + B, where  A and  B are rational numbers.
     Solution: ( + )² = 7 + 2
      
  3. Write  ( + )⁴  as  A + B, where  A and  B are rational numbers .
     Solution: ( + )⁴ = (7+ 2)² = 89 + 28.
      
  4. Find a rational number  S  such that ( + )⁴  + S( )  is a rational number.
      
     Substituting from parts b and c, we have:
                   ( + )⁴ + S( + )² = (89 + 28) + S(7 + 2).
     We can re-arrange this as:
                   ( + )⁴  + S( + )² = (89 + 7S) + (28 + 2S).
     To solve the problem, we must choose  S  so that   disappears from the answer.
     To do this, we need to have  28 + 2S = 0.  Therefore  S = -14.
      
  5. Find rational numbers  S and  T such that ( + )⁴  + S( + )²  + T = 0.
      
     Taking  S = -14  and substituting into the second long equation in part d, we have:
                   ( + )⁴  - 14( + )² = (89 + 7(-14)) + (28 + 2(-14)), or simply:
                   ( + )⁴  - 14( + )² = -9.     (Note that  89 + 7(-14) = 89 - 98 = -9.)
     Adding  9  to both sides of the equation, we get:               ( + )⁴  - 14( + )² + 9 = 0.
     In other words,   T = 9.
      
     (This shows that  +  is also algebraic, since we now have the equation:
                     ( + )⁴  - 14( + )²  + 9 = 0. )
      
     As an alternative, we can combine parts  d  and  e,   thus solving simultaneously for  S  and  T,
     as follows:
      
     Substituting from parts b and c, the equation  ( + )⁴  + S( + )²  + T = 0
     is the same as
                   (89 + 28) + S(7 + 2) + T = 0.
     We can re-arrange this as:
                   (89 + 7S + T) + (28 + 2S) = 0.
     So, we have these equations:
                   89 + 7S + T = 0
                   28 + 2S = 0.
     From the second equation we see that  S = -14. Substituting into the first equation, we have:
                    89 - 98 + T = 0, or  T = 9.
     (Once again, we see that  +  is also algebraic, since we have obtained the equation:
                     ( + )⁴  - 14( + )²  + 9 = 0. )

  8.

  1. (Exercise 10.1.13)   No, the irrational numbers do not form a subfield of the real numbers.
     Since the sum of two elements of a subfield is supposed to again be an element of the subfield,
     it's sufficient to note that the sum of two irrational numbers need not be irrational, as shown
     in Exercise 10.1.11. If we want further corroboration of this answer, we can also note that
     the product of two irrational numbers need not be irrational, as shown in Exercise 10.1.12.
       
  2. (Exercise 10.2.1) To six decimal places, is approximately equal to 1.414214… .
     (The dots just indicate that the decimal expansion of doesn't end at that stage.)
     Therefore, is approximately equal to 1.414224…. So, one choice of a rational number
     (or terminating decimal) in this range is 1.414220 = 1.41422.
       
  3. (Exercise 10.4.11)
                             (1 + i)² = (1 + i)·(1 + i) = 1 + 2i + i² = 2i. (FOIL; then recall that  i² = -1).
     Therefore:
                             (1 + i)⁴ = (1 + i) ²·(1 + i) ² = 2i·2i = 4 i² = -4,
     and finally:
                             (1 + i)⁵ = (1 + i) ⁴·(1 + i) = -4·(1 + i) = -4 - 4i.   

     

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