Section 2-2: Remainder and Factor Theorems
 
 
 Two theorems are important when dealing with synthetic substitution.
   1)  The remainder Theorem
  2)  The factor Theorem
 
 The remainder theorem:  When a polynomial P(x) is divided by x - a , the remainder is P(a)!
 Remember basic arithmetic:  Dividend = Divisor  x Quotient + Remainder
    If we apply this to synthetic division, we get the following result:
 
Divide: x3 + 5x2 + 5x - 3 by x + 2  keep in mind it must be in the form x-(-2)!

    Use synthetic division:  -2|            1        5        5        -3
                                                                    -2       -6         2
                                                           1        3       -1        -1
                                                            x2 + 3x - 1            -1
                                                            Quotient        Remainder
                    The correct form of the answer:  x2 + 3x - 1 - 1/(x + 2)


The factor theorem:  For a polynomial P(x), x - a is a factor iff P(a) = 0.
        This says that if using synthetic division, the divisor is a factor (also the quotient) when the last number is zero!  If the last number isn't zero, what you divided by is not a factor and neither is the quotient!
 
 Sample problems!
    1)  Find the quotient and remainder of  x3 + 8x2 -5x - 84 divided by x + 5
 

   2)  Determine if x - 3 is a factor of x3 + 5x2 - 17x - 21
 

   3)  Show that x + 2 is a factor of 2x3 + 3x2 - 8x - 12 and then factor completely.
 
 

   4)  Show that  x + 3 and x - 4 are factors of x4 - 2x3 - 13x2 + 14x + 24 and then factor completely.
 Answers are at the bottom of this page!


 
  
 
 
 
 

Answers
1)  -5|        1        8        -5        -84
                          -5      -15       100
                  1       3      -20         16
                x2 + 3x - 20 + 16/(x + 5)
 
2)  3|        1        5        -17        -21
                          3         24         21
                1        8           7          0
                Since the remainder is 0, the divisor is a factor!
3)  -2|        2        3        -8        -12
                          -4         2          12
                  2      -1       -6            0
                  (2x2 - x - 6)(x + 2) = (2x + 3)(x - 2)(x + 2)
4)  -3|        1        -2        -13        14        24
                            -3        15        -6        -24
                  1        -5         2          8           0
       4|        1        -5        2        8
                             4       -4       -8
                  1       -1        -2        0
                    (x2 - x - 2)(x + 3)(x - 4) = (x -2)(x + 1)(x + 3)(x - 4)