Section 2-6:  Solving Polynomial Equations
 
We will review three different ways of solving Polynomial Equations.  We will refresh your memory on:
       1)  Factoring by grouping
     2)  Using the quadratic form
     3)  The rational root theorem.
Grouping:

                    Sample problems:

                            1)  x3 + 2x2 - 9x - 18 = 0
                                  x2( x + 2) - 9(x + 2) = 0  group the first two and last two!
                                  (x2 - 9)(x + 2) = 0  Factor out the common factor!
                                  (x - 3)(x + 3)(x + 2) = 0
                                        The solutions are x = 3, x = -3 and x = -2

                            2)  x3 - 4x2 - 9x + 36 = 0
                                    x2(x - 4) - 9(x - 4) = 0  Group the first two and last two!
                                    (x2 - 9)(x - 4) = 0  Take out the common factor!
                                    (x - 3)(x + 3)(x -4) = 0
                                        The solutions are x = 3, x = -3 and x = 4


Quadratic form

                    Sample problems

                       1)  x4 - 2x2 -3 = 0  This is not a quadratic, but can look like one!

                              (x2)2 - 2(x2) - 3 = 0  Now looks like a quadratic.
                               let y = x2  basic substitution
                               y2 - 2y - 3 = 0   Now it is a quadratic equation!!
                                (y - 3)(y + 1) = 0    Factoring
                                    This means y = 3 and y = -1
                                    But we don't want to find what y is.  Now find x.
                                    3 = x2   or -1 = x2
     By taking the square root!
 

                        2)    x4 + 5x2 + 6 = 0
                                y = x2 is the substitution again!!
                                y2 + 5y + 6 = 0
                                (y + 3)(y + 2) = 0
                                y = -3 or y = -2
                                -3 = x2 or -2 = x2
   By taking the square root!

Rational Root Theorem
    The rational root theorem says that any rational roots must be factors of the constant divided by the positive factors of the leading coefficient!   By using synthetic division, you can find enough roots to factor the polynomial to linear factors and a quadratic.  Then, you can factor the quadratic by any method you choose.

                Sample problems

                    1)  x4 - 11x2 - 18x - 8 = 0
                                The possible zeros are the divisors of 8, divided by the divisors of 1.  Thus the possible zeros are: 1, 2, 4, 8, -1, -2, -4, -8.  Use synthetic division to find a zero

                                -1|        1        0        -11        -18        -8
                                                    -1            1         10          8
                                            1       -1        -10          -8          0

                                    Now use this last row minus the zero at the end.  Again, find a zero by using synthetic division.

                                -2|        1        -1        -10        -8
                                                      -2           6          8
                                            1        -3          -4         0

                                    Now factor the polynomial that remains!
                                        x2 - 3x - 4 = (x - 4)(x + 1)
                                The solutions are x = -1, x = -2, x = 4, and x = -1 again!
                                -1 is a double root!!

This should give you enough to go on for the problems in the book.  Good luck!