Section 2-1:  Polynomials
 
 
 Polynomials are expressions whose exponents are non-negative integers and the coefficients are real numbers.  The leading term is the term with the highest power.  The coefficient of the leading term is the leading coefficient.  The power of x in the leading term is called the degree of the polynomial!  Here are some examples of polynomials:
     Degree Name Example
        0 constant 11
        1 linear 4x + 7
        2 quadratic 3x2 + 4x + 5
        3 cubic 4x3 + 3x + 7
        4 quartic 3x4 + 2x2 + 4
        5 quintic x5 + 3x + 1
Any value of x such that P(x) = 0 is a root of the equation and a zero of the function.

Review of synthetic substitution
To use synthetic substitution, do the following:
                    1)  Write the polynomial in descending order.
                    2)  Make sure any missing terms are replaced with zero!
                    3)  Bring down the leading coefficient.
                    4)  Multiply by the divisor and add to the next term.
                    5) Repeat the process until the last coefficient is reached.
 
 Example
Use synthetic division to find P(2) for P(x) = 2x3 - 9x2 + 27
    2|        2        -9        0        27
                          4     -10      -20
               2        -5     -10         7   The value of P(2) = 7, the last number!!

Sample Problems
1)  State whether each is a polynomial function.  If yes, find the degree and the zeros.
          Problem                 Polynomial?            Degree                 zeros
    a)  f(x) = 5 - 4x                    yes                        1                        5/4
    b)  f(x) = 3x2 - 6x                yes                        2                        0, 2
            c)  f(x) = x - 3/x                  no
     d)  f(x) = 18                        yes                         0                        none
            e)  f(x) = (x-3)/(x+2)          no

2)  Use synthetic substitution to find P(3) for each function:
                     a)  P(x) = 4x3 - 5x2 + 3
                                        3|                4                -5                0            3
                                                                             12             21          63
                                                           4                 7              21          66
            Therefore, P(3) = 66
 

                   b)  P(x) = 3x4 + 2x3 - 5x - 2
 
                                        3|                3            2            0            -5            -2
                                                                         9          33           99          282
                                                           3          11          33           94          280
                Therefore, P(3) = 280



3)  If 4 is a zero of f(x) = 3x3 + kx - 2, find the value of k.
            Since 4 is a zero, the value of f(4) = 0.  Use synthetic substitution.
                                    4|        3        0        k              -2
                                                       12      48           192 + 4k
                                               3      12      48 + k    190 + 4k
            Now 190 + 4k = 0, solve for k
                               4k = -190
                                 k = -190/4 = -47.5

That's it for this first section.  On to synthetic division!!