Real Zeros of Polynomial Functions, Section 3.3 #30, 58, 80.

  1. Divide by synthetic division -3x2/(x + 2).
    -2 | -3 0 0 0 0
    | 6 -12 24 -48
    _______________
    -3 6 -12 24 -48

    So our quotient is -3x3 + 6x2 - 12x + 24 - 48/(x + 2).

  1. f(x) = 4x3 - 12x2 - x + 15.
    1. List the possible rational zeros of f,
    2. use a graphing utility to graph f so that some of the possible zeros in part (a) can be disregarded, and
    3. determine all the real zeros of f.
      1. {p/q}: {+ 1/4, + 1/2, + 3/4, + 1, + 5/4, + 3/2, + 5/2, + 3, + 15/4, + 5, + 15/2, + 15}
      3. -1, 3/2, 5/2.

  1. Find the rational zeros of the polynomial f(x) = x3 - 3/2 x2 - 23/2 x + 6 = 1/2 (2x3 - 3x2 - 23x + 12). We had to rewrite the equation to get integer coefficients in order to utilize the rational zero test.
    {p/q}: {+ 1/2, + 1, + 3/2, + 2, + 3, + 4, + 6, + 12}
    Graphing the function allows us to eliminate many of the potential rational zeros.
    We then use synthetic division to verify the actual rational zeros.
    The rational zeros are x = {-3, 1/2, 4}.

The Fundamental Theorem of Algebra, Section 3.4 #20, 30, 42.

In 1799 Gauss, at the age of 22, proved the Fundamental Theorem of Algebra:

"Every polynomial with complex coefficients has a complete set of roots in C; that is every polynomial with complex coefficients is a product of linear factors such as (x - c)."

Note: Complex zeros occur in conjugate pairs whenever a polynomial has real coefficients.

  1. 2s3 - 5s2 + 12s - 5; p = {+1, +5}, q = {+1, +2}, p/q = {+1/2, +1, +2.5, +5}.
Graphing eliminates seven of these potential rational zeros and we try 1/2 using synthetic division.

1/2

|

2

-5

12

-5

 

 

1

-2

-5

 

_______________________________________________

 

 

2

-4

10

0

Synthetic division gives use a quotient of 2s2 - 4s + 10, which can easily be factored using the quadratic equation.

s =

(-(-4)+sqrt( (-4)2 -4(2)(10) )) / (2(2) )

=

( 4 + sqrt( 16 - 80 )) / 4

=

1 + sqrt( -64)/4

=

1 + 2i
2s3 - 5s2 + 12s - 5 = (s - 1/2 )(s - (1 + 2i) )(s - (1 - 2i) ).
  1. c1 = 4, c2 = 3i, c3 = -3i, find a polynomial with integer coefficients having these roots.
    f(x) = (x - 4)(x - 3i)(x + 3i) = (x - 4)(x2 + 9) = x3 - 4x2 + 9x - 36.
  1. f(x) = x3 + x2 + 9x + 9, given r1 = 3i.
    We know complex zeros of real polynomials come in conjugate pairs,
    so r2 = -3i. Using synthetic division we can quickly find the third root:

3i

|

1

1

9

9

 

 

3i

3i - 9

-9

 

_______________________________________________

 

 

1

1 + 3i

3i

0

-3i

|

1

1 + 3i

3i

 

 

 

-3i

-3i

 

 

__________________________________________

 

 

1

1

0

 

Which leaves us with the equation x + 1 = 0, so r3 = -1.

Essential Trigonometry, Math Links

by Jack Tompkins