Sunday, October 10, 2010

Well, here ya go.

2.3: Real Zeros of Polynomial Functions

To find the real zeros of polynomial functions, one can use both long division as well as synthetic division.

Here is an example of how to use long division:

Divide f(x) = 6x^3 - 19x^2 + 16x - 4 by x - 2.

 |  | \nx | - | 2 |  | 6 x^2 | - | 7 x | + | 2\n6 x^3 | - | 19 x^2 | + | 16 x | - | 4\n6 x^3 | - | 12 x^2 |  |  |  | \n |  | -7 x^2 | + | 16 x |  | \n |  | -7 x^2 | + | 14 x |  | \n |  |  |  | 2 x | - | 4\n |  |  |  | 2 x | - | 4\n |  |  |  |  |  | 0

Now you know that both x - 2, 2x - 1, and 3x - 2 are the factors of f(x) (2x - 1 and 3x - 2 being the factors of 6x^2 - 7x + 2). You then set these values equal to zero and solve for x, giving you the zeros of the f(x) equation.

Here is a video that will better explain long division of polynomials:

Here is a video that will go through the process of synthetic division of polynomials.

Note 1: The end of the video about synthetic division also goes over the remainder theorem at the end. The remainder theorem reads as follows:

If a polynomial f(x) is divided by x - k, the remainder is r = f(k).

Note 2: When using synthetic division with polynomials that have 0 as some of their coefficients, you must include the 0 in the problem in order to ensure you get the right answer.

The Rational Zero Test

If the polynomial f(x) = ax^n + ax^n-1 + ... + ax^2 + ax + a has integer coefficients, every rational zero of f has the form:

rational zero = p/q

where p and q have no common factors other than 1, p is a factor of the constant term, and q is a factor of the leading coefficient a.


1 comment:

  1. As you're studying, don't forget about the Remainder Theorem.

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