Module 3: Polynomial Functions

 

 

A lot of new information has been introduced in this lesson. Here is a summary of how the information fits together.

 

Example Mathematical Justification
In order to sketch f(x) = 2x3 + x2 – 13x + 6, you need to create an equivalent factored form. The factors correspond to x-intercepts.

Test for factors:


 

f(1) = −4


 

f(−1) = 18


 

f(2) = 0

 

x − 2 is a factor of f(x) because f(2) = 0.

The remainder theorem tells you that f(a) = the remainder when f(x) is divided by x a.

 

The factor theorem tells you to look for values of a such that f(a) = 0.

 

The integral zero theorem tells you to try factors of the constant term of f(x). (In this example, 6.)
 Perform division to determine a second factor.
f(x) = (x − 2)(2x2 + 5x − 3) This equivalent factored form can be written using f(x) = (x a)Q(x) + R, where Q(x) is the quotient and R is the remainder.
f(x) = (x − 2)(x + 3)(2x − 1) The equivalent factored form of 2x2 + 5x − 3 is (x + 3)(2x − 1).

The graph of f(x) will have x-intercepts at 2, −3, and  Since f(x) is a cubic polynomial function with a positive leading coefficient, its graph will start in quadrant 3 and end in quadrant 1.

 

This is a sketch of the graph of f(x) = 2x cubed + x squared – 13x + 6. x-intercepts are labelled at –3, one half, and 2.

 

These are the values that make each factor equal to zero, thus making f(x) = 0.