Review
Let p(x) be a polynomial function with real coefficients. If p(s) = 0
 s is a zero for the polynomial function p(x).
 s is a solution to the equation p(x) = 0
 x  s is a factor of p(x).
 The point (s , 0) is an x intercept of the graph of p(x).
Example: let p(x) = x^{3}  2·x^{2} + 2·x  4
 p(2) = 2^{3}  2*2^{2} + 2*2  4
= 8  8 + 4  4
= 0
2 is a zero of p(x).
 x = 2 is a solution of p(x) = 0
 p(x) can be written in factored form as
p(x) =(x  2)·(x^{2} + 2)
 The graph of p(x) below shows an x intercept at x = 2.
TUTORIAL
Example  Problem 1: The graph below is that of a polynomial function p(x) with real coefficients. The degree of p(x) is 3 and the zeros are assumed to be integers. Find p(x).
Solution to Problem 1:
 The graph has 2 x intercepts: 1 and 2. The x intercept at 1 is of multiplicity 2. p(x) can be written as follows
p(x) = a(x + 1)^{2}(x  2) , a is any real constant not equal to zero.
 To find a we need to use more information in the graph. The y intercept is at (0 , 2), which means that p(0) = 2
 Solve the above equation for a to obtain
 p(x) is given by
p(x) = (x + 1)^{2}(x  2)
Example  Problem 2: A polynomial function p(x) with real coefficients and of degree 5 has the zeros: 1, 2(with multiplicity 2) , 0 and 1. p(3) = 12.
Find p(x).
Solution to Problem 2:
 p(x) can be written as follows
p(x) = ax(x + 1)(x  2)^{2}(x  1) , a is any real constant not equal to zero.
 p(3) = 12 gives the following equation in a.
a(3)(3 + 1)(3  2)^{2}(3  1) = 12
 Solve the above equation for a to obtain
 p(x) is given by
p(x) = 0.5x(x + 1)(x  2)^{2}(x  1)
 The graph of p(x) is shown below.
Check the intercepts and the point (3 , 12) on the graph of p(x) found above.
More references and links to polynomial functions.
Factor Polynomials.
Polynomial Functions.
Multiplicity of Zeros and Graphs Polynomials.
Find Zeros of Polynomial Functions  Problems
Graphs of Polynomial Functions  Self Test.
