
Example 1:
a) Factor polynomial P given by
P (x) =  x^{3}  x^{2} + 2x
b) Determine the multiplicity of each zero of P.
c) Determine the sign chart of P.
d) Graph polynomial P and label the x and y intercepts on the graph obtained.
Solution to Example 1

a) Factor P as follows
P (x) =  x^{3}  x^{2} + 2x
=  x (x^{2} + x  2)
=  x (x + 2)(x  1)
b) P has three zeros which are 2, 0 and 1 and are all of multiplicity one.
c) The three zeros of P will split the number line into four intervals given by:
( ∞ , 2) , ( 2 , 0) , (0 , 1) and (1 , +∞)
select one value of x within each interval and evaluate polynomial P for this value to determine the sign of P.
P(3) = 12 > 0 , P(1) =  2 < 0 , P(1/2) = 5/8 >0 , P(2) =  8 < 0
Using the signs of P with each interval, the sign chart is as follows:
Example 2:
a) Factor polynomial P given by
P (x) = x^{4}  2 x^{2} + 1
b) What is the multiplicity of each zero of P?
c) Determine the sign chart of P.
d) Graph polynomial P and label the x and y intercepts on the graph obtained.
e) What is the range of polynomial P?
Solution to Example 2
 a) Factor P as follows
P (x) = x^{4}  2 x^{2} + 1
= (x^{2}  1)^{2}
= ((x  1)(x + 1))^{2}
= (x  1)^{2} (x + 1)^{2}
 b) Polynomial P has zeros at x = 1 and x = 1 and both has multiplicity 2.
 c) Polynomial P(x) is a perfect square and therefore positive or zero for all real values of x. P(x) is equal to zero at the two zeros 1 and 1 and positive everywhere else. The sign chart is as follows:
 d) The x intercepts are at (1,0) and (1,0) and the y intercept is at (0,1). The graph of P touches the x axis at x = 1 and x = 1 and opens up since P(x) is positive, cutting the y axis at (0,1). See graph below.
 e) Using the graph of P above, the range of P is given by the interval
[ 0 , + ∞)
Example 3:
a) Show that x =  3 is a zero of polynomial P given by
P (x) = x^{4} + 5 x^{3} + 5 x^{2}  5 x  6
b) Show that (x  1) is a factor of P.
c) Factor P and determine the multiplicity of each zero of P.
d) Determine the sign chart of P.
e) Graph polynomial P and label the x and y intercepts on the graph obtained.
Solution to Example 3
 a) Calculate P(3)
P (3) = (3)^{4} + 5 (3)^{3} + 5 (3)^{2}  5 (3)  6
= 0
Hence 3 is a zero of P and x + 3 is a factor of P(x).
 b) Since (x + 3) is a factor of P(x), the division of P(x) by (x + 3) must give a remainder equal to zero. Hence
P(x) / (x + 3) = x^{3} + 2 x^{2}  x  2
We now divide x^{3} + 2 x^{2}  x  2
by (x  1)
(x^{3} + 2 x^{2}  x  2) / (x  1) = x^{2} + 3x + 2 , remainder equal to zero which proves that (x  1) is a factor of x^{3} + 2 x^{2}  x  2 and therefore of P(x).
 c) Using the above, P(x) may be written as follows
P(x) = (x + 3)(x  1)(x^{2} + 3x + 2)
We now factor the quadratic term x^{2} + 3x + 2 included in P(x). Hence
P(x) = (x + 3)(x  1)(x + 1)(x + 2)
P has zeros at x = 3, 2, 1 and 1 and are all of multiplicity one.
 d) The sign chart is shown below
 e) Using the information on the zeros and the sign chart, the graph of P is as shown below with x and y intercepts labeled.
Example 4:
x = 1 is a zero of multiplicity 2 of polynomial P defined by
P (x) = x^{5} + x^{4}  3 x^{3}  x^{2} + 2 x.
Construct a sign chart for P and graph it.
Solution to Example 4
 If x = 1 is a zero of multiplicity 2, then (x  1)^{2} is a factor of P(x) and a division of P(x) by (x  1)^{2} yields a remainder equal to 0. Hence
P (x) / (x  1)^{2}
= (x^{5} + x^{4}  3 x^{3}  x^{2} + 2 x) / (x  1)^{2}
= x^{3} + 3 x ^{2} + 2 x
P(x) is now factored as follows
P(x) = (x  1)^{2} (x^{3} + 3 x ^{2} + 2 x)
= x (x  1)^{2} (x^{2} + 3 x + 2 )
= x (x  1)^{2} (x + 1)(x + 2)
P(x) has 4 zeros at x = 2, 1, 0 and 1 and the zero at x= 1 is of multiplicity 2.
The sign chart is shown below
Use the sign chart and the zeros of P to grpah P as shown below.
More references and links to Graphing Functions.
