PROJECT 3
Roots and Graphs of Polynomials
Roots and their Multiplicity
We begin by considering the polynomial
p( x)  a n x n  a n1 x n1    a1 x  a0
Recall that if ( x  c) is a factor of p, then c is a root (zero) of p and that the degree, n, of p tell us the
maximum number of roots. If we use other theorems, which we will neither state nor prove here, we
can also find out that every polynomial can be factored as
p( x)  a( x  c1 )( x  c 2 )( x  c3 )  ( x  c n ) , where each c i is a complex number.
From experience, we know that each of the numbers c i are not necessarily all different. To illustrate
this point, let us look at the polynomial
f ( x)  3x 3  3x 2  3x  3 .
We can factor f as:
f ( x)  3( x  1) 2 ( x  1) .
If a factor x  c occurs m times in the factorization, then x is a root of multiplicity m of f(x). In the
preceding example, 1 is a root of multiplicity 2 and 1 is a root of multiplicity 1.
Example 1:
Find the roots of the polynomial f ( x)  ( x 1)(x  4) 3 ( x  1) 2 , and state the multiplicity of each.
Solution:
We see from the factored form that the roots are 1, 4 and 1. From the power of their
corresponding factors, we see that 1 is a root of multiplicity 1, 4 is a root of multiplicity 3, and 1
is a root of multiplicity 2.
The multiplicity of a root has an effect on the shape of the graph near the root. A root with
multiplicity 2 will have behavior similar to x 2 , a root of multiplicity 3 will have behavior similar
to x 3 , a root of multiplicity 4 will have behavior like x 4 , etc. (only close to the root).
Descartes’ Rule of Signs
We can use Descartes’s rule of signs to give us a bit more information about our polynomial. We will
assume the p(x) is a polynomial with real coefficients and nonzero constant term. The rule follows:
i. The number of positive real roots of p(x) is equal to the number of variations of sign in p(x) or is
less than that number by an even integer.
ii. The number of negative real roots of p(x) is equal to the number of variations of sign in p(x) or
is less than that number by an even integer.
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Project 3
Roots and Graphs of Polynomials
Example 2:
Discuss the number of positive and negative roots of the polynomial:
f ( x)  2 x 5  7 x 4  3x 2  5
Solution:
We note that three are 3 variations of sign I f(x), hence the function has 3 or 1 positive real roots.
Since f ( x)  2x 5  7 x 4  3x 2  5 has two variations of sign, we can conclude that f(x) has 2 or
no negative real roots. Since the degree of f is 5, we can conclude that:
f
has
3
positive real roots,
2
negative real roots, and
0
complex roots
or
f
has
3
positive real roots,
0
negative real roots, and
2
complex roots
or
f
has
1
positive real roots,
2
negative real roots, and
2
complex roots
or
f
has
1
positive real roots,
0
negative real roots, and
4
complex roots
The Rational Root Test
Although we can find all the roots of a quadratic polynomial using the quadratic formula and there are
analogous more complicated formulas for finding the roots of cubics and quartics, it is not possible to
find all the roots of any polynomial of degree 5 or greater. We can though find all the rational roots
of any polynomial with integer coefficients. To do this we use the Rational Root Test. It is stated as
follows:
r
(in lowest terms) is a root of the polynomial a n x n  a n1 x n1    a1 x  a0
s
where the coefficients a n , , a1 , a 0 are integers with a n  0 and a 0  0 , then r is a factor of the
If a rational number
constant term a 0 and s is a factor of the leading coefficient a n .
Example 3:
Find all the possible rational roots of 3x 3  8 x 2  x  20 .
Solution:
The factors of the constant term 20 are 1, 2, 4, 5, 10, and 20. The factors of the leading
coefficient 3 are 1 and 3. If we form all possible ratios of these two sets of numbers, we will
have all the possible rational roots. The possible roots are
1
2
4
5 10
20
 1,  2,  4,  5,  10 ,  20 ,  ,  ,  ,  ,  , 
3
3
3
3
3
3
To determine which are actual roots, you would need to plug them into the polynomial.
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Project 3
Roots and Graphs of Polynomials
Turns of Polynomials
The degree of a polynomial also determines the shape of the graph, as shown in Figure 1 below. These
graphs correspond to a positive coefficient for x n ; a negative coefficient flips the graph over. Notice
that the graph of the quadratic “turns around” once, the cubic “turns around” twice, and the quartic
(fourth degree) “turns around” three times. An n th degree polynomial “turns around” at most n  1
times (where n is a positive integer), but there may be fewer turns.
Quadratic
(n = 2)
Cubic
(n = 3)
Quintic
Quartic
(n = 5)
(n = 4)
Figure 1: Graphs of typical polynomials of degree n
Long-Run or End Behavior of Polynomial Functions
As x gets very large (in other words as x approaches  or as x approaches ), the graph of a
polynomial function closely resembles the graph of its highest degree term. More specifically:
if p(x) is an odd-degree polynomial function then
if the leading coefficient is positive,
then as x  , p( x)   and as x  , p( x)   ;
if the leading coefficient is negative,
then as x  , p( x)   and as x  , p( x)   .
if p(x) is an even-degree polynomial function then
if the leading coefficient is positive,
then as x  , p( x)   and as x  , p( x)   ;
if the leading coefficient is negative,
then as x  , p( x)   and as x  , p( x)   .
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Project 3
Roots and Graphs of Polynomials
Finding the Formula for a Polynomial from its Graph
Given the graph of a polynomial, it is often possible to find a formula for a function that will have the
same graphical behavior as the polynomial. To do this we use what we know about basic polynomial
graphs and specific information from the given graph such as roots and other specific points.
Example 3:
Find possible formulas for the polynomials whose graphs are in Figure 2
4
(i)
(ii)
1
x
2
2
(iii)
g(x)
f(x)
3
2
h(x)
x
12
3
2
Solution:
(i.) This graph appears to be a parabola, turned upside down and moved up by 4, so
f ( x)   x 2  4
The minus sign turns the parabola upside down and the +4 moves it up by 4. You should notice that
this formula does five the correct x-intercepts since 0   x 2  4 has solutions x  2 .
You can also solve this problem by looking at the x-intercepts first, which tell you f(x) must have
factors of ( x  2) and ( x  2) , so f ( x)  k ( x  2)( x  2) where k is some constant. To find k, use the
fact that the graph has a y-intercept of 4, so f (0)  4 , giving
4  k (0  2)(0  2)
so k  1 . Therefor f ( x)  ( x  2)( x  2) , which multiplies out to f ( x)   x 2  4 . Note that
x4
also fits the requirements, but its “shoulders” are sharper. There are many possible
4
answer to these questions.
This looks like a cubic with factors ( x  3) , ( x  1) , and ( x  2) , one for each intercept:
g ( x)  k ( x  3)( x  1)( x  2) .
Since the y-intercept is 12,
12  k (0  3)(0  1)(0  2)
so k  2 , and
g ( x)  2( x  3)( x  1)( x  2) .
This also looks like a cubic with zeros at x  2 and x  3 . Notice that at x  2 the graph
of h(x) touches the x-axis but does not cross it, whereas at x  3 the graph crosses the xaxis. We say that x  2 is a double zero or root with multiplicity 2, but that x  3 is a
single zero or root with multiplicity 1.
To find a formula for h(x), first imagine the graph of h(x) to be slightly lower down, so that
the graph has one x-intercept near x  3 and two near x  2 , say at x  1.9 and x  2.1 .
Then
h( x)  k ( x  3)( x  1.9)( x  2.1) .
Now move the graph back to its original position. The zeros at x  1.9 and x  2.1 move
toward x  2 giving
h( x)  k ( x  3)(x  2) 2 .
f ( x)  4 
(ii.)
(iii.)
Thus the double zero leads to a repeated factor, ( x  2) 2 . Notice that when x  2 , the factor
( x  2) 2 is positive, and when x  2 , ( x  2) 2 is still positive. This reflects the fact that
22
x
Project 3
Roots and Graphs of Polynomials
h(x) does not change sign near x  2 . Compare this with the behavior near the single zero,
where h does change sign.
You cannot find k, as no coordinates are given for points off of the x-axis. Inserting any
positive value of k will stretch the graph but not change the zeros and therefore will still
work.
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Project 3
Roots and Graphs of Polynomials
Problems for Roots and Graphs of Polynomials
In Contemporary Precalculus: A Graphing Approach by Thomas W. Hungerford complete problems 1-12,
14, 15, 44-47 in section 4.3 page 239.
Also complete the following problems:
1.
Assume that each of the graphs below is a polynomial. For each graph:
a. What is the minimum possible degree of the polynomial?
b. Is the leading coefficient of the polynomial positive or negative?
(i)
2.
3.
(iii)
(ii)
(iv)
(v)
Describe the end or long-run behavior of each of the functions below.
a. f ( x)  3x 3  70x 2  20
b.
f ( x)  20x 4  3x 3  x 2  1000
c.
f ( x)  3x 4  20x 3  5x 2  x  20
d.
f ( x)  x 4  x 5
e.
f ( x)  4 x 4  5x 5  6 x 6
Use the graph of y  h(x) below to determine the factored form of h( x)  x 4  x 3  3x 2  12x .
y
y = h(x)
5
4.
5.
6.
5
x
Find possible formulas for the following polynomials given:
a. f is a third degree polynomial with f(3) = 0, f(1) = 0, f(4) = 0, and f(2) = 5.
b. g is a fourth degree polynomial, g has a “double zero” at x = 3, g (5) = 0, g(1) = 0, and g (0) = 3.
Find the real zeros of the following polynomials:
a. y  x 2  5x  6
b.
y  x 4  6x 2  9
c.
y  4x 2 1
d.
y  4x 2  1
e.
y  2 x 2  3x  3
f.
y  3x 5  7 x  1
Suppose you wish to pack a cardboard box inside a wooden crate. In order to have room for the
packing materials, you need to leave a 0.5-ft space around the front, back, and sides of the box and a 1ft space around the top and bottom of the box. If the cardboard box is x feet long, (x + 2) feet wide,
and (x  1) feet deep, find a formula in terms of x for the amount of packing material needed.
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Project 3
7.
Roots and Graphs of Polynomials
Find a possible formula for each of the polynomial functions whose graphs are shown below.
a.
1
2
 1 27 
  , 
 2 16 
9.
1
c.
g(x)
x
1
8.
b.
f(x)
2
h(x)
d.
x
j(x)
10
1
x
2
1
2
4
3
2
If f ( x)  x and g ( x)  ( x  2)( x  1)( x  3) , find all x for which f ( x)  g ( x) . (Note : When
f ( x)  g ( x) , then g ( x)  f ( x)  0 .)
For each of the following polynomial functions, state the degree of the polynomial, the number of
terms in the polynomial, and describe its long-run or end behavior.
a. y  2 x 3  3x  7
b. y  ( x  4)( 2 x  3)(5  x)
2
c. y  1  2x 4  x 3
10. Suppose f is a polynomial function of degree n, where n is a positive even integer. For each of the
following statements, write true is the statement is always true, false otherwise. If the statement is
false, give an example that illustrates why it is false.
a. f is even.
b. f has an inverse.
c. f cannot be an odd function.
d. If f (x)   as x   , then f (x)   as x   .
11. Suppose the following statements about f(x) are true:
 f(x) is a polynomial function
 f(x) = 0 at exactly four different values of x

f (x)   as x  
For each of the following statements, write true if the statement must be true, never true is the
statement is never true, or neither if it is sometimes true and sometimes not true.
a. f(x) is odd.
b. f(x) is even.
c. f(x) is a fourth degree polynomial.
d. f(x) is a fifth degree polynomial.
e. f ( x)   as x  
f. f(x) is invertible.
12. For the polynomial graphed below,
find
an equation that represents is.
(2,
128)
g(x)
4
2
x
3
13. Find all the rational roots for the polynomial 2 x 6  3x 5  7 x 4  6 x 3 . (Hint: The Rational Root Test
can only be used on polynomials with non-zero constant terms. You will need to factor the polynomial
to find one that has a non-zero constant term.)
25
1
x