Polynomials
Dr. Philippe B. Laval
Kennesaw State University
September 26, 2000
Abstract
Handout on polynomials
1
Definition and Examples
Definition 1
1. A polynomial function is a function of the form
f (x) = a0 + a1 x + a2 x2 + a3 x3 + ... + an xn
where an 6= 0, a0 , a1 , an are real numbers, they are called the coefficients
of the polynomial. n is any positive integer.
2. The above form is called the standard form of a polynomial.
3. The highest power of x is called the degree of the polynomial.
4. The term an xn is called the leading term of the polynomial, an is the
leading coefficient.
5. The zeros (or the roots) of a polynomial f are the values of x for which
f (x) = 0.
Example 2 f (x) = 2 − 5x is a polynomial function. It is also a linear function. Linear functions are polynomials. The leading term is −5x, the leading
coefficient is −5.
Example 3 f (x) = 5 − 2x + x2 is a polynomial function. It is also a quadratic
function. Quadratic functions are polynomials. The leading term is x2 , the
leading coefficient is 1.
Example 4 f (x) = −x3 + 7x + 6 is a third degree polynomial. The leading
term is −x3 , the leading coefficient is −1.
Example 5 f (x) = 5 is a constant polynomial, its degree is 0.The degree of
any constant polynomial is 0.
1
2
Graph
2.1
Practice
In this section, we show a few examples. The graphs shown were obtained using
computer software. You can also graph them on your calculator. You may have
to specify a larger viewing window in the y direction as the y values tend to
increase very quickly. As you look at these graphs, try to answer the following
questions:
1. What seems to determine the end behavior of a polynomial?
2. What seems to determine how many x-intercepts the polynomial has?
3. What seems to determine how many local maxima and minima the polynomial has?
We will address these questions in the sections which follow. However, you
should try to answer them without reading ahead. Then, you can read ahead
to check your answers.
Example 6 f (x) = x3 − 7x − 6
This is a third degree polynomial. Its graph is shown on Figure 1.
20
10
-4
-2
0
2
x
4
-10
-20
Figure 1: Third degree polynomial
Example 7 f (x) = −x3 + 7x + 6
This is also a third degree polynomial, very similar to the previous example.
Its graph is shown on Figure 2. The main difference is the sign of the leading
coefficient. It is negative, while in the previous example it was positive. Compare
this graph with the previous one.
2
20
10
-4
0
-2
2
4
x
-10
-20
Figure 2: Third degree polynomial
Example 8 f (x) = x5 + 2x4 − 23x3 − 28x2 + 76x + 80
This is a Þfth degree polynomial. Its graph is shown on Figure 3
300
200
100
-6
-4
0
-2
2
x
4
-100
-200
-300
Figure 3: Fifth degree polynomial
Example 9 f (x) = x4 − x3 − 13x2 + x + 12
This is a fourth degree polynomial. Its graph is shown on Figure 4.
Example 10 f (x) = −x4 + x3 + 13x2 + x + 12
This is also a fourth degree polynomial. Its graph is shown on Figure 5.
3
200
150
100
50
-4
-2
0
2
x
4
-50
Figure 4: Fourth degree polynomial
-4
-2
2
x
4
0
-100
-200
-300
-400
Figure 5: Fourth degree polynomial
4
2.2
End Behavior of a Polynomial
This is entirely determined by the leading term. Remember, in the case of
a quadratic function, the sign of the coefficient of x2 determined whether the
graph opened up or down. For polynomials in general, the sign of the leading
coefficient as well as the degree of the polynomial determine the shape of their
graph. For the discussion that follows, let us assume that the leading coefficient
is an . When n is even, both ends point in the same direction. The sign of an
determine whether they point up or down. When n is odd, the ends point in
opposite directions. The sign of an determine which end points down and which
one points up. Thus, there are four possible cases.
2.2.1
case 1: n is even and an > 0
The general shape of the graph at the ends will look like Figure 6. An example
of such a graph is Figure 4. In this case, we say that y approaches ∞ as
x approaches ±∞. In this case, we write y → ∞ as x → ±∞.
25
20
15
10
5
-4
-2
0
2
x
4
Figure 6: n even, an > 0
2.2.2
case 2: n is even and an < 0
The general shape of the graph at the ends will look like Figure 7. An example
of such a graph is Figure 5. Here, we have y → −∞ as x → ±∞.
2.2.3
case 3: n is odd and an > 0
The general shape of the graph at the ends will look like Figure 8. An example
of such a graph is Figure 1. Here, y → −∞ as x → −∞, and y → ∞ as x → ∞.
5
-4
-2
2
x
4
0
-5
-10
-15
-20
-25
Figure 7: n even, an < 0
100
50
-4
-2
0
2
x
4
-50
-100
Figure 8: n odd, an > 0
2.2.4
case 4: n is odd and an < 0
The general shape of the graph at the ends will look like Figure 9. An example
of such a graph is Figure 2. Here, y → ∞ as x → −∞, and y → −∞ as x → ∞.
2.2.5
Conclusion
It is not difficult to see why a polynomial behaves like its term of highest degree.
We simply have to factor the leading term. Let f (x) be a polynomial of degree
6
100
50
-4
-2
0
2
x
4
-50
-100
Figure 9: n odd, an < 0
n. Then, we have:
f (x) = an xn + an−1 xn−1 + ... + a1 x + a0
µ
¶
a1 x
a0
an−1 xn−1 an−2 xn−2
= an xn 1 +
+
+
...
+
+
an xn
an xn
an xn an xn
µ
¶
a1
a0
an−1 an−2
+
+
...
+
+
= an xn 1 +
an x
an x2
an xn−1 an xn
As x → ±∞, all the fractions containing x to some power in the denominator
an−2
an−1
+
+ ... +
will get smaller and smaller. Therefore, as x → ±∞, 1 +
an x
an x2
a1
a0
→ 1. It follows that f (x) → an xn as x → ±∞.
+
an xn−1 an xn
2.3
Local Extrema of a Polynomial
A polynomial of degree n can have at most n − 1 local extrema, it may have
fewer. Recall that in the case of a quadratic function, we found that it either had
1 max or 1 min. Knowing the number of max and min is helpful is determining
the shape of the graph of a polynomial. However, Þnding the max and min of
a polynomial given its formula is beyond the scope of this class. It is usually
learned in a Calculus class.
7
Example 11 Select the graph from the four below that best represents the graph
of f (x) = 2x4 − 9x3 + 11x2 − 4
The degree of f (x) is even (4) and the leading coefficient is positive, this reduces
the choices to graphs 2 and 3. Since the degree of f (x) is 4, it has at most 3
local extrema, this rules out graph 2. Thus, graph 3 is the correct choice.
2.4
X-intercepts
A polynomial of degree n can have at most n x-intercepts, it may have fewer.
Recall, a parabola (which is a polynomial of degree 2) can have 2, 1 or 0 xintercepts. Knowing the number of x-intercepts is helpful is determining the
shape of the graph of a polynomial. For polynomials of degree 2, one can use
the quadratic formula to Þnd the x-intercepts. If the degree of the polynomial
is higher, it is more difficult. If the degree is 3, there is a formula, but it is
8
complicated and few people remember it. If the degree is 4 or higher, there are
no formulas like the quadratic formula. We will see in the next section how to
Þnd the x-intercepts using other methods.
2.5
Y-intercept
When a polynomial is in standard form, its y-intercept is the constant term.
3
Real Zeros of a Polynomial
If f (x) is a polynomial, then we said earlier that c was a root of f if f (c) =
0. Geometrically, when we are Þnding the real zeros of a polynomial, we are
Þnding its x-intercepts. Finding the zeros of a polynomial is also closely related
to factoring the polynomial. Thus, Þnding the zeros of a polynomial is very
important in the study of this polynomial. We study how to Þnd the zeros of a
polynomial, and see how it relates to factoring it.
3.1
Dividing Polynomials
Polynomials can be divided in a way similar to the way numbers are divided.
The process is illustrated below as we show how to divide 6x2 − 26x + 12 by
x − 4.
6x − 2
x − 4 ) 6x2 − 26x + 12
6x2 − 24x
−2x + 12
−2x + 8
4
In this process, we use the following terminology:
• 6x2 − 26x + 12 is called the dividend
• x − 4 is called the divisor
• 6x − 2 is called the quotient
• 4 is called the remainder
and we write
6x2 − 26x + 12 = (x − 4) (6x − 2) + 4
If the remainder had been 0, the polynomial would have been factored. It turns
out that this process can be repeated for any polynomials. It is called the
division algorithm. It says the following:
9
Theorem 12 (Division Algorithm) If P (x) and D (x) are polynomials with
D (x) 6= 0, then there exist unique polynomials Q (x) and R (x) such that
P (x) = D (x) Q (x) + R (x)
where R (x) is either 0 or of degree less than the degree of D (x). P (x) is called
the dividend, D (x) is called the divisor, Q (x) is called the quotient, and R (x)
is called the remainder.
Remark 13 If the divisor is a polynomial of degree 1, then the remainder will
be a constant (since its degree has to be less than that of the divisor).
Theorem 14 (Factor Theorem) Let P (x) be a polynomial. P (c) = 0 (i.e.
c is a zero of P ) if and only if x − c is a factor of P (x).
Remark 15 If (x − c)2 is a factor of P (x), we say that c is a double zero. If
(x − c)3 is a factor of P (x), we say that c is a triple zero. And so on.
This theorem is very useful. Consider the examples below.
Example 16 Factor f (x) = 2x2 + 5x − 3
By the theorem, we know that if c is a zero of f , x−c will be a factor. Since this is
a second degree polynomial, we know how to Þnd all its zeros. They are given by
1
1
the quadratic formula. The zeros are −3 and . Thus x+3 and x− are factors
2
¶
µ2
1
. In
of f . Be careful though, we cannot conclude that f (x) = (x + 3) x −
2
1
fact it is not. We can only conclude that x + 3 and x − are factors of f .
2
¶
µ
1
are of degree 2, the only thing we
However, since both f and (x + 3) x −
2
are missing is a constant. In fact,
¶
µ
3
5
1
= x2 + x −
(x + 3) x −
2
2
2
1
f (x)
=
2
Thus,
µ
¶
1
f (x) = 2 (x + 3) x −
2
Example 17 Factor f (x) = x3 − 7x + 6 completely, given that 1 is a root of
f (x).
If 1 is a root, then x − 1 is a factor of f (x). Thus, there exists a polynomial
Q (x) such that
f (x) = (x − 1) Q (x)
f (x)
Q (x) =
x−1
10
We divide f (x) by x − 1, we obtain
¡
¢
f (x) = (x − 1) x2 + x − 6
= (x − 1) (x + 3) (x − 2)
Example 18 Find a polynomial F (x) of degree 4 that has zeros −3, 0, 1, and
5.
By the factor theorem, x − 0 (= x), x − 1, x − 5, and x + 3 must be factors.
Thus
F (x) = x (x + 3) (x − 1) (x − 5)
is such a polynomial. It is not the only one though. The answer is no. However,
since we want a polynomial of degree 4, any other solution will be a constant
multiple of what we found. This brings us to the following remark.
Remark 19 If we know all the zeros of a polynomial, we do not know the
polynomial exactly, we know it up to a constant. To know the polynomial exactly,
we must also be given its value at a point. In the example above, we were asked to
Þnd a polynomial of degree 4, knowing 4 (that is all) of its zeros. We found that
F (x) = x (x + 3) (x − 1) (x − 5) was such a polynomial. All the polynomials
which satisfy the given condition are of the form
F (x) = Cx (x + 3) (x − 1) (x − 5)
where C is a constant.
Example 20 Find all the polynomials of degree 3 having −1, 1, and 2 as zeros.
Since the polynomial is of degree 3, we are given all its zeros. So, we know all
its linear factors. The polynomials we want are of the form
f (x) = C (x + 1) (x − 1) (x − 2)
where C is a constant.
Example 21 Find all the polynomial of degree 5 having −1, 1, and 2 as zeros.
The linear factor of the polynomials we want are (x + 1) (x − 1) (x − 2). However, since we want polynomials of degree 5, the polynomials we want are of the
form
f (x) = g (x) (x + 1) (x − 1) (x − 2)
where g (x) is a polynomial of degree 2.
3.2
Rational Zeros Theorem
The factor theorem is very useful in factoring a polynomials, when we know
some of its zeros. For example, if we have to factor a polynomial of degree 3,
and we know one of its zeros, say c. Then, x − c is a factor. The remaining
factor must be a polynomial of degree 2. We know how to factor a polynomial
of degree 2 since we know how to Þnd its zeros. But, how did we happen to
know one of the zeros of this polynomial to begin with? The next theorem gives
us a method for Þnding all the rational zeros of a polynomial.
11
Theorem 22 (Rational zeros theorem) Let f (x) = an xn +an−1 xn−1 +...+
a1 x + a0 be a polynomial with integer coefficients. Then, every zero of f is of
the form
p
q
where
• p is a divisor of a0
• q is a divisor of an
Remark 23 In the above theorem,
p
is assumed to be in lowest terms.
q
Example 24 Use the rational zeros theorem to Þnd the zeros of P (x) = 2x3 +
x2 − 13x + 6, then factor f.
In this case, a0 = 6, therefore p = ±1, ±2, ±3, ±6. Similarly, an = a3 = 2,
p
1
2
3
6
1
therefore q = ±1, ±2. The possible values for
are ± , ± , ± , ± , ± ,
q
1
1
1
1
2
2
3
6
± , ± , ± . If we simplify, and eliminate the duplicates, we see that the
2
2
2
p
1
3
possible values for
are: ±1, ±2, ±3, ±6, ± , ± . We check each value.
q
2
2
With ScientiÞc Notebook, this amounts to doing


−1


18
 −2 


20 
 −3  

 


 −6   0 



−312 
 1  

 −   25 

 2  
 3   2 

 −  
21 



P 2 =
−4 
 1  




0 
 2  




30 
 3  




396 
 6  


 1  
  0 




9
 2 
−
3
2
2
1
Thus, the zeros are −3, 2 and . Hence,
2
µ
¶
1
P (x) = C x −
(x − 2) (x + 3)
2
Since P (0) = 6, it follows that
µ ¶
1
(−2) (3)
6 = C −
2
6 = 3C
12
Or, C = 3. Thus,
P (x) = (2x − 1) (x − 2) (x + 3)
4
Sign of a Polynomial
The method of Þnding the sign of a polynomial function is similar to the method
used for a quadratic function. First, we Þnd the zeros of the polynomial. The
zeros of the polynomial divide the real line into intervals. The sign of the
polynomial is constant in each interval. To Þnd it, we pick a number in each
interval, and Þnd the sign of the polynomial at that number.
Example 25 Find the sign of f (x) = x3 − 7x + 6.
We found above that f (x) = (x − 1) (x + 3) (x − 2). Thus its zeros are −3, 1
and 2. Three numbers determine four intervals. We Þnd the sign of f in each
interval.
Interval
Test point Value of f at the test point Sign of f
−4
−
(−∞, −3)
f (−4) = −30
(−3, 1)
0
f (0) = 6
+
−
(1, 2)
1.5
f (1.5) = −1. 125
(2, ∞)
3
f (3) = 12
+
Therefore, f is positive in (−3, 1) ∪ (2, ∞). It is negative everywhere else, that
is in (−∞, −3) ∪ (1, 2).
5
Problems
1. Problems assigned in the book
(a) # 1, 5, 11, 27, 28, 29, 30, 37, 39, 65,68 on pages 230 - 232
(b) # 1, 3, 5, 7, 9, 21, 23, 25, 27, 29, 45, 47, 49, 95 on pages 245 - 248
2. Find the degree, leading term and leading coefficient of each polynomial
below.
(a) f (x) = 3x2 + 9x − 10x3 − 4
(b) f (x) = −x3 + 7x2 + 36x − 240
(c) f (x) = −2x4 + x3 + 13x2 + 8x − 1
¢
¡
(d) f (x) = (x − 1) x3 − x2 − x − 1
3. Using your calculator, sketch the graph of the polynomials below. For each
graph you obtained, try to Þnd the coordinates of the x and y-intercepts
as well as the coordinates of the max and min. Graphing polynomials
with a calculator usually involves ”playing” with the viewing window.
(a) f (x) = 3x2 + 9x − 10x3 − 4
13
(b) f (x) = −x3 + 7x2 + 36x − 240
(c) f (x) = −2x4 + x3 + 13x2 + 8x − 1
(d) f (x) = x4 − 2x3 + 1
4. Select the graph that best represents the given polynomial. Explain your
choice. In making your choice, you will have to look at the degree of the
polynomial, its maximum number of x-intercepts, its y-intercept and the
number of max and min it can have.
(a) f (x) = −x3 + x2 + x − 1
(b) g (x) = x3 − 6x2 + 9x + 5
(c) h (x) = x4 − 3x3 + 3x2 − x
14
5. Select the graph from the graphs below that best represent the given
polynomial.
(a) f (x) = x3 − 2x2
(b) f (x) = x4 − x3 − x6 + x
(c) f (x) = 2x4 − 5x3 + 4x2 − x
(d) f (x) = 2x4 + x3 − 1
6. Find the quotient and the remainder in each case below:
15
x3 + 2x2 + 2x + 1
x+2
3
x − 8x + 2
(b)
x+3
6x3 + 2x2 + 22x
(c)
2x2 + 5
¡ 2
¢
x + 3x + 1 (x + 1)
(d)
(x + 1)
(a)
7. Find a polynomial of the speciÞed degree that has the given zeros.
(a) Degree 3, zeros: −1, 1, 3.
(b) Degree 4, zeros: −2, 0, 2, 4
(c) Degree 4, zeros: −1, 1, 3, 5
8. Find all the zeros of the given polynomials, then factor them and study
their sign.
¡
¢
(a) (x − 1)2 x2 − x − 6
(b) x3 − x2 − 14x + 24 given that 2 is a zero.
(c) x4 − x3 − 7x2 + 13x − 6 given that 1 is a double zero.
(d) x4 − 5x2 + 4 given that −1 and 1 are zeros.
16