Download Algebra for College Students, 6e

§ 11.2
Zeros of Polynomial Functions
Zeros of Polynomial Functions
We used a shortcut called “Synthetic Division” for long
division in Chapter 6 to divide a polynomial by a
binomial of the form x - c .
We will use synthetic division now and a result known
as the “Factor Theorem” in solving a polynomial
equation. We will first review the rules in synthetic
division and then will look at the Factor Theorem.
Blitzer, Algebra for College Students, 6e – Slide #2 Section 11.2
Synthetic Division of Polynomials
The method of synthetic division is just a shortcut for
long division. With this process -we can save both steps
and paper by writing down only what is necessary from
the long division problem and by also compacting the
form.
Synthetic division is quick and can be extremely useful.
Note that this process works only when you can express
the divisor in the form x – c.
Blitzer, Algebra for College Students, 6e – Slide #3 Section 11.2
Synthetic Division of Polynomials
Synthetic Division
To divide a polynomial by x – c:
STEPS
1) Arrange polynomials in descending
powers, with a 0 coefficient for any missing
terms.
2) Write c for the divisor, x – c. To the
right, write the coefficients of the dividend.
3) Write the leading coefficient of the
dividend on the bottom row.
4) Multiply c (in this case, 3) times the
value just written on the bottom row. Write
the product in the next column in the
second row.
Blitzer, Algebra for College Students, 6e – Slide #4 Section 11.2
EXAMPLES
x  3 x3  4 x 2  5x  5
3 1 4 5 5
3 1 4 5 5
1
3 1 4 5 5
3
1
Synthetic Division of Polynomials
CONTINUED
Synthetic Division
To divide a polynomial by x – c:
STEPS
EXAMPLES
5) Add the values in this new column,
writing the sum in the bottom row.
3 1 4 5 5
3
1 7
6) Repeat this series of multiplications and
additions until all columns are filled in.
3 1 4 5 5
3 21 48
1 7 16 53
7) Use the numbers in the last row to write
the quotient, plus the remainder above the
divisor. The degree of the first term of
the quotient is one less than the degree of
the first term of the dividend. The final
value in this row is the remainder.
1x 2  7 x  16 
x  3 x3  4 x 2  5x  5
Blitzer, Algebra for College Students, 6e – Slide #5 Section 11.2
53
x3
Synthetic Division of Polynomials
EXAMPLE
Use synthetic division to divide x 2  6 x  6 x3  x 4 by 6  x.
SOLUTION
The divisor must be in the form x – c. Thus, we write 6 + x as
x – (-6). This means that c = -6. Rearranging the terms of the
dividend in descending order and writing a 0 for the missing
constant-term we can express the division as follows:
x   6 x 4  6 x3  x 2  6 x  0
Blitzer, Algebra for College Students, 6e – Slide #6 Section 11.2
Synthetic Division of Polynomials
CONTINUED
Now we are ready to set up the problem so that we can use
synthetic division.
6
This is c in
x – (-6).
1 6
1 6
0
Use the coefficients
of the dividend
x 4  6 x3  x 2  6 x  0
in descending
powers of x.
We begin the synthetic division process by bringing down 1.
This is followed by a series of multiplications and additions.
1) Bring down 1.
6
1 6
1
1 6
2) Multiply: -6(1) = -6.
0
6
1 6
6
1
Blitzer, Algebra for College Students, 6e – Slide #7 Section 11.2
1 6
0
Synthetic Division of Polynomials
CONTINUED
3) Add: -6 + -6 = -12.
6
1 6
6
1  12
1 6
4) Multiply: -6(-12) = 72.
0
1 6 1 6
 6 72
1  12 73
1 6 1 6
 6 72
1  12
0
6) Multiply: -6(73) = -438.
5) Add: 1 + 72 = 73.
6
6
0
6
1 6 1
6 0
 6 72  438
1  12 73
Blitzer, Algebra for College Students, 6e – Slide #8 Section 11.2
Synthetic Division of Polynomials
CONTINUED
7) Add: -6 + -438 = -444.
8) Multiply: -6(-444) = 2664.
6
6
1 6 1
6 0
 6 72  438
1  12 73  444
1 6 1
6 0
 6 72  438 2664
1  12 73  444
9) Add: 0 + 2664 = 2664.
6
1 6 1
6 0
 6 72  438 2664
1  12 73  444 2664
Blitzer, Algebra for College Students, 6e – Slide #9 Section 11.2
Synthetic Division of Polynomials
CONTINUED
The numbers in the last row represent the coefficients of the
quotient and the remainder. The degree of the first term of the
quotient is one less than that of the dividend. Because the degree
of the dividend, x 2  6 x  6 x 3  x 4, is 4, the degree of the quotient
is 3. This means that the 1 in the last row represents 1x 3 .
6
Thus,
1 6 1
6 0
 6 72  438 2664
1  12 73  444 2664
2664
x  12 x  73x  444 
x6
x  6 x 2  6 x  6 x3  x 4
3
2
Blitzer, Algebra for College Students, 6e – Slide #10 Section 11.2
Zeros of Polynomial Functions
The Factor Theorem
Let f(x) be a polynomial.
a. If f(c)=0, then x - c is a factor of f(x).
b. If x – c is a factor of f(x), then f(c)= 0.
This theorem is extremely useful because it can be used in solving a
polynomial equation. It simply says that if you substitute c into the
polynomial for x and get a result of 0, then x – c has to be a factor of
the polynomial. Also, whenever the polynomial has a factor of x – c,
if you substitute c for x in the polynomial – you will get a result of 0.
Blitzer, Algebra for College Students, 6e – Slide #11 Section 11.2
Using the Factor Theorem
EXAMPLE
Solve the equation x 3  2 x 2  x  2  0. given that -1 is a zero
of the polynomial.
SOLUTION
Since -1 is a zero of the polynomial, this means that x + 1 is a
factor. We’ll use synthetic division to divide f(x) by x + 1.
1
1
1
Proposed
Solution
2
1
3
1
3
2
2
2
0
x 2  3x  2
x  1 x3  2 x 2  x  2
Remainder
Blitzer, Algebra for College Students, 6e – Slide #12 Section 11.2
Using the Factor Theorem
CONTINUED
Equivalently,
x 3  2 x 2  x  2  x  1x 2  3x  2
 ( x  1)( x  2)( x  1)
3
2
Now we can solve the polynomial equation x  2 x  x  2  0.
The solutions are -1,2,1 and the solution set is {-1,2,1}.
Blitzer, Algebra for College Students, 6e – Slide #13 Section 11.2
Using the Factor Theorem
Based on the Factor Theorem, the following statements are useful
in solving polynomial equations:
1. If f(x) is divided by x – c and the remainder is zero, then c is a
zero of f and c is a root of the polynomial equation f(x) = 0.
2. If f(x) is divided by x – c and the remainder is zero, then x – c
is a factor of f(x).
Blitzer, Algebra for College Students, 6e – Slide #14 Section 11.2
Zeros of Polynomial Functions
The Rational Zero Theorem
n
n 1
If f ( x)  an x  an 1 x      a1 x  a0 has integer coefficients and p/q where
p/q is reduced to lowest terms) is a rational zero of f, then p is a factor of the
constant term in the polynomial and q is a factor of the leading coefficient of
the polynomial .
To use this theorem – list all the integers that are factors of the constant term, a0
Then list all the integers that are factors of the leading coefficient, a n . Finally,
list all possible rational zeros:
__Factors of the constant term
Possible Rational Zeros =
Factors of the leading coefficient
Blitzer, Algebra for College Students, 6e – Slide #15 Section 11.2
Using the Rational Zero Theorem
EXAMPLE
List all possible rational zeros of
f ( x)  15x3  14 x 2  3x  2
Solution
The constant term is -2 and the leading term is 15. Factors of the
constant term are plus and minus 1 and 2. Factors of the leading
term are plus and minus 1,3,5, and 15. Then by the Rational Zero
Theorem, possible rational zeros are:
 1,2
1 2 1
2
1
2
 1,2, , , , ,
,
 1,3,5,15
3 3 5 5 15 15
There are 16 possible rational zeros. The actual solution set is {-1,-1/3,2/5}
which contains three of the possible 16 zeros.
Blitzer, Algebra for College Students, 6e – Slide #16 Section 11.2
Zeros of Polynomial Functions
Properties of Polynomial Equations
1. If a polynomial equation is of degree n, then counting
multiple roots separately, the equation has n roots.
2. If a + bi is a root of a polynomial equation (b is not 0),
then the imaginary number a – bi is also a root. Imaginary
roots, if they exist, occur in conjugate pairs.
Blitzer, Algebra for College Students, 6e – Slide #17 Section 11.2
Zeros of Polynomial Functions
Descarte's Rule of Signs
1.
(a)
(b)
2.
(a)
(b)
Let f(x) be a polynomial with real coefficients.
The number of positive real zeros of f is either
the same as the number of sign changes of f(x) or
less than the number of sign changes of f(x) by a positive
even integer
The number of negative real zeros of f is either
The same as the number of sign changes of f(-x) or
Less than the number of sign changes of f(-x) by a
positive even integer. If f(-x) has only one variation in
sign, then f has exactly one negative real zero.
Blitzer, Algebra for College Students, 6e – Slide #18 Section 11.2
Using Descartes’s Rule of Signs
EXAMPLE
Determine the number of positive and negative real zeros of
f ( x)  x  2 x  5 x  4
3
SOLUTION
2
We first note that since all the terms are positive, there are no
sign changes in f(x). That means that there are no positive real
zeros.
We next find f(-x). We get:
f ( x)  ( x) 3  2( x) 2  5( x)  4
  x3  2 x 2  5x  4
Now count the sign changes. There are three here, a change from a negative
to a positive between the first two terms, then a change from a positive to a
negative between the next two terms and finally a change from a negative to
a positive between the last two terms. Then the number of nonnegative real zeros
is equal to the number of sign changes or is less than this number by an even integer.
So there are either 3 negative real zeros or there are 3 – 2 = 1 negative real zeros.
Blitzer, Algebra for College Students, 6e – Slide #19 Section 11.2
Using Descarte's Rule of Signs
Continued
Determine the number of positive and negative real zeros of
f ( x)  x  2 x  5 x  4
3
2
Since there are three roots to this equation, and there are no positive
roots and either 1 or 3 negative roots – it follows that there must be
three negative roots.
We would now use the Rational Root Theorem to find that possibilities for
rational roots would be +1,-1,+2,-2,+4,-4. But since there are no positive roots,
we would just use synthetic division to check -1,-2, and -4. Using synthetic
division we get a 0 remainder when we divide by x+1, so -1 is a root. As it
turns out, -1 is the only real root and the other two roots are imaginary numbers
in a conjugate pair. Complete the synthetic division process to verify this.
Blitzer, Algebra for College Students, 6e – Slide #20 Section 11.2