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Zeros of a Polynomial Function
An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we will study more methods that help us find the real zeros of a polynomial, and thereby factor the polynomial.
Rational Zeros of Polynomials:
The next theorem gives a method to determine all possible candidates for rational zeros of a polynomial function with integer coefficients.
Rational Zeros Theorem:
If the polynomial P x ( ) = a xn n^ + an − 1 x n −^1 + ...+ a x 1 + a 0 has integer
coefficients, then every rational zero of P is of the form
p q
where p is a factor of the constant coefficient a 0 and q is a factor of the leading coefficient an
Example 1: List all possible rational zeros given by the Rational Zeros Theorem of P ( x ) = 6 x^4 + 7 x^3 - 4 (but don’t check to see which actually are zeros).
Solution:
Step 1: First we find all possible values of p , which are all the factors of a 0 (^) = 4. Thus, p can be ±1, ±2, or ±4.
Step 2: Next we find all possible values of q , which are all the factors of an = 6. Thus, q can be ±1, ±2, ±3, or ±6.
Step 3: Now we find the possible values of p q by making combinations
of the values we found in Step 1 and Step 2. Thus, p q will be of
the form factors of 4factors of 6. The possible (^) qp are
Example 1 (Continued):
Step 4: Finally, by simplifying the fractions and eliminating duplicates, we get the following list of possible values for p q.
Now that we know how to find all possible rational zeros of a polynomial, we want to determine which candidates are actually zeros, and then factor the polynomial. To do this we will follow the steps listed below.
Finding the Rational Zeros of a Polynomial:
- Possible Zeros: List all possible rational zeros using the Rational Zeros Theorem.
- Divide: Use Synthetic division to evaluate the polynomial at each of the candidates for rational zeros that you found in Step 1. When the remainder is 0, note the quotient you have obtained.
- Repeat: Repeat Steps 1 and 2 for the quotient. Stop when you reach a quotient that is quadratic or factors easily, and use the quadratic formula or factor to find the remaining zeros.
Example 2: Find all real zeros of the polynomial P ( x ) = 2 x^4 + x^3 – 6 x^2 – 7 x – 2.
Solution:
Step 1: First list all possible rational zeros using the Rational Zeros Theorem. For the rational number (^) qp to be a zero, p must be a factor of a 0 = 2 and q must be a factor of a n = 2. Thus the possible rational zeros, p q , are
Example 2 (Continued):
Step 4: At this point the quotient polynomial, 2 x^2 – 3 x – 2, is quadratic. This factors easily into ( x – 2)(2 x + 1), which tells us we have
zeros at x = 2 and
x = − , and that P factors as
2 x^4 + x^3 – 6 x^2 – 7 x – 2 = ( x + 1)(2 x^3 – x^2 – 5 x – 2) = ( x + 1) ( x + 1)(2 x^2 – 3 x – 2) = ( x + 1)^2 (2 x^2 – 3 x – 2) = ( x + 1)^2 ( x – 2)(2 x + 1)
Step 5: Thus the zeros of P ( x ) = 2 x^4 + x^3 – 6 x^2 – 7 x – 2 are x = –1, x = 2,
and
x = −.
Descartes’ Rule of Signs and Upper and Lower Bounds for Roots:
In many cases, we will have a lengthy list of possible rational zeros of a polynomial. A theorem that is helpful in eliminating candidates is Descartes’ Rule of Signs.
In the theorem, variation in sign is a change from positive to negative, or negative to positive in successive terms of the polynomial. Missing terms (those with 0 coefficients) are counted as no change in sign and can be ignored. For example,
has two variations in sign.
Descartes’ Rule of Signs: Let P be a polynomial with real coefficients
- The number of positive real zeros of P ( x ) is either equal to the number of variations in sign in P ( x ) or is less than that by an even whole number.
- The number of negative real zeros of P ( x ) is either equal to the number of variations in sign in P (– x ) or is less than that by an even whole number.
Example 3: Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros P ( x ) = 6 x^3 + 17 x^2 – 31 x – 12 can have. Then determine the possible total number of real zeros.
Solution:
Step 1: First we will count the number of variations in sign of
P x ( ) = 6 x^3^ + 17 x^2 − 31 x − 1 2.
Since there is only one variation, P ( x ) has one positive real zero.
Step 2: Now we will count the number of variations in sign of
P ( − x ) = − 6 x^3 + 17 x^2 + 31 x − 1 2.
P (– x ) has two variations in sign, thus P ( x ) has two or zero negative real zeros.
Step 3: Finally by combining our findings in the previous steps, P ( x ) has either one or three real zeros.
Another theorem useful in eliminating candidates for real zeros of a polynomial is the Upper and Lower Bounds Theorem.
We say that a is a lower bound and b is an upper bound for the roots of a polynomial equation if every real root c of the equation satisfies a ≤ c ≤ b.
Example 5: Find all rational zeros of P ( x ) = 6 x^4 – 23 x^3 + 3 x^2 + 32 x + 12, and then find the irrational zeros, if any and graph the polynomial. If appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes’ Rule of Signs, the quadratic formula, or other factoring techniques.
Solution:
Step 1: First we will use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros P ( x ) = 6 x^4 – 23 x^3 + 3 x^2 + 32 x + 12 can have.
Counting the number of variations in sign of P ( x ) and P (– x ), we see that P ( x ) = 6 x^4 – 23 x^3 + 3 x^2 + 32 x + 12 has zero or two positive real zeros and zero or two negative real zeros, making a total of either zero, two or four real zeros.
Step 2: Next, using the Rational Zeros Theorem, we list the possible rational zeros of P , and then begin testing candidates for zeros.
The possible rational zeros of P are
1 1 1 2 4 3 , , , , 1, , , 2, 3, 4, 6, 12 6 3 2 3 3 2
We will check the positive candidates first, beginning with the smallest.
Example 5 (Continued):
is not 6
a zero
is not 3
a
zero
is not 2
a zero
… (we continue testing candidates until) …
6 11 19 6 0 P 2 0
So 2 is a zero, and P ( x ) = ( x – 2)(6 x^3 – 11 x^2 – 19 x – 6).
Step 3: We continue by factoring the quotient. Again use the Rational Zeros Theorem to list the possible rational zeros, and test candidates for zeros.
The possible rational zeros of the quotient are
1 1 1 2 3 , , , , 1, , 2, 3, 6 6 3 2 3 2
From Step 1 we know we cannot have just one positive real zero, so we will continue testing positive candidates. Also, in
Step 2, we already tested
, , , , 1, , and 6 3 2 3 3 2
and
found they were not zeros, so we do not need to test them again, but we do need to test 2 again.
Example 5 (Continued):
P is of even degree and its leading coefficient is positive, therefore it has the following end behavior:
y → ∞ as x → ∞ and y → ∞ as x → – ∞
Thus, the graph of P ( x ) = 6 x^4 – 23 x^3 + 3 x^2 + 32 x + 12 is completed below.
Example 6: Find all rational zeros of P ( x ) = 2 x^4 + 7 x^3 + 9 x^2 + 21 x + 9, and then find the irrational zeros, if any and graph the polynomial. If appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes’ Rule of Signs, the quadratic formula, or other factoring techniques.
Solution:
Step 1: Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros P ( x ) = 2 x^4 + 7 x^3 + 9 x^2 + 21 x + 9 can have.
P x ( ) = 2 x^4^ + 7 x^3^ + 9 x^2 + 21 x + 9
P ( x ) has no positive real zeros, and zero, two or four negative real zeros, making a total of either zero, two or four real zeros.
Step 2: Use the Rational Zeros Theorem to list the possible rational zeros of P , and synthetic division to begin testing candidates for zeros.
The possible rational zeros of P are
1 3 9 , 1, , 3, , 9 2 2 2
Since in Step 1 we determined there are no positive real zeros, we can begin testing the negative candidates, starting with the value closest to zero.
1 2 7 9 21 9 2 1 3 3 9
P
Example 6 (Continued):
P has the following end behavior:
y → ∞ as x → ∞ and y → ∞ as x → – ∞
Up to this point we have studied polynomials in the real number system, and we have seen that an n th-degree polynomial can have at most n real zeros. In this section we will study polynomials in the complex number system where an n th-degree polynomial has exactly n zeros, and so can be factored into exactly n linear factors.
Fundamental Theorem of Algebra:
In 1799 the German mathematician C. F. Gauss proved the Fundamental Theorem of Algebra. This Theorem forms the basis for much of our work in factoring polynomials and solving polynomial equations.
Fundamental Theorem of Algebra:
Every polynomial P x ( ) = a xn n^ + an − 1 x n −^1 + ... + a x 1 + a 0 ( n ≥ 1, an ≠ 0 )
with complex coefficients has at least one complex zero.
Because any real number is also a complex number, the theorem applies to polynomials with real coefficients as well.
The conclusion of the Fundamental Theorem of Algebra is that for every polynomial P ( x ), there is a complex number c 1 such that
rom the Factor Theorem (Section 5.2), this tells us x – c is a factor of P ( x ). Thus we
where Q ( x ) has degree n – 1. If the quotient Q ( x ) has degree ≥ 1 we can repeat the procedure of obtaining a factor and a quotient with degree 1 less than the previous quotient until we arrive at the complete factorization of P ( x ). This process is summarized by the next theorem.
Complete Factorization Theorem:
If P ( x ) is a polynomial of degree n > 0, then there exist complex numbers a, c 1 , c 2 ,... c (^) n (with a ≠ 0) such that
P c ( 1 )= 0
F 1
can write
P x ( ) = ( x − c 1 ) Q x ( )
P x ( ) = a x ( − c 1 )( x − c 2 ) ... ( x − cn )
Theorem the numbers c 1 , c 2 ,... c (^) n are the zeros of P. If the factor x – c appears k times in the complete
Zeros Theorem:
Every polynomial of degree n ≥ 1 has exactly n zeros, provided that a zero of
In the Complete Factorization These zeros need not all be different. factorization of P ( x ), then we say that c is a zero of multiplicity k.
The next theorem follows from the Complete Factorization Theorem.
multiplicity k is counted k times.
Example 8: Find a polynomial with integer coefficients that satisfies the given conditions that P has degree 5, zeros 0, 2, 3 i , and –3 i , with 2 a zero of multiplicity 2, and with P (2) = 26.
We are given that the zeros are 0, 2, 3 i , and –3 i , with 2 a zero of multiplicity 2, thus by the factor theorem, the required polynomial has the form
Solution:
Step 1:
2 P x = a x − 0 x − 2 x − 3 i x − − 3 i
Note that the factor x – 2 is squared because 2 is a zero multiplicity 2.
with
Example 8 (Continued):
Step 2: Next we expand and simplify the polynomial.
( ) ( )
( )
2
2
(^2 )
5 4 3 2
2 9 Difference of squares
4 11 36 18 Expand
P x a x x x i x i
ax x x i x i
ax x x
a x x x x x
Step 3: Now we must determine what a is. We know that
( ) ( )
2 25 4 2^4 11 2^3 36 2^2 18 2
P a a
So a = − 12.
nd the
(Note: If we are given no information about P other than its zeros a their multiplicity, we can choose any number for a. Usually a = 1 is simplest choice.)
Step 4: Thus
P x ( (^) ) 12 ( x^5 4 x^4 11 x^3^36 x^218 x ) (^1 5 2 4 113 18 ) = 2 x − x + 2 x − x + x
Example 9: Find all zeros of the polynomial P ( x ) = x^4 + x^3 + 3 x^2 – 5 x.
Solution:
x in previous sections. Since x is a common factor, we have
Step 1: We begin by factoring the polynomial P ( x ) = x^4 + x^3 + 3 x^2 – 5 using the tools we studied
( )
4 3 2
3 2
3 5 Factor
P x x x x x
x x x x x
Example 10 (Continued):
Step 3: Now we expand and simplify the equation.
1 2 1 2 (^2 5 ) 2 2
2 5 3 2 2 2 2 2 5 3 2 2 2
1 1 E
1 Difference of
2 2
Squares
P x a x x x i x i
a x x x i x i
a x x x i x i
a x x x i
a x x x x
xpand (^2 5 ) = a x + 2 x − 2 x − 1 − i x − 1 + i Regroup
(^4 13 2 ) 2 3 2 2
Expand
= a x + x − x + x − E xpa dn
Step 4: Finally we must make all coefficients integers. The lowest common denominator of the coefficients is 2, so we set a = 2 and get
1 3 2 13 2 2 4 3 2
x − 3 x + x − 2
Note that any other polynomial that satisfies the given requirements must
nomial factors completely into linear factors if we use complex e complex numbers, then a polynomial with real coefficients
Linear and Quadratic Factors Theorem:
very p lynomial with real coefficients can be factored into a product of linear and irreducible quadratic factors with real coefficients.
P x = 2 x^4 +
= 2 x + x − 6 x + 13 x − 4
be an integer multiple of this one.
We have seen that a poly numbers. If we do not us can always be factored into linear and quadratic factors. A quadratic polynomial with no real zer s is called o irreducible over the real numbers. Such a polynomial cannot be factored without using complex numbers.
E o
Example 11: Given the polynomial P ( x ) = x^4 + 2 x^2 – 63
real coefficients. fficients.
Solution (a):
(a) Factor P into linear and irreducible quadratic factors with (b) Factor P completely into linear factors with complex coe
( )( )
( )( )(^ )
4 2
2 2
2
Factor like a quadratic
uadratic Equation
of squar
Q
or difference es formula
P x x x
x x
x x x
The factor x^2 – 9 is irreducible since it has only the imaginary zeros ± 3 i.
Solution (b):
( ) ( )( )( )
( )( )(^ )(^ )
7 7 3 3 Quadratic Equation
x
x x x i x i
P x = x − 7 x + 7
