Download Rational Zeros of a Polynomial Function and Factoring Over R and more Study notes Calculus in PDF only on Docsity!
SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS
Assume f (^) ( x ) is a nonconstant polynomial with real coefficients written in standard form.
PART A: TECHNIQUES WE HAVE ALREADY SEEN
Refer to: Notes 1.31 to 1. Section A.5 in the book Notes 2.
Refer to
Factoring (Notes 1.33)
Methods for Dealing with Quadratic Functions (Book Section A.5: pp.A49-51)
a) Square Root Method (Notes 1.31, 2.45) b) Factoring (Notes 1.33) c) QF (Notes 1.34, 2.45) d) CTS (Completing the Square) (Book Section A.5: p.A49)
Bisection Method (for Approximating Zeros) (Notes 2.20 to 2.21)
Synthetic Division and the Remainder Theorem (for Verifying Zeros) (Notes 2.33)
PART B: RATIONAL ZERO TEST
Rational Zero Test (or Rational Roots Theorem)
Let (^) f (^) ( x ) be a polynomial with integer (i.e., only integer) coefficients written in standard form: an x n^ + an 1 x n ^1 + ... + a 1 x + a 0
(^ each constant ai Z ;^ a^ n ^ 0;^ a 0 ^ 0;^ n^ Z +)
If f (^) ( x ) has rational zeros, they must be in the list of ±
p q
candidates, where:
p is a factor of a 0 , the constant term, and q is a factor of an , the leading coefficient.
Note: We require a 0 0. If a 0 = 0 , try factoring out the GCF first.
Example
Factor f (^) ( x ) = 4 x^3 5 x^2 7 x + 2 completely, and find all of its real zeros.
Solution
Since the GCF = 1 , and Factoring by Grouping does not seem to help, we resort to using the Rational Zero Test. We will now list the candidates for possible rational zeros of f (^) ( x ).
p (factors of the constant term, 2): ±1, ± 2 q (factors of the leading coefficient, 4): ±1, ± 2, ± 4
Note: You may omit the ± symbols above if you use them below.
List of ±
p q
candidates:
Simplified:
±1, ±
Redundant
Method 2
We may also use the Synthetic Division process and see if we get a 0 remainder.
We do not get a 0 remainder, so 1 is not a zero of f (^) ( x ).
Method 3
Observe from both previous methods that we can compute f (^) ( ) 1 by simply adding up the coefficients of
f (^) ( x ) in standard form. This does not work in general for other values of k , though.
Let’s try k = 2.
Let’s use the Synthetic Division / Remainder Theorem method:
We do get a 0 remainder, so 2 is a zero of f (^) ( x ).
This turns out to be the key that cracks the whole problem.
Incidentally, this is the same f (^) ( x ) that we saw in Notes 2.33-2.35.
Now we know how our “little bird” got its info!
By the Factor Theorem, (^) ( x (^2) ) must be a factor of f (^) ( x ).
We can find q x ( ) , the other (quadratic) factor, by using the last row of the
table.
f (^) ( x ) = (^) ( x (^2) ) (^) ( 4 x^2 + 3 x (^1) )
Factor (^) q x ( ) completely over the reals:
f (^) ( x ) = (^) ( x - (^2) ) ( 4 x - (^1) ) ( x + 1 )
Remember that we always know how to break down a quadratic (use the QF if you have to), although its zeros may or may not be real. We no longer have to rely on rational zeros at this point.
The zeros of f (^) ( x ) are the zeros of these factors:
Observe that all three are rational and appeared in our list of candidates for rational zeros. Any one of these three could have been used to start cracking the problem.
Below is a graph of f (^) ( x ) = 4 x^3 5 x^2 7 x + 2. Where are the x -intercepts?
Note: If we can get a graph of f (^) ( x ) beforehand, then we may be able to
choose our guesses for rational zeros more wisely.
PART C: FACTORING OVER VARIOUS SETS
Recall our Venn diagram from Notes P.03:
Factoring over Z (the Integers)
Example
In our Example in Part B, we factored:
4 x^3 5 x^2 7 x + 2 = (^) ( x - (^2) ) ( 4 x - (^1) ) ( x + (^1) )
This is an example of factoring over Z , because we only use integers as coefficients (including constant terms within factors).
Factoring over Q (the Rationals)
Example
Let’s factor a 4 out of the second factor in the previous Example.
4 x^3 5 x^2 7 x + 2 = ( x - 2 ) ( 4 x - 1 ) ( x + 1 ) Factored over Z
= 4 ( x - 2 ) x -
^
( x^ +^1 ) ^ Factored over^ Q
The “Factored over Z ” expression is also an example of factoring over Q ,
but this new factorization over Q immediately identifies
as a zero.
Factoring over R (the Reals)
Example
x^2 (^3) is prime (or irreducible) over Z and Q ; it cannot be factored further (nontrivially; breaking out a 1 or a 1 doesn’t count) using only integer or rational coefficients. However, it can be factored over R.
x^2 3 = ( x + 3 ) ( x 3 )
Note that 3 and 3 are immediately identified as zeros.
We have factored completely over Z (and Q ). Let us now factor completely over R.
x^5 + x^3 6 x = x x ( 2 + 3 ) ( x^2 2 ) Reminder
= x x (^2 + 3 ) ( x + 2 ) ( x - 2 ) Factored over R
(^ x^2 +^3 ) is a quadratic that is irreducible over the reals. It therefore
yields no real zeros. (We need more theorems to show this.)
From the other three factors, we obtain 0, - 2, and 2 as real zeros.
Example
Factor f ( x ) = x^5 + x^3 6 x over C and find all of its real zeros.
Solution
We continue with our work from the previous Example.
We can factor ( x^2 + 3 ) further over C.
x^5 + x^3 6 x = x x ( 2 + 3 ) ( x + 2 ) ( x - 2 ) Factored over R
= x x + i ( 3 ) ( x - i 3 ) ( x + 2 ) ( x - 2 ) Factored over C
We immediately obtain the five complex zeros of f ( x ) :
0, - i 3, i 3 , - 2, and 2
PART D: COMPLEX CONJUGATE PAIRS OF ZEROS
Complex Conjugate Pairs Theorem
Let a , b R.
If f (^) ( x ) is a polynomial with (only) real coefficients, then:
a + bi is a zero of f (^) ( x ) a bi is a zero of f (^) ( x ).
In our last Example in Part C, if we know that i 3 is a zero of f (^) ( x ) , then we can
conclude that i 3 must also be a zero.
Technical Note: The theorem requires real coefficients. Observe that x i has i as a zero but not i.
Note: There is a Conjugate Pairs Theorem for a quadratic polynomial f (^) ( x ) with
(only) rational coefficients. Consider f (^) ( x ) = ax^2 + bx + c , where a , b , c Q and
a 0. As an example, 2 + 3 is a zero of such an f (^) ( x ) 2 3 is. The structure of the QF implies this.
PART F: THE LINEAR FACTORIZATION THEOREM (LFT)
The Linear Factorization Theorem (LFT)
If f (^) ( x) is a nonconstant polynomial in standard form with real coefficients, then it must have a factorization into linear factors of the form:
f (^) ( x) = an ( x − c 1 ) ( x − c 2 ) (^) ( x − cn)
( an ∈R;^ an ≠^0 ;^ each^ ci ∈C)
Note: The zeros of f (^) ( x) are then c 1 , c 2 ,…, cn. Note: There may be repetitions of a zero ci , based on the multiplicity of ci. Note: an is the leading coefficient of f (^) ( x).
Technical Note: The LFT is proven using the FTA and the Factor Theorem. See p.193 of the textbook.
Technical Note: This helps explain the Complex Conjugate Pairs Theorem in Notes 2.56.
Example
Let f (^) ( x) = x^5 − 8 x^4 + 16 x^3.
x^5 − 8 x^4 + 16 x^3 = x^3 ( x^2 − 8 x + (^16) )
= x^3 ( x − (^4) )
2
It may be said that f (^) ( x) has 5 zeros: 0, 0, 0, 4, and 4.
f (^) ( x) has only 2 distinct zeros: 0 and 4. They are both repeated zeros: The multiplicity of 0 is 3, and the multiplicity of 4 is 2.
You can think of x as (^) ( x − (^0) ).
Recall our Examples in Notes 2.52 and 2.53.
4 x^3 − 5 x^2 − 7 x + 2 = ( x - 2 ) ( 4 x - 1 ) ( x + 1 ) ← Factored over Z
= 4 ( x - 2 ) x -
⎠⎟^
( x^ +^1 ) ←^ Factored over^ Q
The second factorization is in “LFT Form.” The zeros can be immediately read off (watch out for signs, though).
PART G: FACTORING OVER R
“Factoring Over R” Theorem
Let f ( x) be a nonconstant polynomial in standard form with real coefficients.
Its complete factorization over R (the reals) consists of:
- Linear factors,
- Quadratic factors that are irreducible over R (i.e., have no real zeros), or
- Some product of the above, possibly including repeated factors, and
- Maybe a nonzero constant factor.
One consequence: A 3rd- or higher-degree polynomial f ( x) with real coefficients must
be factorable (reducible) over the reals. Knowing how to factor such an f ( x) may pose a
problem, however!
Note: We will need this theorem when we do Partial Fraction Decompositions in Section 2.7.
Technical Note: See the Proof on p.193 of the textbook. Consider the “LFT Form” of
f ( x). If all the zeros of f ( x) are real, then it can be factored accordingly. If there exists
an imaginary zero ci , then its conjugate ci must also be a zero by the Complex
Conjugate Pairs Theorem, and the product ( x − ci) ( x − ci) of their corresponding factors
must have real coefficients (see p.193); this product would be a quadratic factor of f ( x)
with real coefficients that is irreducible over R (because its zeros are not real). Keep pairing off complex conjugate pairs of imaginary zeros until the remaining factors have only real coefficients.
Parity
Two integers have the same parity ⇔ They are both even or both odd.
Descartes’s Rule of Signs
We want information about z+^ and z−^ , where:
z+^ is the number of positive real zeros of f (^) ( x) , and
z−^ is the number of negative real zeros of f (^) ( x).
Let v+^ be the number of variations in sign in f (^) ( x) (written in standard form).
Let v−^ be the number of variations in sign in f (^) ( −x) (written in standard form).
Then,
0 ≤ z+^ ≤ v+^ , where z+^ has the same parity as v+^ , and 0 ≤ z−^ ≤ v−^ , where z−^ has the same parity as v−^.
Warning: A zero of multiplicity k is counted k times here. For example, f (^) ( x) = x^3 + 3 x^2 + 3 x + 1 , which factors as (^) ( x + (^1) )
3 , is said to have 3 real zeros: − 1 , − 1 , and − 1.
Example
Based on Descartes’s Rule of Signs, give the possible values of z+^ and z− for f (^) ( x) = 4 x^3 − 5 x^2 − 7 x + 2. (We’ve used this f (^) ( x) in previous sections.)
Solution
Find possible values for z+^ :
We see that v+^ = 2 , an even integer. Therefore, 0 ≤ z+^ ≤ 2 , where z+^ is also even. The possible values for z+^ are then 0 and 2.
Tip: Observe that we start with 2 and count down by twos; we stop before reaching negative numbers. This is similar to listing possible numbers of turning points for polynomial graphs in Section 2.2.
Find possible values for z−^ :
We see that v−^ = 1 , an odd integer. Therefore, 0 ≤ z−^ ≤ 1 , where z−^ is also odd. The only possible value for z−^ is 1.
(In other words, f (^) ( x) must have exactly one negative real zero.)
PART I: UPPER AND LOWER BOUND RULES FOR ZEROS
a is a lower bound for the real zeros of f , and b is an upper bound for them ⇔ All the real zeros of f lie in the interval ⎡⎣ a, b⎤⎦.
It is easier to demonstrate the Upper and Lower Bound Rules rather than to state them in general.
We require that the leading coefficient of f (^) ( x) , an , be positive, and that all the coefficients be real.
Example and Demonstration
Show that all the real zeros of f (^) ( x) = 4 x^3 − 5 x^2 − 7 x + 2 must lie in the interval
⎡⎣ − 1 , 3 ⎤⎦.
Solution
Use Synthetic Division to divide f (^) ( x) by x − 3 :
Because 3 > 0 , and all the entries in the last row are nonnegative, 3 is an upper bound for the real zeros of f.
Use Synthetic Division to divide f (^) ( x) by x − (^) (− 1 ) :
Because - 1 < 0 , and the entries in the last row alternate between nonnegative and nonpositive entries, − 1 is a lower bound for the real zeros of f.
In fact, because we get a 0 remainder, − 1 must be a zero of f.
Therefore, all the real zeros of f (^) ( x) must lie in the interval ⎡⎣ − 1 , 3 ⎤⎦.
Note: These rules can be used (possibly in conjunction with Descartes’s Rule of Signs and/or a graph) to shrink the list of candidates for zeros resulting from the Rational Zero Test. The information obtained from these rules can also help us use the Intermediate Value Theorem (see Notes 2.20-2.21 on Section 2.2) more effectively in attempting to locate where zeros may be.
