Download Zeros and Roots of Polynomial Functions 1 New Theorem`s

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Zeros and Roots of Polynomial Functions
Finding a Root (zero or x-intercept) of a polynomial is identical to the process of factoring
a polynomial.
Theorems used to find the Zero’s (or Roots) of a Polynomial Function:
Z er o0 s
n-Zero’s Theorem: If p .x/ is of Degree n, then it has at most n
The Remainder Theorem: If
The Factor Theorem: If
p .x/
D q .x/ C r then p .c/ D r
x c
p .x/
D q .x/ C 0 then p .x/ D .x
x c
c/ q .x/
Also since p .c/ D 0 (by the remainder theorem) we also have that
x D c is an x-intercept of p .x/
Intermediate Value Theorem: If p .a/ p .b/ D
1 where p .x/ is a continuous function
on the interval of x 2 .a; b/ then there exists some x D c where
c 2 .a; b/ and p .c/ D 0
1
New Theorem’s
Theorem 1 Rational Root (zero) Theorem
Given that p .x/ D an x n C an 1 x n 1 C
C ax C a0 where all ak are Real , then any
Rational Roots of p .x/ will be of the form of (factors of a0 / .factors of an /
This theorem is only useful if the Roots (zeros) of the polynomial are rational to begin with.
If they are irrational (or imaginary roots) this theorem will not help us, and we will have to
use another "trick" to locate them.
The rationale can be inductively shown by solving a simple polynomial equation:
4x 4
37x 2 C 9 D 0
To factor this we need to find the Factors of .4/ .9/ D 36 that add up to 37: Which would
be 36 and 1:
0 D 4x 4
36x 2
1x 2 C 9
0 D 4x 2 x 2
9
0 D x2
4x 2
0 D .x
thus:
9
1 x2
1
3/ .x C 3/ .2x
x D f3; 3;
1 1
; g
2 2
9
1/ .2x C 1/
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9
1; 3; 9
a0
D we get
D f1; 3; 9; 1=2; 3=2; 9=2; 1=4; 3=4; 9=4g:
an
4
1; 2; 4
Notice that the roots found are from this list . / :
Setting up all the factors of
Example 2 List the possible Rational Roots of f .x/ D 2x 3
First write
21x 2 C 49x C 30
a0
30
1; 2; 3; 5; 6; 10; 15; 30
D
and their respected factors D
an
2
1; 2
Next write all combinations of these factors:
1 2 3 5 6 10 15 30
; ; ; ; ; ; ;
1 1 1 1 1 1 1 1
1 2 3 5 6 10 15 30
; ; ; ; ; ; ;
2 2 2 2 2 2 2 2
After simplifying and removing repeats we get our final list
f1; 2; 3; 5; 6; 10; 15; 30; 1=2; 3=2; 5=2; 15=2g
If this function had a rational root, it will have to be one of these numbers, positive or
negative.
2
Descarte’s Rule of Sign:
To help us find the possible rational roots it would be helpful to know how many of them
might be positive or negative. Using Descarte’s Rule will help. To find what is the
Maximum number of Positive Roots in a Polynomial (with no missing powers), one just
has to count the number of Sign Changes:
Example 3 How many possible positive roots exist for f .x/ D 6x 5 C 11x 4
33x 2 C 11x C 6
33x 3
There are two sign changes here, C11x 4 to 33x 3 and 33x 2 to C11x; thus there are
at most TWO Positive Roots. Knowing that the Number of Roots is equal to the Degree of
the polynomial we can deduce that there will be at most THREE Negative Roots.
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The Search for Zeros
Example 4 Find all the Zeros of f .x/ D 2x 4
7x 3
4x 2 C 27x
18
Using everything that we have available, it would be better to analyze this first to know
what we are looking for
n-Zeros Theorem. f .x/ has at most 4 real roots
Descarte’s Rule of Sign:
f .x/ has Three Changes in Sign, thus there are Three Positive roots, and One Negative
Root, for the total of 4 zer o0 s
Rational Root Theorem: List Possible Roots
a0
1; 2; 3; 6; 9; 18
18
D
D
D
an
2
1; 2
1 3 9
f1; 2; 3; 6; 9; 18; ; ; g
2 2 2
The Remainder Theorem: Find f .c/ using the above numbers and synthetic division zero
is of interest!
1j
2
2
2j
2
2
7
2
9
4
9
5
7
4
11
4
22
18
27
5
22
27
36
9
18
22
40
18
18
0
Factor Theorem: Since the remainder is Zero, that means three things:
1. x D
2 is a Zero of f .x/
2. .x C 2/ is a Factor of f .x/
3. The last line q .x/ D 2x 3
11x 2 C 18x
9 is also a factor of f .x/
Now we can use Step 4 and 5 over again using q .x/ to find more factors. But we
don’t need to search for any more Negative Zero’s because we found the ONE that existed (Descarte)
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1.
1j
2
11
2
9
2
2. x D 1 is a Zero, .x
18
9
9
9
9
0
1/ is a Factor, and q .x/ is now: 2x 2 9xC9 (quadratic=factor!)
3.
2x 2
2
9x C 9
2x
6x 3x C 9
2x .x 3/ 3 .x 3/
.x 3/ .2x 3/
3
x D f3; g
2
Putting it all together:
Zeros:
xD
2; 1; 3;
3
2
Factored:
f .x/ D 2x 4 7x 3
D .x C 2/ .x
4x 2 C 27x 18
1/ .x 3/ .2x 3/
Corollary 5 Given p .x/ is an odd degree polynomial then there exist at least one real
root.
This is just a combination of Polynomial End Behavior and Intermediate Value Theorem.
Now I have been stating over and over that the Real Roots are AT MOST so many. The
reason for this is simple: there may be complex or Imaginary roots involved.
Example 6 Find the Roots of p .x/ D 2x 3
Here we can start with x D
x 2 C 18x
9 given that
1
is a zero of p .x/
2
1
and "reduce" our polynomial into one less degree.
2
1
2
2
1
18
9
2
1
0
0
18
9
0
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So upon setting up q .x/ D 0 we get:
q .x/ D 2x 2 C 0x C 18
0 D 2x 2 C 18
2x 2 D
18
2
x D 9
x D f3i; 3ig
So the Roots of p .x/ D 2x 3
x 2 C 18x
1
; 3i; 3i :
2
9 are x D
Notice that there are Three Roots and the Degree of the Polynomial is Three...
Theorem 7 Complex Root Theorem
If p .x/ is a polynomial function with Real Coefficients and a C bi is a Root, then a
must also be a Root of p .x/ :
bi
This theorem in short states that if the Coefficients are real, then all Complex Roots come
in Conjugate Pairs! If there is one, there will be another, its conjugate.
Example 8 If p .x/ is a polynomial with real coefficients and its roots are f 1; 3; 2g then
find a polynomial f .x/ that is equal to a constant multiple of p .x/.
NOTE: f .x/ D k p .x/ ; but with out more information we will never find the
value for k; but we must account for its existence.
By the factor theorem we can take each root .c/ and write he binomial associated with it
.x .c// :
xD
1 becomes .x C 1/ D 0
x D 3 becomes .x
3/ D 0
x D 2 becomes .x
2/ D 0
So
p .x/ D k .x C 1/ .x
3/ .x
2/
Example 9 Given p .x/ is a polynomial function with real coefficients and p .x/0 s roots
are f 3i; 2 C ig ; find f .x/ D k p .x/ :
Since both of these roots are not real, they must match up with a Conjugate Pair:
f 3i; 3i; 2 C i; 2
ig
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in which the Factors would be:
f .x/ D .x . 3i// .x .3i// .x .2 C i// .x
f .x/ D .x C 3i/ .x 3i/ .x .2 C i// .x .2
.2
i//
i//
Multiplying out the first two factors (remembering rule of imaginary numbers) we get:
h
i
f .x/ D x 2 C 9 .x .2 C i// .x .2 i//
now the other two complex factors are a little more tricky, so I will use a simple substitution
to help. Let A D .2 C i/ and B D .2 i/ then:
.x
.2 C i// .x .2 i//
becomes
.x A/ .x B/
x2
x
2
Bx
Ax C AB
.A C B/ x C AB
So all we need to find is the value of .A C B/ and .A B/
A C B D .2 C i/ C .2 i/ D 4
A B D .2 C i/ .2 i/ D 4 C 1 D 5
Thus
.x
x
2
x
2
.2 C i// .x
.2
i//
.A C B/ x C AB
.4/ x C .5/
and
h
i
f .x/ D x 2 C 9 .x
.2 C i// .x
.2
i//
becomes
f .x/ D x 2 C 9
x2
4x C 5
It should be clear that any Polynomial can be broken into exactly the same number of
Unique Product of Linear Factors (real and complex) as the degree of the polynomial.
Thus any p .x/ D .x cn / .x cn 1 / .x cn 2 /
.x c1 / k where the c’s can be any
complex number and n is the degree of p .x/. Remember that a complex number is also
Real: 5 D 5 C 0i