Download KYMATYC Conference, 2015, Power Point presentation

Finding Zeros
of
Polynomial Functions
Taylor Johnson
([email protected])
Elizabethtown Community
&
Technical College
Tools for Searching for Zeros
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Remainder Theorem
Factor Theorem
Intermediate Value Theorem
Synthetic Division
Rational Zero Theorem
Signs of Roots (Descartes’s Rule)
Product of Roots
Sum of Roots
Boundaries: upper and lower
(1) Remainder Theorem
A polynomial function f with real
coefficients divided by ( x – c ) yields
the remainder = f(c).
The division algorithm states:
f(x) = (x – c) * q(x) + r
(E1)
Substitute x = c into (E1) to yield:
f(c) = (c – c) * q(c) + r = r.
(2) Factor Theorem
Let f(x) be a polynomial function with
real coefficients.
(a) If f(c) = 0, then (x – c) factors f(x).
(b) If (x – c) factors f(x), then f(c) = 0.
Recall from the Remainder Theorem
that f(c) = r and is zero, thus (x – c) is a
factor.
(3) Intermediate Value Theorem (IVT)
Given polynomial function f with real
coefficients and distinct real numbers a and b,
if the sign of f(a) is the opposite of the sign of
f(b), then at least one number c exists
between a and b for which f(c) = 0, a root of
the function f.
Example (IVT)
Let f(x) = x2 + 7x – 60.
f(0) = –60
f(8) = 64 + 56 – 60 = 60
The Remainder Theorem locates two
points on opposite sides of the x-axis.
Divide and conquer:
f(4) = 16 + 28 – 60 = –16
f(6) = 36 + 42 – 60 = 18
f(5) = 25 + 35 – 60 = 0
Example (IVT) - (cont)
f(0) =
= –60
f(5) = 25 + 35 – 60 = 0 root at x = 5
f(8) = 64 + 56 – 60 = 60
This example illustrates how to combine
these three theorems to find roots of a
function.
(4) Synthetic Division
This technique abbreviates long division of
polynomials for divisors of the form (x – c). The
“dividend” is the polynomial coefficients,
including zeros, in row one. The “divisor” is the
root c.
Let f(x) = x2 – 7x + 12 as an example, let c = 5
and start from left coefficient with 0 in row 2
and the sum of row one and row 2 in row 3.
5 | 1 -7
0
1
12
Working left to right by coefficient record the
product of row 3 value and the divisor on
row 2 under the next coefficient and record
the sum of row 1 and row 2 into row 3, until a
sum appears in row 3 under the rightmost
coefficient.
5 | 1 -7 12
0 5 -10
1 -2
2
Row three is the result of the division. The
rightmost number in row 3 is the remainder
and the other numbers are the coefficients of
the quotient which is q(x) = (x – 2). The
result follows:
(x2 – 7x + 12)/(x – 5) = (x – 2) + 2/(x – 5)
The result follows:
(x2 – 7x + 12)/(x – 5) = (x – 2) + 2/(x – 5)
Notice that
f(5) = 5*5 – 7 *5 + 12 = 25 – 35 + 12 = 2
as expected from the Remainder Theorem.
Performing the division
(x2 – 7x + 12)/(x – 4) = (x – 3)
using synthetic division appears in this form:
4 | 1 -7 12
0 4 -12
1 -3
0
The zero remainder indicates the divisor is a root
of f.
(5) Rational Zero Theorem (RZT)
If f(x) = anxn + an-1xn-1 + … + a1x + a0 with integer
coefficients and a rational zero of p/q, where
(p, q) = 1, then p divides a0 and q divides an.
Substitute x = p/q into
anxn + an-1xn-1 + … + a1x + a0 = 0
and subtract a0 on both sides to yield
an(p/q)n + an-1(p/q)n-1 + … + a1(p/q) = –a0
an(p/q)n + an-1(p/q)n-1 + … + a1(p/q) = –a0
Since p is a factor of each term on the left side,
then it is a factor of a0.
Similarly solving for an and dividing by xn shows
that q is a factor of an.
For example, the function f(x) = 6x2 + 7x – 5 has
an = 6 and a0 = – 5.
In this example f(x) = 6x2 + 7x – 5:
p factors a0 = – 5 into ± 1, ± 5.
q factors an = 6 into ± 1, ± 2, ± 3, ± 6.
Possible rational zeros are:
p
 1,  5

q  1,  2,  3,  6
A leading coefficient of 1 eliminates the options
from the denominator.
In this example g(x) = x3 – 12x2 + 41x – 42:
p factors a0 = – 42
into ± 1, ± 2, ± 3, ± 6, ± 7, ± 14 , ± 21, ± 42.
q factors an = 1 into ± 1 and has no impact.
The p factors are the possible rational zeros.
Checking x = 1 or x = –1 is fairly simple:
g( 1) = 1 – 12 + 41 – 42 = –12
g(-1) = – 1 – 12 – 41 – 42 = – 94
Neither one or minus one is a root by the Factor
Theorem.
Using synthetic division and the list of
potential rational zeros the search for
possible factors may begin.
Let x = 2 and perform synthetic division:
2 | 1 -12 41 -42
0
2 -20 42
1 -10 21
0
With a remainder of zero start a list of roots
with 2. Row three is the quotient:
g(x) = x2 – 10x + 21
After translating row three into the
quotient, notice that it factors and reveals
two more roots:
g(x) = x2 – 10x + 21
g(x) = ( x – 3 ) ( x – 7 )
The additional roots are 3 and 7.
The search uncovered the three roots expected
in a function of degree three, namely
{ 2, 3, 7 }
The remaining topics present some number
sense to help narrow the list of possible roots.
(6) Descartes’s Rules of Signs
Let f(x) = anxn + an-1xn-1 + … + a1x + a0 be a
polynomial with real coefficients. Count
the number of sign changes in the function
by adding one when contiguous terms have
opposite signs.
Counting positive real zeros
The number of positive real zeros is either:
(a) the same as the number of sign
changes in f(x)
or
(b) less than the number of sign
changes by an even number.
Counting positive real zeros
Examples
f(x) = + x – 3
\/
One sign change
One + root { 3 }
g(x) = + x2 – 10x + 21 Two sign changes
\/ \/
Two + roots { 3, 7 }
Counting positive real zeros
More Examples
Two sign changes
Two pos. real roots
or no real roots.
g(x) = + x2 – 10x + 25 = -4
(x - 5)2 = -2
Sq Root Prop
x = 5 +/- 2i
Roots { 5 - 2i, 5 + 2i }
g(x) = +x2–10x+29
\/\ /
Counting positive real zeros
More Examples (2)
g(x) = + x3 – 12x2 + 41x – 42
Three
sign chg
\ / \ /\ /
+ roots { 2, 3, 7 }
h(x) = + x3 + 8x2 + 1x – 42 One sign change
\ /
One positive root
Now is the time to investigate negative
roots.
Counting positive real zeros
More Examples (2)
g(x) = + x3 – 12x2 + 41x – 42
Three sign chg
\ / \ /\ /
+ roots { 2, 3, 7 }
h(x) = + x3 + 8x2 + 1x – 42
\ /
One sign change
One positive root
Now is the time to investigate negative
roots.
Counting negative real zeros
The number of negative real zeros is either:
(a) the same as the number of sign
changes in f(–x)
or
(b) less than the number of sign
changes by an even number.
Counting negative real zeros
Examples
Recall from last positive real zero example:
h(x) = + x3 + 8x2 + 1x – 42
\ /
One sign change
One positive root
h(–x) = – x3 + 8x2 – 1x – 42 Two sign changes
\ /\ /
Two neg. roots
Or perhaps no negative roots. Hmm.
Counting negative real zeros
Examples (cont)
h(x) = + x3 + 8x2 + 1x – 42
Notice that this function will have the
following possible rational roots:
± 1, ± 2, ± 3, ± 6, ± 7, ± 14 , ± 21, ± 42
Time to start the synthetic division tests!
Or to stop and think about this.
(7) Product of Roots
Let f(x) = (x – r1)(x – r2) … ( x – rn) be the
factorization of the function of degree n.
Let f(x) = anxn + an-1xn-1 + … + a1x + a0 be a
polynomial with real coefficients.
Notice that (-1)n· a0 is equivalent to the
product of the roots from expanding the
factorization. Both are the constant terms
of two representations of the function.
(7) Product of Roots (cont.)
Recall that the function
f(x) = + x3 + 8x2 + 1x – 42
has one positive root and possibly two
negative roots.
The product of the roots is (-1)3· -42 = 42.
The prime factors of 42 are 2, 3, and 7.
Or perhaps 1, 6, and 7.
f(x) = + x3 + 8x2 + 1x – 42
Since 7 is a factor in both cases, check for 7
as a root using synthetic division:
Let x = -7 (2 out 3 chances) and perform
synthetic division:
-7 | 1
8 1 -42
0 -7 -7 42
1 1 -6
0
And the quotient factors to yield 2 and -3.
The solution set is { 2, -3, -7 }.
(8) Sum of Roots
Let f(x) = (x – r1)(x – r2) … ( x – rn) be the
factorization of the function of degree n.
Let f(x) = anxn + an-1xn-1 + … + a1x + a0 be a
polynomial with real coefficients.
The expansion of the factorization produces
n terms of degree n-1 with a root as
coefficient and their sum is the opposite of
an-1.
(8) Sum of Roots (cont)
Again, recall that the function
f(x) = + x3 + 8x2 + 1x – 42
has one positive root and possibly two
negative roots.
The product of the roots is (-1)3· -42 = 42.
The prime factors of 42 are 2, 3, and 7.
The sum of the roots is – 8 = 2 – 3 – 7.
No synthetic division required.
Another example from a previous exercise:
g( x) = + x3 + 12x2 + 41x + 42
0 sgn chg
0 pos roots
g(– x) = + x3 – 12x2 + 41x – 42
\ / \ /\ /
3 sgn chg
3 neg roots
– 12 = – 7 – 3 – 2 and (-7)(-3)(-2) = -42(-1)3.
No synthetic division required.
(9) Root Upper Boundary
Let f(x) be a polynomial with real
coefficients and a positive leading
coefficient a.
If p(a) > 0 and synthetic division yields row
three with all positive values, then a is an
upper bound for the roots of the
polynomial.
(9) Boundaries (cont.)
The leading coefficient of the quotient
function q(x) is the same as p(x). Since a > 0
and all numbers on row three are positive,
then q(a) > 0.
The proof begins by selecting a root b and
stating that
p(x) = (x – b)· q(x).
Let x equal a, then
p(a) = (a – b)· q(a).
Dividing by q(a) yields
(a – b) = p(a)/q(a) > 0.
Since a > b and p(a) and q(a) are both
positive conclude that
a > b.
An example follows.
Example (from earlier)
p(x) = + x3 + 8x2 + 1x – 42
has these roots: { –7, –3, 2 }.
Given a = 3 > 0, p(3) = 60 > 0. and all row
three numbers are positive,
3| 1
8 1 -42
0 3 33 102
1 11 34 60
the assumptions are satisfied.
Thus 3 is an upper bound of the roots.
Example (from earlier)
g(x) = + x3 – 12x2 + 41x – 42
has these roots: { 2, 3, 7 }.
Given a = 8 > 0, p(8) = 30 > 0. but not all
row three numbers are positive,
8| 1 -12 41 -42
0 8 32 72
1 -4 9 30
so not all assumptions are satisfied.
The process does not guarantee least UB.
Increase the tested boundary to 13:
Given a = 13 > 0, p(13) = 660 > 0. and all
row three numbers are positive,
13| 1 -12 41 -42
0 13 13 702
1 1 54 660
so all assumptions are satisfied.
The process does identify 13 as an upper
bound of the roots. The sum coefficient
greater than largest root distorts result.
(9) Root Lower Boundary
Let f(x) be a polynomial with real
coefficients and a negative leading
coefficient a.
If p(a) not equal to 0 and synthetic division
yields row three with values of alternating
signs, then a is a lower bound for the roots
of the polynomial.
The proof is similar to the upper bound.
Example of Lower Bound (earlier)
g(x) = + x3 + 12x2 + 41x + 42
has these roots: { –2, – 3, – 7 }.
Given a = -13 < 0, p(-13) = -660 ≠ 0, and all
row three numbers have alternate signs,
-13| 1 12 41
42
13 -13 13 -702
1 -1 54 -660
so all assumptions are satisfied.
The process does not guarantee least UB.
One Last Example
f(x) = x3 – x – 60
This is what I call the Cardano polynomial.
It led him to speculate about the imaginary
unit.
To find the roots set f(x) = 0 and begin:
x3 – x – 60 = 0
Add 60 on both sides and factor left side:
x3 – x = 60
x(x2 – 1) = 60
(x – 1)x(x + 1) = 60
Three consecutive numbers with a product
of 60 are 3, 4, and 5. Thus x = 4 is a root.
f(x) = x3 – x – 60
f(-x) = – x3 + x – 60
The rules of signs indicate one positive root
and two negative roots (or none).
f(x) = x3 + 0x2 – x – 60
The sum of the roots is zero and the
product is (-60)(-1)3 = 60.
The positive solution is 4.
Regardless of whether signed or not, the
other two solutions are a conjugate pair,
say (r + s) and (r – s ).
The sum of the roots is:
4 + (r + s) + (r – s ) = 0
4 + 2r = 0
r = -2
The product of the roots is:
4(r + s)(r – s ) = 60
Divide by 4 and *
r2 – s2 = 15
Let r = -2
4 – s2 = 15
Solve for s
s2 = -11
s = [√(11)]i
The roots are { 4, (-2- [√(11)]i), (-2+ [√(11)]i)}
Thank You
Thank you for attending.
If you have questions about this
presentation, please contact me.
Taylor Johnson
Elizabethtown Community
& Technical College
[email protected]
(270) 706 - 8564