Download 3.3-The Theory of Equations Multiplicity

3.3-The Theory of Equations
Multiplicity
If the factor x – c occurs k times in the complete factorization of a polynomial, then c is called a root of
the polynomial with a multiplicity of k. The multiplicity refers to the number of times a root or solution
occurs.
Example: Given f ( x) = ( x − 2)( x + 3) 2 determine the roots and their multiplicity.
Solution:
The factor (x-2) gives the root x = 2 which occurs once. Therefore its multiplicity is k = 1.
The factor (x+3) gives the root x = -3 which occurs twice. Therefore its multiplicity is k = 2.
Example: Given f ( x) = ( x + 5) 3 ( x − 4) 2 ( x + 1) 4 determine the multiplicity.
Solution:
The factor (x+5) occurs three times. Its multiplicity is k = 3.
The factor (x-4) occurs twice. Its multiplicity is k = 2.
The factor (x+1) occurs four times. Its multiplicity is k = 4.
Nth Root Theorem
Any polynomial function of degree n has n roots. This allows us to predict how many roots to expect
when solving an equation.
Example: The function f ( x ) = 2 x 3 − 5 x + 1 is a 3rd degree polynomial, so it has 3 roots.
Example: The function f ( x) = 2 x 5 − 5 x 2 + 13 is a 5th degree polynomial, so it has 5 roots.
Example: Determine the number of zeros of the polynomial function f ( x) = x 5 + 4 x 4 + 4 x 3 and then find
the roots and the multiplicity of each root.
Solution: The function is a 5th degree polynomial so there are 5 roots. We can find the roots by factoring.
x 5 + 4x 4 + 4x 3 = 0
x 3 ( x 2 + 4 x + 4) = 0
x 3 ( x + 2) 2 = 0
x = 0, k = 3
x = −2, k = 2
x = 0 is a root with multiplicity k=3 and x = -2 is a root with multiplicity k=2. Taking multiplicity into
account there are 2+3 = 5 roots as expected.
Example: Given f ( x ) = x 5 − 6 x 4 + 9 x 3 determine the roots and their multiplicity.
Solution: Find the roots by factoring the equation 0 = x 5 − 6 x 4 + 9 x 3 .
x5 − 6x 4 + 9x3 = 0
x 3 ( x 2 − 6 x + 9) = 0
x 3 ( x − 3) 2 = 0
The roots are x = 0 with a multiplicity of 3 and x = 2 with a multiplicity of 2.
Conjugate Pairs Theorem:
If a complex number is a root of a polynomial equation, then its complex conjugate is also a root of the
equation. Consequently, complex roots will always occur in pairs.
Example: Find the roots of the polynomial equation x 2 + 4 = 0 .
Solution: Solve by using the square root property.
x2 + 4 = 0
x 2 = −4
x = ± −4
x = ±2i
Notice that 2i and -2i are complex conjugates.
The Linear Factorization Theorem:
Just as an nth degree polynomial has n roots, an nth degree polynomial also has n linear factors. An nth
degree polynomial can be expressed as the product of a non-zero constant and n linear factors. This is
called the linear factorization theorem. This can help us to find a polynomial equation given its roots.
Example: Find a polynomial equation given the roots x = 4, x = 7
Solution: Working backward from the roots, we first find the factors that those solutions are derived
from. Then we multiply the factors together to find the polynomial equation.
x = 4, x = 7
x − 4 = 0, x − 7 = 0
( x − 4)( x − 7) = 0
x 2 − 11x + 28 = 0
Example: Find a polynomial equation with the root 3i.
Solution: If 3i is a root, then -3i is also a root according to the complex conjugate theorem. We can find
the polynomial equation by multiplying the corresponding factors (x-3i) and (x+3i) together.
x = 3i
x = −3i
x − 3i = 0
x + 3i = 0
( x − 3i )( x + 3i ) = 0
x 2 − 9i 2 = 0
x 2 − 9(−1) = 0
x2 + 9 = 0
Example: Find a polynomial equation with the root 2-3i.
Solution: If 2-3i is a root, then 2+3i is also a root according to the complex conjugate theorem. We can
find the polynomial equation by multiplying the corresponding factors together.
x = 2 − 3i
x = 2 + 3i
x − ( 2 − 3i ) = 0
x − ( 2 + 3i ) = 0
( x − (2 − 3i ))( x − (2 + 3i )) = 0
x 2 − x(2 + 3i ) − x(2 − 3i ) + (2 + 3i )(2 − 3i ) = 0
x 2 − 2 x − 3 xi − 2 x + 3 xi + (4 − 9i 2 ) = 0
x 2 − 4 x + (4 + 9) = 0
x 2 − 4 x + 13 = 0
Example: Find a polynomial equation with the roots 3, and 1+2i.
Solution: If 1+2i is a root, then 1-2i is also a root according to the complex conjugate theorem. We can
find the polynomial equation by multiplying the corresponding factors together.
x=3
x−3= 0
x = 1 + 2i
x − (1 + 2i ) = 0
x = 1 − 2i
x − (1 − 2i ) = 0
( x − 3)( x − (1 + 2i))( x − (1 − 2i)) = 0
( x − 3)[ x 2 − x(1 + 2i) − x(1 − 2i) + (1 + 2i)(1 − 2i )] = 0
( x − 3)[ x 2 − x − 2 xi − x + 2 xi + (1 − 4i 2 )] = 0
( x − 3)[ x 2 − 2 x + (1 + 4)] = 0
( x − 3)[ x 2 − 4 x + 5] = 0
x 3 − 4 x 2 + 5 x − 3 x 2 + 12 x − 15 = 0
x 3 − 7 x 2 + 17 x − 15 = 0
Sign Changes & Descartes Rule
Descartes rule of signs is a method for determining the number of positive, negative, and imaginary roots.
It does not help use to actually find those roots. To use Descartes rule, we need to determine the number
of sign changes in a polynomial. A sign change occurs when the signs alternate from one term to the next.
Example: The polynomial function P( x) = x 3 − 7 x 2 + 17 x − 15 has three sign changes. The first term is
positive, the second is negative, the third is positive, and the fourth is negative.
Example: The polynomial function P( x) = 5x 4 − 2 x 3 − x 2 + 7 x + 1 has two sign changes. The first term
is positive, the second is negative, the third is also negative, the fourth is positive, and the fifth is also
positive.
Descartes Rule of Signs
According to Descartes rule of signs the number of positive real roots is equal to the number of sign
changes of P(x) or less than that by an even number. The number of negative real roots is equal to the
number of sign changes of P (-x) or less than that by an even number.
Example: Find the number of positive, negative and imaginary roots of the polynomial equation
x 3 − 7 x 2 + 17 x − 15 = 0
Solution: The polynomial P( x) = x 3 − 7 x 2 + 17 x − 15 has three sign changes; therefore there are either 3
or 1 positive real roots. To find the number of sign changes of P (-x) we must first find P (-x)
P(− x) = (− x) 3 − 7(− x) 2 + 17(− x) − 15
P(− x) = − x 3 − 7 x − 17 x − 15
The polynomial P (-x) has no sign changes; therefore there are no negative roots. To determine the
number of imaginary roots it is helpful to make a table to organize the results we have so far.
Positive Negative imaginary
3
0
0
1
0
2
After filling in the possible combinations of positive and negative roots, the imaginary roots can be found
by using the n-root theorem which states that the total number of roots must be equal to the degree of the
polynomial. Consequently, each row must add up to n.
Example: Find the number of positive, negative and imaginary roots of the polynomial equation
5x 4 − 2 x 3 − x 2 + 7 x + 1 = 0
Solution: The polynomial P( x) = 5x 4 − 2 x 3 − x 2 + 7 x + 1has two sign changes; therefore there are either
2 or 0 positive real roots. To find the number of sign changes of P (-x) we must first find P (-x)
P(− x) = 5(− x) 4 − 2(− x) 3 − (− x) 2 + 7(− x) + 1
P(− x) = 5 x 4 + 2 x 3 − x 2 − 7 x + 1
The polynomial P (-x) has two sign changes; therefore there are two or 0 negative roots. To determine the
number of imaginary roots it is helpful to make a table to organize the results we have so far.
Positive Negative imaginary
2
2
0
2
0
2
0
2
2
0
0
4
After filling in the possible combinations of positive and negative roots, the imaginary roots can be found
by using the n-root theorem which states that the total number of roots must be equal to the degree of the
polynomial. Consequently, each row must add up to n.
3.3-Applications
Example: The instantaneous growth rate of a population is the rate at which it is growing at every instant
in time. The instantaneous growth rate “r” of a colony of bacteria “t” hours after the start of an
experiment is given by the function
r = 0.01t 3 − 0.08t 2 + 0.11t + 0.20
for 0 ≤ t ≤ 7 . Find the times for which the instantaneous growth rate is zero.
Solution: To simplify the function, begin by factoring out 0.01. This results in integral coefficients.
r = 0.01t 3 − 0.08t 2 + 0.11t + 0.20
r = 0.01(t 3 − 8t 2 + 11t + 20)
Use the Rational Zeros Theorem to find possible rational zeros.
Possible Rational Zeros = ±
1,2,4,5,10,20
= ±1,2,4,5,10,20
1
5
1
1
-8
11
20
5
-15
-20
-3
-4
0
Therefore, t = 5 is a zero. The quotient is then t 2 − 3t − 4 . Solving this equation gives:
t 2 − 3t − 4 = 0
(t − 4)(t + 3) = 0
t=4
t = −3
Therefore the times when the instantaneous growth rate is zero in the interval 0 ≤ t ≤ 7 are t = 4,
and t = 5.
Example: The manager of a retail store has figured that her monthly profit “P” (in thousands of dollars) is
determined by her monthly advertising expense “x” (in tens of thousands of dollars) according to the
formula
P = x 3 − 20 x 2 + 100 x
for
0 ≤ x ≤ 4 . For what value of x does she get $147,000 in profit?
Solution: Let P = 147. Remember that P is in thousands of dollars.
147 = x 3 − 20 x 2 + 100 x
Then, we will need to solve the equation.
x 3 − 20 x 2 + 100 x − 147 = 0
Use the Rational Zeros Theorem to find possible rational zeros.
Possible Rational Zeros = ±
1,3,7,21,49,147
= ±1,3,7,21,49,147
1
3
1
1
-20
100
-147
3
-51
147
-17
49
0
Therefore, x = 3 is a zero. The quotient is then t 2 − 17t + 49 . Solving this equation using the quadratic
formula gives us x=13.31, and x=3.68.
The only solutions in the interval 0 ≤ x ≤ 4 are x=3 and x=3.68
Example: The growth rate of a population is the rate at which it is growing at every instant in time. The
growth rate of a colony of bacteria t hours after the start of an experiment is given by the function
‫ = )ݐ(ݎ‬0.01‫ ݐ‬ଷ − 0.08‫ ݐ‬ଶ + 0.11‫ ݐ‬+ 0.20
1. What is the maximum number of possible times at which the growth rate is zero? Explain your
answer and justify using concepts from this weeks work.
2. Is it possible to not have any times at which the growth rate is zero over the interval (0, ∞ ) ?
Explain your answer and justify using concepts from this week’s work.
3. Find the times for which the growth rate is zero in the interval (0, ∞) ?
Solution:
1. This requires use of the N-Root Theorem. Since this is a 3rd degree polynomial, there will be
exactly 3 roots. Therefore, the maximum number of times the graph can cross or touch the x-axis
(zero growth) is 3.
2. Use Descartes Rule of Signs. The function r(t) has 2 sign changes and the function r(-t) has 1 sign
change. This results in the following table.
Positive Negative imaginary
2
1
0
0
1
2
From the table we can see that there are two possibilities. The second possibility indicates that
there are no positive zeros. Consequently, it is possible to not have any zeros in the required
interval.
3. Using the rational zeros test gives the following possible rational zeros.
Possible rational zeros: ± 1,2,4,5,10,20
Use synthetic division to obtain:
4
0.01
0.01
-0.08
0.04
-0.04
0.11
-0.16
-0.05
0.2
-0.2
0
Therefore t= 4 is a zero and the quotient is Q(t ) = 0.01t 2 − 0.04t + 0.05 . Solving this equation by
factoring gives additional zeros of x = -1, and x = 5. Therefore, the growth rate is zero at 4 hours
and at 5 hours.
Example: An observatory is built in the shape of a right circular cylinder with a hemispherical roof. The
heating and air contractor has figured the volume of the structure as 3168π cubic feet. If the height of the
cylindrical walls is 2 feet more than the radius of the building, then what is the radius of the building?
Solution: This problem requires two formulas:
2
π ⋅r3
3
Volume of a Cylinder: π ⋅ r 2 h
Volume of a Hemisphere:
The volume of the observatory is the sum of these formulas:
2
Volume = π ⋅ r 3 + π ⋅ r 2 h
3
Substituting in V = 3168π and h = r + 2 , we have:
2
3158π = π ⋅ r 3 + π ⋅ r 2 ( r + 2)
3
Simplify this equation:
2
3158π = π ⋅ r 3 + π ⋅ r 2 (r + 2)
3
2
3158π = π ⋅ r 3 + π ⋅ r 3 + 2π ⋅ r 2
3
5
3158π = π ⋅ r 3 + 2π ⋅ r 2
3
5
3158 = r 3 + 2r 2
3
9504 = 5r 3 + 6r 2
5r 3 + 6r 2 − 9504 = 0
This leaves us a polynomial equation to solve. Use the Rational Zeros Theorem to find possible rational
zeros.
1,2,4,8,12,16, Κ
1,5
Because there are a lot of factors of 9504, start dividing the ones you have using synthetic division. If
these do not work, find more factors.
Possible Rational Zeros = ±
12
5
5
6
0
-9504
60 792
9504
66 792
0
Therefore, x = 12 is a factor. The quotient is then 5r 2 + 66r + 792 . Solving this equation using the
quadratic formula gives
− 66 ± i 11,484
x=
10
Because these solutions are complex, the only real solution is 12. Therefore, the radius must be 12 feet.