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Theorems about Roots of Polynomial Equations and
The fundamental theorem of Algebra
To solve equations using the Rational Root Theorem.
To use the conjugate Root Theorem.
To use the Fundamental Theorem of Algebra to
solve polynomial equations with complex solutions
Rational Root Theorem.
Conjugate Root Theorem.
Descartes’ Rule of Signs.
Fundamental theorem of Algebra.
Factoring a polynomial can be challenging, but there is a
theorem to help you with that.
Rational Root Theorem
2
21
x
 29 x  10  0
Example:
Factors of the leading
coefficient:
 1,3,7, and  21
Factors of the constant
term:
 1,2,5, and  10
Problem 1: Finding a Rational Root
What are the rational roots of 2 x 3  x 2  2 x  5  0
Leading coefficient 2 (±1,±2) and constant term 5 (±1, ±5)
The only possible rational
factor of constant term
roots have the form
factor of leading coefficient
The only possible roots are  1,5, 1 , 5
2
x
P(x)
1 -1
5
-5
1/2
8 0 240  280 6
-1/2
7
2
5/2
35
2
-5/2

75
2
the only rational root of 2 x 3  x 2  2 x  5  0 is  1
Note: the Rational Root Theorem does not necessarily give the
zeros of the equation. It provides a list of first guesses to
test as roots
Your turn
What are the rational roots of x 3-5 x 2 -2 x  10  0?
Leading coefficient 1 (±1) and constant term 10 (±1, ±2,±5±10)
factor of constant t erm
The only possible rational
roots have the form
factor of leading coefficient
x
P(x)
1
4
-1
2
-2
5
16  6
18
0
-5
1 2 5 10
 , , ,
1 1 1
1
10
-10
 230 940  1470
The only rational root of x 3  5 x 2  2 x  10  0 is 5
Once you find one root, use synthetic division to factor the
polynomial. Continue finding roots and dividing until you have a
second degree polynomial. Use the Quadratic Formula to find
the remaining roots.
Problem 2: Using the Rational Root Theorem
What are the rational roots of 15 x 3  32 x 2  3 x  2  0 ?
Leading coefficient 15 (±1,±3,±5,±15) and constant term 2(±1, ±2)
The only possible rational roots have the form
 1,  2 , 
factor of constant term
factor of leading coefficient
1
2
1
2
1
2
, , , ,
,
3
3
5
5
15
15
Test each possible rational root in 15 x 3  32 x 2  3 x  2  0
until you find a root
Test: x=1
x2
1513  3212  31  2  12  0
152 3  322 2  32   2  0
So 2 is a root, now factor the polynomial using synthetic division
2
15
15
-32
3
2
30
-4
-2
-2
-1
0
15 x 2  2 x  1


P x    x  2  15 x 2  2 x  1
5 x  13 x  1  0
5x  1  0
1
x5
3 x-1  0
x
1
3
1 1
The rational roots of 15 x  32 x  3 x  2  0 are 2 , ,
5 3
3
2
Remember steps to find rational roots.
1. Get the constant term factors and the leading coefficient
of the polynomial.
factor of constant term
2. Find all possible rational roots factor of leading coefficient
3.Test each possible rational root until you find a root.
4. Factor the polynomial until you get a quadratic.
(you can use synthetic division, quadratic formula,
a calculator, or any other method).when using
TI-84 or 85 store the polynomial in using the
1
Y= menu. Store the root to be tested inY
x(enter
the
number then press the key STO).
Use VARS Y-VARS 1: Function Y1 to evaluate.
Your turn:
What are the rational roots of 2 x 3  x 2  7 x  6  0
Answer: 2,-1,-3/2
Did you remember what is a conjugate number?
a  bi and a  bi are conjugates
a  b and a  b are conjugates
If a complex number or an irrational number is a
root of a polynomial equation with rational coefficients,
so is its conjugate.
Conjugate Root Theorem: If P(x) is a polynomial with
rational coefficients, then the irrational roots of P(x)=0
that have a form a  b occur in conjugate pairs. That is,
If a  b is an irrational root with a and b rational then
a  b is also a root.
If P(x) is a polynomial with real coefficients, then the
complex roots of P(x)=0 occur in conjugate pairs. That
is a  bi is a complex root with a and b real, then a  bi
is also a root.
Problem 3:Using the Conjugate Root Theorem to identify Roots.
A quartic polynomial P x  has rational coefficients. If 2 and 1  i are roots
of P(x)  0 what are the other two roots?
Answers:
Remember all rational numbers are real numbers.
The other two roots are - 2 and 1  i
Your turn
A cubic polynomial P(x) has real coefficients.
5
If 3-2i and are two roots of P(x)  0, What is the additional root?
2
Answer
3  2i
Problem 4: Using the Conjugates to construct A Polynomial
A. What is a third deg ree polynomial function y  P(x) with rational
coefficients so that P(x)  0 has the roots -4 and 2i?
Answer: Since 2i is a root, then -2i is also a root
P(x)   x  2i  x  2i  x  4  multiply


P(x)  x 2  4  x  4 
P(x)  x 3  4 x 2  4 x  16  0
B. What quartic polynomial equation has roots 2-3i, 8, 2?
Answer:
Descartes’ Rule of Signs Theorem: Let P(x) be a polynomial
with real coefficients written in standard form.
*The number of positive real roots of P(x) is either equal
to the number of sign changes between consecutive
coefficients of P(x) or is less than that by an even number.
*The number of negative real roots of P(x)=0 is either
equal to the number of sign changes between consecutive
coefficients of P(-x) or is less than that by an even number.
In both cases, count multiple roots according to their
multiplicity
This theorem implies that two tests must be done on a
polynomial function to determine both positive and negative
real roots.
Problem 5: Using the Descartes’ Rule of Signs.
What the Descartes' Rule of Signs tell you about the real roots of
x3  x2  1  0
There are two sign changes, + to – and - to +.
Therefore, there are either 0 or 2 positive real roots.
To find the negative real root you need to plug in -x
into the polynomial P  x    x 3   x 2  1  0
P  x    x 3  x 2  1  0
There is only one negative real root
Graph the function and recall that cubic functions have
zero or two turning points. Because the graph already
shows two turning points, it will not change the direction
again. So there are no positive real roots.
Your turn
A.What does Descartes' Rule of Signs tell you about the real roots of
2 x4  x3  3 x2  1  0
Answers: there are 3 or 1 positive real roots and 1 negative
root. The graph confirm one negative and one positive real
Root
B.Can you confirm real and complex roots graphically? Explain
Real roots can be confirmed graphically because they are
x-intercepts. Complex roots cannot be confirmed graphically
because they have an imaginary component.
Take a note: the degree of a polynomial equation tells
you how many roots the equation has. That is the
result of the Fundamental theorem of Algebra provided by
the German mathematician Carl Friedrich Gauss
(1777-1855)
The fundamental Theorem of Algebra: If P(x) is a
polynomial of degree n≥1, then P(x)=0 has exactly
n roots, including multiple and complex roots
Problem 6: Using the Fundamental Theorem of Algebra
What are all the roots of x 5  x 4  3 x 3  3 x 2  4 x  4  0
I need to find the zeros(5) and using the rational
root and factor theorem, synthetic division and factoring
Using the rational root the possible rational roots are  1,2,4
Evaluate the polynomial function for x  1
P( 1 )  0 , 1 is a root and ( x  1 ) is a factor
Then use synthetic division
1 1 -1 - 3
3
-4
1
0 -3
0
1
0 -3 0
-4
x4  3x2  4
4
-4
0

x 5  x 4  3 x 3  3 x 2  4 x  4   x  1 x 4  3 x 2  4
The factors are -4 and 1
x2  4  0
x2  4
x  2
Answers:




  x  1 x 2  4 x 2  1
x2  1  0
x 2  1
x2   1
x  i
 x  1 x  2  x  2  x  i  x  i  so the roots are 1,2 ,2 ,i ,i
Your turn
What are all the roots of the equation x 4  2 x 3  13 x 2  10 x?
Since there is no constant term, make the equation
equal to zero and factor x from the polynomial.
CF, GC, SD and Factoring.
Answers:
0,1,-5,2
Problem 7: Finding all the zeros of a Polynomial Function
What are the zeros of f(x)  x 4  x 3  7 x 2  9 x  18
Step 1: use the graphing calculator to find any real roots
Step 2: Factor out the linear factors
(x-3)(x+3).Use Synthetic division twice.
-3
1 1 -7
-3 6
- 9 - 18
3 18
1 - 2 -1 - 6

3
1 - 2 -1 - 6
3 3 6
1 1
0
x 4  x 3  7 x 2  9 x  18  x  3 x  3 x 3  2 x 2  x  6


  x  3x  3 x 2  x  2
2
0

Step3: use the quadratic formula to find the complex roots
x2  x  2  0
 1  12  412   1   7

21
2
Step 4:
the four zeros are  3,3,
1 i 7 1 i 7
,
2
2
Your turn
What are all the zeros of the function g ( x )  2 x 4  3 x 3  x  6 ?
Answer:
1 i 3
 1,2 ,
2
The Fundamental Theorem of Algebra: Here are
equivalent ways to state the fundamental theorem
of Algebra. You can use any of these statements to
prove the others.
*Every polynomial equation of degree n≥1 has exactly
n roots, including multiple and complex roots.
*Every polynomial of degree n≥1 has n factors.
*Every polynomial function of degree n≥1 has at least
one complex zero.
Classwork odd
Homework even
TB pg 316 exercises 9-41 and pg322 exercises 8-37