theorems about roots of polynomial equations
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Theorems About Roots of Polynomial Equations. Rational Roots. POLYNOMIALS and THEOREMS Theorems of Polynomial Equations. There are 5 BIG Theorems to know about Polynomials Rational Root Theorem Fundamental Theorem of Algebra Irrational Root Theorem Imaginary Root Theorem Descartes Rule. - PowerPoint PPT PresentationTRANSCRIPT
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Theorems About Roots of Polynomial Equations
Rational Roots
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POLYNOMIALS and THEOREMSTheorems of Polynomial Equations
• There are 5 BIG Theorems to know about Polynomials
1) Rational Root Theorem2) Fundamental Theorem of Algebra3) Irrational Root Theorem4) Imaginary Root Theorem5) Descartes Rule
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• All the mentioned theorems give some clues about the roots of a function that is the form of a polynomial.
• In advance Algebra we study only the first two. You will see the other theorems in Pre calculus.
• The first theorem… which is the, Fundamental Theorem of Algebra state:
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For f(x) where n > 0, there is at least one zero in the complex number system
Complex → real and imaginary
As a reminder:
Real zeros → can see on graph
Imaginary → cannot see on graph
f(x) of degree n will have exactly n zeros (real and imaginary)
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Next
• THE Rational Root Theorem is explained by the next few slides.
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Consider the following . . .
x3 – 5x2 – 2x + 24 = 0
This equation factors to:(x+2)(x-3)(x-4)= 0
The roots therefore are: -2, 3, 4
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Take a closer look at the original equation and our roots:
x3 – 5x2 – 2x + 24 = 0
The roots therefore are: -2, 3, 4
What do you notice?
-2, 3, and 4 are all factors of the constant term, 24
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Spooky! Let’s look at another
• 24x3 – 22x2 – 5x + 6 = 0
• This equation factors to:• (x+1)(x-2)(x-3)= 0 2 3 4
• The roots therefore are: -1/2, 2/3, 3/4
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Take a closer look at the original equation and our roots:
• 24x3 – 22x2 – 5x + 6 = 0
• This equation factors to: (x+1)(x-2)(x-3)= 0 2 3 4
The roots therefore are: -1, 2, 3 2 3 4
The numerators 1, 2, and 3 are all factors of the constant term, 6.
The denominators (2, 3, and 4) are all factors of the coefficient of the leading term, 24.
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This leads us to the Rational Root Theorem
For a polynomial,If is a root of the polynomial, then p is a factor of the constant term, andq is a factor of the coefficient of the leading term,
011
1 ...)( axaxaxaxp nn
nn
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Example (RRT)1. For polynomial 03x3xx 23
Possible roots are ___________________________________
Here p = -3 and q = 1
Factors of -3Factors of 1 ±3, ±1
±1
2. For polynomial 012x4x9x3 23
Possible roots are ______________________________________________
Here p = 12 and q = 3
Factors of 12Factors of 3 ±12, ±6 , ±3 , ± 2 , ±1 ±4
±1 , ±3
Or ±12, ±4, ±6, ±2, ±3, ±1, ± 2/3, ±1/3, ±4/3
Or 3,-3, 1, -1
Wait a second . . . Where did all of these come from???
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Let’s look at our solutions
414
111
212
313
616
12112
±12, ±6 , ±3 , ± 2 , ±1, ±4 ±1 , ±3
34
34
31
31
32
32
133
236
4312
Note that + 2 is listed twice; we only consider it as one answer
Note that + 1 is listed twice; we only consider it as one answer
That is where our 9 possible answers come from!
Note that + 4 is listed twice; we only consider it as one answer
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Let’s Try One
Find the POSSIBLE roots of 5x3-24x2+41x-20=0
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Let’s Try One5x3-24x2+41x-20=0
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That’s a lot of answers!
• Obviously 5x3-24x2+41x-20=0 does not have all of those roots as answers.
• Remember: these are only POSSIBLE roots. We take these roots and figure out what answers actually WORK.
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• Step 1: find p and q
• p = -3• q = 1
• Step 2 : by RRT, the only rational root is of the form…
• Factors of pFactors of q
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• Step 3 : factors
• Factors of -3 = ±3, ±1Factors of 1 = ± 1
• Step 4 : possible roots
• -3, 3, 1, and -1
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• Step 5: Test each root
• Step 6: synthetic division
X X³ + X² – 3x – 3
-3
3
1
-1
(-3)³ + (-3)² – 3(-3) – 3 = -12
(3)³ + (3)² – 3(3) – 3 = 24
(1)³ + (1)² – 3(1) – 3 = -4
(-1)³ + (-1)² – 3(-1) – 3 = 0
THIS IS YOUR ROOT BECAUSE WE ARE LOOKINGFOR WHAT ROOTS WILL MAKE THE EQUATION =0
-1 1 1 -3 -3
0
1 -3
3
0
-1
0
1x² + 0x -3
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• Step 7: Rewrite
• x³ + x² - 3x - 3 = (x + 1)(x² – 3)
• Step 8: factor more and solve
• (x + 1)(x² – 3)• (x + 1)(x – √3)(x + √3)
• Roots are -1, ± √3
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Notes
• You may also use the synthetic division/ substitution to evaluate the polynomial.
• You may also use the graph of the polynomial to find the first root.
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Let’s Try One
Find the roots of 2x3 – x2 + 2x - 1
Take this in parts. First find the possible roots. Then determine which root actually works.
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Let’s Try One
2x3 – x2 + 2x - 1
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Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
• Step 1: find p and q
• p = -6• q = 1
• Step 2: by RRT, the only rational root is of the form…
• Factors of pFactors of q
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Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
• Step 3 – factors
• Factors of -6 = ±1, ±2, ±3, ±6 Factors of 1 = ±1
• Step 4 – possible roots
• -6, 6, -3, 3, -2, 2, 1, and -1
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Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
• Step 5 – Test each root • Step 6 – synthetic division
X x³ – 5x² + 8x – 6
-6
6
3
-3
2
-2
1
-1
THIS IS YOUR ROOT
3 1 -5 8 -6
-6
1 2
6
-2
3
0
1x² + -2x + 2
-450
78
0
-102
-2
-50
-2
-20
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Using the Polynomial Theorems FACTOR and SOLVE x³ – 5x² + 8x – 6 = 0
• Step 7 – Rewrite
• x³ – 5x² + 8x – 6 = (x - 3)(x² – 2x + 2)
• Step 8– factor more and solve
• (x - 3)(x² – 2x + 2)
• Roots are 3, 1 ± i
Quadratic Formula
i1x
X= 3
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Irrational Root Theorem
• For a polynomial
• If a + √b is a root,• Then a - √b is also a root• Irrationals always come in pairs. Real
values might not.
CONJUGATE ___________________________Complex pairs of form a + √ b and a - √ b
011
1 ...)( axaxaxaxp nn
nn
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Example (IRT)1. For polynomial has roots 3 + √2
Other roots ______3 - √2 Degree of Polynomial ______2
2. For polynomial has roots -1, 0, - √3, 1 + √5
Other roots __________√3 , 1 - √5 Degree of Polynomial ______6
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Example (IRT)1. For polynomial has roots 1 + √3 and -√11
Other roots ______ _______1 - √3
Degree of Polynomial ______4
√11
Question: One of the roots of a polynomial is
Can you be certain that is also a root?
24 24
No. The Irrational Root Theorem does not apply unless you know that all the coefficients of a polynomial are rational. You would have to haveas your root to make use of the IRT.
24
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Write a polynomial given the roots5 and √2
• Another root is - √2 • Put in factored form
• y = (x – 5)(x + √2 )(x – √2 )
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Decide what to FOIL firsty = (x – 5)(x + √2 )(x – √2 )
X -√2
x √2
X2 -X √2
X √2 -2
(x² – 2)
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FOIL or BOX to finish it up (x-5)(x² – 2)
y = x³ – 2x – 5x² + 10
Standard Formy = x³ – 5x² – 2x + 10
x2 -2
x -5
X3 -2x
-5x2 10
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Write a polynomial given the roots-√5, √7
• Other roots are √5 and -√7• Put in factored form• y = (x – √5 )(x + √5)(x – √7)(x + √7)• Decide what to FOIL first
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y = (x – √5 )(x + √5)(x – √7)(x + √7)
Foil or use a box method to multiply the binomials
X -√7
x √7
X2-X √7
X √7 -7
(x² – 7)
X -√5
x √5
X2-X √5
X √5 -5
(x² – 5)
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y = (x² – 5)(x² – 7)
FOIL or BOX to finish it upy = x4 – 7x² – 5x² + 35Clean upy = x4 – 12x² + 35
x2 -5
x2
-7
X4 -5x2
-7x2 35
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Complex Root Theorem
• For a polynomial
• If is a root,• Then is also a root• Complex roots come in pairs. Real values
might not.
CONJUGATE ___________________________Complex pairs of form and
011
1 ...)( axaxaxaxp nn
nn
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Descartes Rule