6.5 Theorems About Roots of Polynomial Equations

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6.5 Theorems About Roots of Polynomial Equations

• Rational Roots

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POLYNOMIALS and THEOREMSTheorems of Polynomial
Equations

• There are 4 BIG Theorems to know about
  Polynomials
• Rational Root Theorem
• Irrational Root Theorem
• Imaginary Root Theorem
• Descartes Rule

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Consider the following . . .

• x3 5x2 2x 24 0
• This equation factors to
• (x2)(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 all go into the last term, 24!

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Spooky! Lets look at another

• 24x3 22x2 5x 6 0
• This equation factors to
• (x1)(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
• (x1)(x-2)(x-3) 0
• 2 3 4
• The roots therefore are -1, 2, 3
• 2 3 4
• What do you notice?
• The numerators 1, 2, and 3 all go into the last
  term, 6!
• The denominators (2, 3, and 4) all go into the
  first term, 24!

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This leads us to the Rational Root Theorem

• For a polynomial,
• If p/q is a root of the polynomial,
• then p is a factor of an
• and q is a factor of ao

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Example (RRT)
1. For polynomial
Here p -3 and q 1
Factors of -3 Factors of 1
3, 1 1
?
Or 3,-3, 1, -1
Possible roots are _______________________________
____
2. For polynomial
Here p 12 and q 3
Factors of 12 Factors of 3
12, 6 , 3 , 2 , 1 4 1 , 3
?
Possible roots are _______________________________
_______________
Or 12, 4, 6, 2, 3, 1, 2/3, 1/3, 4/3
Wait a second . . . Where did all of these come
from???
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Lets look at our solutions
12, 6 , 3 , 2 , 1, 4 1 , 3
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
Note that 4 is listed twice we only consider
it as one answer
That is where our 9 possible answers come from!
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Lets Try One

• Find the POSSIBLE roots of 5x3-24x241x-200

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Lets Try One

• 5x3-24x241x-200

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Thats a lot of answers!

• Obviously 5x3-24x241x-200 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 p
• Factors of q

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• Step 3 factors
• Factors of -3 3, 1
• Factors 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
-1
1 1 -3 -3
-3 3 1 -1
(-3)³ (-3)² 3(-3) 3 -12
(3)³ (3)² 3(3) 3 24
3
0
-1
(1)³ (1)² 3(1) 3 -4
1
-3
0
0
(-1)³ (-1)² 3(-1) 3 0
THIS IS YOUR ROOT BECAUSE WE ARE LOOKING FOR
WHAT ROOTS WILL MAKE THE EQUATION 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 v3)(x v3)
• Roots are -1, v3

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Lets 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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Lets 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 p
• Factors of q

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Using the Polynomial Theorems FACTOR and SOLVE
x³ 5x² 8x 6 0

• Step 4 possible roots
• -6, 6, -3, 3, -2, 2, 1, and -1

• Step 3 factors
• Factors of -6 1, 2, 3, 6
• Factors of 1 1

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Using the Polynomial Theorems FACTOR and SOLVE
x³ 5x² 8x 6 0

• Step 6 synthetic division

• Step 5 Test each root

X x³ 5x² 8x 6
-450 78 0 -102 -2 -50 -2 -20
-6 6 3 -3 2 -2 1 -1
3
1 -5 8 -6
THIS IS YOUR ROOT
6
-6
3
1
2
-2
0
1x² -2x 2
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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
X 3
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Irrational Root Theorem

• For a polynomial
• If a vb is a root,
• Then a - vb is also a root
• Irrationals always come in pairs. Real values do
  not.

CONJUGATE ___________________________
Complex pairs of form a v b and a - v b
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Example (IRT)
1. For polynomial has roots 3 v2
2
3 - v2
Other roots ______
Degree of Polynomial ______
2. For polynomial has roots -1, 0, - v3, 1 v5
6
v3 , 1 - v5
Other roots __________
Degree of Polynomial ______
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Example (IRT)
1. For polynomial has roots 1 v3 and -v11
1 - v3
v11
Other roots ______ _______
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Degree of Polynomial ______
Question One of the roots of a polynomial is Can
you be certain that is also a root?

No. The Irrational Root Theorem does not apply
unless you know that all the coefficients of a
polynomial are rational. You would have to
have as your root to make use of the IRT.
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Write a polynomial given the roots5 and v2

• Another root is - v2
• Put in factored form
• y (x 5)(x v2 )(x v2 )

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Decide what to FOIL first y (x 5)(x v2 )(x
v2 )
X -v2


x v2
X2
-X v2
(x² 2)
X v2
-2
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FOIL or BOX to finish it up (x-5)(x² 2) y
x³ 2x 5x² 10 Standard Form y x³ 5x²
2x 10
x2 -2


x -5
X3
-2x
-5x2
10
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Write a polynomial given the roots-v5, v7

• Other roots are v5 and -v7
• Put in factored form
• y (x v5 )(x v5)(x v7)(x v7)
• Decide what to FOIL first

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y (x v5 )(x v5)(x v7)(x v7) Foil or
use a box method to multiply the binomials
X -v7
X -v5




x v7
x v5
X2
-X v7
X2
-X v5
X v7
-7
X v5
-5
(x² 7)
(x² 5)
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y (x² 5)(x² 7) FOIL or BOX to finish it
up y x4 7x² 5x² 35 Clean up y x4 12x²
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x2 -5


x2 -7
X4
-5x2
-7x2
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