Theorems About Roots of Polynomial Equations Rational Roots

Theorems About Roots of Polynomial Equations

Rational Roots

POLYNOMIALS and THEOREMSTheorems of Polynomial Equations

• There are 5 BIG Theorems to know about Polynomials

1) Rational Root Theorem

2) Fundamental Theorem of Algebra

3) Irrational Root Theorem

4) Imaginary Root Theorem

5) Descartes Rule

• 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:

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)

Next

• THE Rational Root Theorem is explained by the next few slides.

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

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

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

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.

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,

and

q is a factor of the coefficient of the leading term,

011

1 ...)( axaxaxaxp nn

nn

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???

Let’s look at our solutions

41

4

11

1

21

2

31

3

61

6

121

12

±12, ±6 , ±3 , ± 2 , ±1, ±4 ±1 , ±3

3

4

3

43

1

3

13

2

3

2

13

3

23

6

43

12

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

Let’s Try One

Find the POSSIBLE roots of 5x3-24x2+41x-20=0

Let’s Try One5x3-24x2+41x-20=0

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.

• 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

• Step 3 : factors

• Factors of -3 = ±3, ±1

Factors of 1 = ± 1

• Step 4 : possible roots

• -3, 3, 1, and -1

• 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

• 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

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.

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.

Let’s Try One

2x3 – x2 + 2x - 1

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

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

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

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

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

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

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 have

as your root to make use of the IRT.

24

Write a polynomial given the roots5 and √2

• Another root is - √2

• Put in factored form

• y = (x – 5)(x + √2 )(x – √2 )

Decide what to FOIL firsty = (x – 5)(x + √2 )(x – √2 )

X -√2

x √2

X2 -X √2

X √2 -2

(x² – 2)

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

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

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)

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

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

Descartes Rule