6.5 & 6.6 Theorems About Roots and the Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra

• If P(x) is a polynomial of degree n, there are exactly n complex roots.

• This means that some of the roots might be imaginary and some might be real.

Descarte’s Rule of Signs

• Descarte’s Rule of Signs is a method for finding the number and sign of real roots of a polynomial equation in standard form.

• The number of

positive

real roots of a polynomial equation, with real coefficients, is equal to the number of sign changes (from positive to negative or vice versa) between the coefficient of the terms of P(x), or is less than this number by a multiple of two.

• The number of

negative

real roots of a polynomial equation, with real coefficients, is equal to the number of sign changes between the coefficient of the terms of P(-x), or is less than this number by a multiple of two.

Descarte’s Rule of Signs

Determine the possible number of positive and negative real roots of x 4 – 4x 3 + 7x 2 – 6x – 18 = 0.

For positive real roots, count the sign changes in P(x) = 0.

There are 3 sign changes. So, there are either 1 or 3 real positive roots.

For negative real roots, count the sign changes in P(-x) = 0.

P(-x) = x 4 + 4x 3 + 7x 2 + 6x – 18 There is only one sign change. So, there is only one negative real root.

Descarte’s Rule of Signs

Determine the possible number of positive and negative real roots of x 5 – 6x 4 + 2x 3 – 17x 2 + 18 = 0.

For positive real roots, count the sign changes in P(x) = 0.

There are 4 sign changes. So, there are either 4 or 2 or 0 real positive roots.

For negative real roots, count the sign changes in P(-x) = 0.

P(-x) = -x 5 – 6x 4 – 2x 3 – 17x 2 + 18 There is only one sign change. So, there is only one negative real root.

Rational Root Theorem

• If

f

(x)=a x x n + +a 1 x+a 0 has integer coefficients, then every rational zero of

f

has the following form: • p = factor of constant term a 0 q factor of leading coefficient a n

Rational Root Theorem

Find rational zeros of

f

(x)=x 3 + 2x 2 – 11x – 12 1. List possible X= ± LC=1 CT=-12 1/1, ± 12/1 ± 2/1, ± 3/1, ± 4/1, ± 6/1, 2. Test: 1 2 -11 -12 1 2 -11 -12 X=1 1 3 -8 x=-1 -1 -1 12 1 3 -8 -20 1 1 -12 0 3. Since -1 is a zero: (x+1)(x 2 +x – 12) =

f(

x) Factor: (x+1)(x-3)(x+4) = 0 x=-1 x=3 x=-4

Rational Root Theorem

Find rational zeros of:

f

(x) = x 3 – 4x 2 – 11x + 30 LC=1 CT=30 x= ± 1/1, ± 2/1, ± 3/1, ± 5/1, ± 6/1, ± 10/1, ± 15/1, ± 30/1 Test: 1 -4 -11 30 1 -4 -11 30 x=1 1 -3 -14 x=-1 -1 5 6 1 -3 -14 16 1 -5 -6 36 X=2 1 -4 -11 30 (x-2)(x 2 – 2x – 15)=0 2 -4 -30 1 -2 -15 0 (x-2)(x+3)(x-5)=0 x = 2 x = -3 x = 5

f(x

)=10x 4

Rational Root Theorem

– 3x 3 – 29x 2 + 5x +12 List: LC=10 CT=12 x = ± 1/1, ± ± 12/1, ± 12/5, ± 2/1, ± 3/2, ± 1/10, ± 3/1, 1/5, ± ± 3/10, ± 4/1, ± 2/5, ± 12/10 6/1, 3/5, ± 6/5, ± with so many –sketch graph on calculator and find reasonable solutions: Check: x= -3/2 x= -3/2, -3/5, 4/5, 3/2 10 -3 -29 5 12 -15 27 3 -12 10 -18 -2 8 0 Yes it works * (x+3/2)(10x (x+3/2)(2)(5x 3 3 – 18x 2 – 2x + 8)* – 9x 2 – x + 4) -factor out GCF (2x+3)(5x factor by 2 3 – 9x 2 – x + 4) -multiply 1st

Example Continued

g

(x) = 5x 3 – 9x 2 – x + 4 LC=5 CT=4 x: ± 1, ± 2, ± 4, ± 1/5, ± 2/5, ± 4/5 *The graph of original shows 4/5 may be: x=4/5 5 -9 -1 4 4 -4 -4 5 -5 -5 0 (2x + 3)(x – 4/5)(5x 2 – 5x – 5)= (2x + 3)(x – 4/5)(5)(x 2 – x – 1)= multiply 2nd factor by 5 (2x + 3)(5x – 4)(x 2 – x – 1)= -now use quad for last *-3/2, 4/5, 1 ± 5 , 1 5 2 2 .

Finding Roots

Use Descarte’s Rule of Signs to identify the possible number of positive and negative real roots. Then use the Rational Roots theorem, to find the roots of x 3 + x 2 – 3x – 3 = 0.

Finding Roots

Use Descarte’s Rule of Signs to identify the possible number of positive and negative real roots. Then use the Rational Roots theorem, to find the roots of x 3 – 4x 2 – 2x + 8 = 0.

Finding Roots

Use Descarte’s Rule of Signs to identify the possible number of positive and negative real roots. Then use the Rational Roots theorem, to find the roots of x 3 – 4x 2 – 2x + 8 = 0.

Irrational Root Theorem

Let a and b be rational numbers and let √b be an irrational number. If a + √b s a root of a polynomial equation with rational coefficients, then the conjugate a √b also is a root.

In other words, irrational roots come in pairs.

If 3 √2 and √5 are roots of a polynomial equation what other roots are there?

3 + √2 and -√5

Imaginary Root Theorem

If the imaginary number a + bi is a root of a polynomial equation with real coefficients, then the conjugate a – bi also is a root. In other words, imaginary roots come in pairs.

If 3 – 2i and 5i are roots of a polynomial equation what other roots are there?

3 + 2i and -5i

Writing a Polynomial Equations from its Roots

Find a third – degree polynomial equation (in standard form) with rational coefficients that has roots 3 and 1 + i.

Writing a Polynomial Equations from its Roots

Find a fourth – degree polynomial equation (in standard form) with rational coefficients that has roots i and 2i.

Now that we know specifically how many roots an equation is supposed to have, we can start to factor large polynomials completely.

How many roots does this equation have?

y



x

5  10

x

4  21

x

3 Here is a graph of the equation. (The scale has been adjusted to fit the whole graph on the screen.)

y

Since we know that this equation should have 5 roots, let’s use the graph to help us factor it.



x

5  10

x

4  21

x

3 Can we identify any of the roots as being single, double, or triple roots?

x = 0 appears to squiggle through the axis. Therefore it is an odd degree root .

(x) 3 y=x 3 (x - 3)(x – 7)

x = 3 appears to go straight through the axis. Therefore it is probably a single root .

x = 7 appears to go straight through the axis. Therefore it is probably a single root .

(x - 3) 1 (x - 7) 1

In order to factor the last problem, we had to use a theorem called the Factor Theorem .

Factor Theorem:

If f(a) = 0, then (x - a) is a factor of f(x)

Translation:

If you see a root of the graph at x = 3, then x – 3 divides evenly into the function that you just graphed.

Factor the polynomial completely.

y



x

5  4

x

4 

x

3  10

x

2  4

x

 8

y=(x + 2) 3 (x – 1) 2

x = -2 appears to squiggle through the

(x + 2) 3

axis. Therefore it is an odd degree root .

x = 1 appears to bounce off the axis. Therefore it is an even degree root .

(x - 1) 2

The don’t appear to be any other roots.

Factor the polynomial completely.

y

  4

x

3  20

x

2  9

x

 45

y=A(x + 1.5)(x – 1.5)(x - 5) y=- 4(x + 1.5)(x – 1.5)(x - 5)

x = -1.5 appears to go straight through the axis. Therefore it is a single root.

(x + 1.5)

x = 1.5 appears to go straight through the axis. Therefore it is a single root.

(x – 1.5)

x = 5 appears to go straight through the axis. Therefore it is a single root.

(x – 5)



y



Factor the polynomial completely.

x

4  5

x

3  2

x

2  20

x

 24 According to the graph, the roots are at x = -2, -3, 2

y

 (

x

 2) 2 (

x

 3)(

x

 2) 



Factor the polynomial completely.

y

 2

x

3  6

x

 4 According to the graph, the roots are at x = -2, 1

y

 2(

x

 2)(

x

 1) 2 

Factor the polynomial completely.

y



x

2  2

x

 2 According to the graph, all of the roots are imaginary.

 How many roots are there?

Write the equation in factored form.

Because the roots are imaginary, it is not factorable.

Factor the polynomial completely.

y



x

3  2

x

2  2

x

According to the graph, What are the roots?

Are all of the roots rational?

Write the equation in factored form.

y



x

(

x

2  2

x

 2 ) Root at x = 0 These roots are irrational. You would need to use the quadratic formula to find the exact value of these roots.

Factor the polynomial completely.

y



x

4  16 According to the graph, the roots are at x = 2 , -2 How many roots are there?

Can you factor this equation?

You will need to use synthetic division to factor, because some of the roots appear to be imaginary.



Factor the polynomial completely.

y



x

4  16 Use synthetic division, to factor: 2 1 0 0 0  16

y

 (

x

 2)(

x

 2) 

x

2  4 

 

If factored form of the equation looks like this, how many real and how many imaginary roots does it have? What are they?

y

 (

x

 2)(

x

 2) 

x

2  4  We already know that the graph shows roots at x = 2 and x = -2. This should be apparent from factored form.

To find the imaginary roots, you must use the quadratic formula on the quadratic part.

x

  0  0 2 2  4(1)(4)    16 2   4 2

i

  2

i

There are 4 roots: x = 2, -2, 2i, -2i   

y

Find all of the roots of the equation.



x

3

(Real & Imaginary)

 27 There is a root at x = 3. What is its multiplicity?

Where are the other roots?

The other 2 must be imaginary.

Use Synthetic division to factor completely. Then find all of the roots.

y



x

3  27 3 1 0 0  27

x

3  27  (

x

 3 ) 

x

2  3

x

 9 

x

  3  3 2  4 ( 1 )( 9 ) 2 The roots are:  

x

  3  3 2 3  3

i

2 , 3  27   3  3 2 3

i

Finding Zeros

• Find all the zeros of y = x 3 – 3x 2 – 9x

Finding Zeros

• Find all the zeros of y = 2x 3 + 14x 2 + 13x + 6

x

3  3

x

 2

Solve:

Solve:

x

4  6

x

3  36

x

 

x

2  12

x

 16

Solve:

2

x

3  5

x

2  10

x

 20 

x

3  5

x

2  3

x

 4

Solve:

x

4  2

x

3  

x

2  8

x

 12