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 SignsDetermine the possible number of positive and negative

real roots of x4 – 4x3 + 7x2 – 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) = x4 + 4x3 + 7x2 + 6x – 18

There is only one sign change. So, there is only one negative real root.

Descarte’s Rule of SignsDetermine the possible number of positive and negative

real roots of x5 – 6x4 + 2x3 – 17x2 + 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) = -x5 – 6x4 – 2x3 – 17x2 + 18

There is only one sign change. So, there is only one negative real root.

Rational Root Theorem

• If f(x)=axxn + +a1x+a0 has integer coefficients, then every rational zero of f has the following form:

• p = factor of constant term a0

q factor of leading coefficient an

Rational Root TheoremFind rational zeros of f(x)=x3 + 2x2 – 11x – 12

1. List possible LC=1 CT=-12 X= ±1/1,± 2/1, ± 3/1, ± 4/1, ± 6/1, ±12/12. Test: 1 2 -11 -12 1 2 -11 -12X=1 1 3 -8 x=-1 -1 -1 12 1 3 -8 -20 1 1 -12 03. Since -1 is a zero: (x+1)(x2 +x – 12) = f(x) Factor: (x+1)(x-3)(x+4) = 0 x=-1 x=3 x=-4

Rational Root TheoremFind rational zeros of: f(x) = x3 – 4x2 – 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)(x2 – 2x – 15)=0

2 -4 -30 (x-2)(x+3)(x-5)=0

1 -2 -15 0

x = 2 x = -3 x = 5

Rational Root Theoremf(x)=10x4 – 3x3 – 29x2 + 5x +12List: LC=10 CT=12 x = ± 1/1, ± 2/1, ± 3/1, ± 4/1, ± 6/1, ±12/1,

± 3/2, ± 1/5, ± 2/5, ± 3/5, ± 6/5, ± 12/5, ± 1/10, ± 3/10, ± 12/10

with so many –sketch graph on calculator and find reasonable solutions:

x= -3/2, -3/5, 4/5, 3/2 Check: 10 -3 -29 5 12 x= -3/2 -15 27 3 -12 10 -18 -2 8 0 Yes it

works * (x+3/2)(10x3 – 18x2 – 2x + 8)*

(x+3/2)(2)(5x3 – 9x2 – x + 4) -factor out GCF (2x+3)(5x3 – 9x2 – x + 4) -multiply 1st

factor by 2

Example Continuedg(x) = 5x3 – 9x2 – 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: 5 -9 -1 4x=4/5 4 -4 -4 5 -5 -5 0

(2x + 3)(x – 4/5)(5x2 – 5x – 5)= (2x + 3)(x – 4/5)(5)(x2 – x – 1)= multiply 2nd factor

by 5 (2x + 3)(5x – 4)(x2 – x – 1)=

-now use quad for last- *-3/2, 4/5, 1± , 1- . 2 2

55

Finding RootsUse 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 x3 + x2 – 3x – 3 = 0.

Finding RootsUse 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 x3 – 4x2 – 2x + 8 = 0.

Finding RootsUse 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 x3 – 4x2 – 2x + 8 = 0.

Irrational Root TheoremLet 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 TheoremIf 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?

345 2110 xxxy

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

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

345 2110 xxxy 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 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

(x - 3)1

(x - 7)1y=x3(x - 3)(x – 7)

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.

84104 2345 xxxxxy

x = -2 appears to squiggle through the axis. Therefore it is an odd degree root.

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

The don’t appear to be any other roots.

(x + 2)3

(x - 1)2

y=(x + 2)3(x – 1)2

Factor the polynomial completely.

459204 23 xxxy

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

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

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

(x + 1.5)

(x – 1.5)

y=A(x + 1.5)(x – 1.5)(x - 5) (x – 5)

y=- 4(x + 1.5)(x – 1.5)(x - 5)

Factor the polynomial completely.

y x 4 5x 3 2x 2 20x 24According to the graph, the roots are at x = -2, -3, 2

y (x 2)2(x 3)(x 2)

Factor the polynomial completely.

y 2x 3 6x 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 2x 2According 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.

xxxy 22 23 According to the graph, What are the roots?

Are all of the roots rational?

Write the equation in factored form.

)22( 2 xxxy

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.

164 xyAccording 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.

164 xyUse synthetic division, to factor:

1600012

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 02 4(1)(4)

2

162

4i

2

2i

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

Find all of the roots of the equation. (Real & Imaginary)

273 xy

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.

273 xy 270013

93)3(27 23 xxxx

2

)9)(1(433 2 x

2

273

2

333 i

The roots are: 3,2

333 ix

Finding Zeros• Find all the zeros of y = x3 – 3x2 – 9x

Finding Zeros• Find all the zeros of y = 2x3 + 14x2 + 13x + 6

Solve:

233 xx

Solve:

1612366 234 xxxxx

Solve:

435201052 2323 xxxxxx

Solve:

1282 234 xxxx