Download 6.6 The Fundamental Theorem of Algebra

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 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 Signs
Determine 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 Theorem
Find 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,
±12/1
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)(x2 +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) = 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
2 -4 -30
1 -2 -15 0
(x-2)(x2 – 2x – 15)=0
(x-2)(x+3)(x-5)=0
x=2
x = -3
x=5
Rational Root Theorem
f(x)=10x4 – 3x3 – 29x2 + 5x +12
List: 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 Continued
g(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 4
x=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± 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 x3 + x2 – 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 x3 – 4x2 – 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 x3 – 4x2 – 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  21x 3
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.
Can we identify any of the roots as
y  x  10 x  21x
5
4
3
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.
y=x3(x - 3)(x – 7)
x = 7 appears to go
straight through the
axis. Therefore it is
probably a single root.
(x)3
(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
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. (x - 1)2
Therefore it is an
even degree root.
y=(x + 2)3(x – 1)2
The don’t appear to
be any other roots.
Factor the polynomial completely.
y  4 x  20 x  9 x  45
3
2
x = -1.5 appears to
go straight through (x + 1.5)
the axis. Therefore it
is a single root.
x = 1.5 appears to
go straight through (x – 1.5)
the axis. Therefore it
is a single root.
x = 5 appears to go
y=A(x + 1.5)(x – 1.5)(x - 5)
(x – 5)
straight through the
y=- 4(x + 1.5)(x – 1.5)(x - 5) axis. Therefore it is a
single root.
Factor the polynomial completely.
y  x  5x  2x  20x  24
4
3
2
According to the graph, the
roots are at x = -2, -3, 2
y  (x  2) (x  3)(x  2)
2
Factor the polynomial completely.
y  2x  6x  4
3
According to the graph,
the roots are at x = -2, 1
y  2(x  2)(x 1)
2
Factor the polynomial completely.
y  x  2x  2
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  2x  2x
3
2
According to the graph, What
are the roots?
Are all of the roots rational?
Write the equation in factored
form.
y  x( x  2 x  2)
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  16
4
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  16
4
Use synthetic division, to factor:
21 0 0 0 16
y  (x  2)(x  2)x  4
2
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  4
2
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.
4i
0  0 2  4(1)(4)   16
 2i

x
2
2
2
There are 4 roots: x = 2, -2, 2i, -2i
Find all of the roots of the equation.
(Real & Imaginary)
y  x  27
3
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.
3 1 0 0  27
y  x  27
3

x  27  ( x  3) x  3x  9
3
 3  3 2  4(1)(9)
x
2
The roots are:
2
 3   27

2
 3  3 3i
x
,3
2

 3  3 3i

2
Finding Zeros
• Find all the zeros of y = x3 – 3x2 – 9x
Finding Zeros
• Find all the zeros of y = 2x3 + 14x2 + 13x + 6
Solve:
x  3x  2
3
Solve:
x  6 x  36 x   x  12 x  16
4
3
2
Solve:
2 x  5 x  10 x  20  x  5 x  3x  4
3
2
3
2
Solve:
x  2 x   x  8 x  12
4
3
2