Download PPT Review 4.1-4.4-0

Pre-Calculus
For our Polynomial Function:
2
y  x  2x 15
The Factors are:
(x + 5) & (x - 3)
The Roots/Solutions are:
x = -5 and 3
The Zeros are at:
(-5, 0) and (3, 0)
 Rearrange the terms to have zero on
one side:
2
2
x  2x  15  x  2x  15  0
 Factor:
(x  5)(x  3)  0
 Set each factor equal to zero and
solve: (x  5)  0 and (x  3)  0
x  5
x 3
The only way that x2 +2x - 15 can = 0 is if x = -5 or x = 3
The points where y = 0 are
called the x-intercepts of the
graph.
The x-intercepts for our graph
are the points...
(-5, 0)
and
2
y  x  2x 15
(3, 0)
If a polynomial has ‘n’ complex roots will its
graph have ‘n’ x-intercepts?
3
y  x  4x
In this example, the
degree n = 3, and if we
factor the polynomial, the
roots are x = -2, 0, 2. We
can also see from the
graph that there
are 3 x-intercepts.
We can find the Roots or Zeros of a polynomial by
setting the polynomial equal to 0 and factoring.
Some are easier to
factor than others!
3
f (x)  x  4x
2
 x(x  4)
 x(x  2)(x  2)
The roots are: 0, -2, 2
Just because a polynomial has ‘n’ complex
roots doesn’t mean that they are all Real!
3
In this example,
however, the degree is
still n = 3, but there is
only one Real x-intercept
or root at x = -1, the
other 2 roots must have
imaginary components.
2
y  x  2x  x  4
EXAMPLE 1
Find the zeros of
x2 0
f  x    x  2  x  3   x  1 x  4 
2
 x  3  0
2
x 1 0
2, -3 (d.r), 1, -4
x40
Solve
by factoring.
Original equation
Add
4x to each side.
Factor the binomial.
or
Zero Product Property
Solve the second equation.
Answer: The solution set is
{0, –4}.
Solve
by factoring.
Original equation
Subtract 5x and
each side.
2 from
Factor the trinomial.
or
Zero Product Property
Solve each equation.
Answer: The solution set is
Check each solution.
Solve each equation by factoring.
a.
Answer:
b.
Answer:
{0, 3}
Solve
by factoring.
Original equation
Add
9 to each side.
Factor.
Zero Product Property
or
Solve each equation.
Answer: The solution set is
{3}.
Check
The graph of the related function,
intersects the x-axis only once. Since the zero of the function is
of the related equation is 3.
3, the solution
Solve
Answer: {–5}
by factoring.
Factor Theorem
1. If f(c)=0, that is c is a zero of f,
then x - c is a factor of f(x).
2. Conversely if x - c is a factor of f(x),
then f(c)=0.
Example – Find the roots/zeros for this
polynomial 10 x3  9 x 2  19 x  6
Given -2 is a zero, use synthetic division to find the remaining
roots.
Don’t forget your
-2
remainder should be zero
10 9 -19 6
-20 22
10 -11
3
-6
The new, smaller
polynomial is:
10 x  11x  3
0
2
This quadratic can be factored into:
Therefore, the zeros to the problem
(5x – 3)(2x – 1)
3 1
x  2, ,
5 2
Example:
Find all the zeros of each polynomial
function
10 x3  9 x 2  19 x  6
If we were to graph this
equation to check you could
see the zero
ZERO
ZERO
ZERO
From looking at
the graph you can
see that there is a
zero at -2
EXAMPLE 3
Find All the Rational Zeros
We have already known that the possible rational
zeros are:
P(1) = 0
P(2) = 0
P(-1) = -24
P(-2) = 0
P(3) = 120
P(-3) = 0
P(4) = 756
P(6) = 7, 200
P(-4) = -300
P(-6) = -4, 704
P(12) = 254,100
P(-12) = -21, 000
So 1, 2, -2, and 3 are rational
roots
Example:
The zeros of a third-degree polynomial
are 2, 2, and -5. Write a polynomial.
(x – 2)(x – 2)(x+5)
First, write the zeros in factored
form
= (x - 4x+ 4)(x+ 5)
Second, multiply the factors out to
find your polynomial
2
ANSWER
x  x  16 x  20
3
2
EXAMPLE 5
Given 1 ± 3i and 2 are roots, find the remaining
roots.
If x = the root then
x - the root is the factor
form.
x  2x  1  3i x  1  3i 
x  2x 1  3i x 1  3i 
x  2 x
2
Multiply the last two
factors together. All i
terms should disappear
when simplified.
 x  3xi  x  1  3i  3xi  3i  9i

 x  2 x  2 x  10
2


-1
Now multiply the x – 2
through
 x  4 x  14 x  20
3
2
2
Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i









f(x)= (x-1)(x-(-2+i))(x-(-2-i))
f(x)= (x-1)(x+2 - i)(x+2+ i)
f(x)= (x-1)[(x+2) - i] [(x+2)+i]
f(x)= (x-1)[(x+2)2 - i2]
Foil
f(x)=(x-1)(x2 + 4x + 4 – (-1))Take care of i2
f(x)= (x-1)(x2 + 4x + 4 + 1)
f(x)= (x-1)(x2 + 4x + 5)
Multiply
f(x)= x3 + 4x2 + 5x – x2 – 4x – 5
f(x)= x3 + 3x2 + x - 5









Note: 2+i means 2 – i is also a zero
F(x)= (x-4)(x-4)(x-(2+i))(x-(2-i))
F(x)= (x-4)(x-4)(x-2-i)(x-2+i)
F(x)= (x2 – 8x +16)[(x-2) – i][(x-2)+i]
F(x)= (x2 – 8x +16)[(x-2)2 – i2]
F(x)= (x2 – 8x +16)(x2 – 4x + 4 – (– 1))
F(x)= (x2 – 8x +16)(x2 – 4x + 5)
F(x)= x4– 4x3+5x2 – 8x3+32x2 – 40x+16x2 –
64x+80
F(x)= x4-12x3+53x2-104x+80

Complex Zeros Occur in Conjugate Pairs = If a + bi
is a zero of the function, the conjugate a – bi is
also a zero of the function
(the polynomial
function must have real coefficients)
EXAMPLES: Find a polynomial with the given zeros
-1, -1, 3i, -3i

2, 4 + i, 4 – i


If we cannot factor the polynomial, but know one of the
roots, we can divide that factor into the polynomial. The
resulting polynomial has a lower degree and might be
easier to factor or solve with the quadratic formula.
2
x
2
3
2
3
2
f (x)  x  5x  2x  10
x  5 x  5x  2x  10

one root is x  5
(x - 5) is a factor
We can solve the resulting polynomial to get the
other 2 roots:
x  2, 2
x 3  5x 2
 2x  10
2x  10
0
EXAMPLE:
Equation
Solve:
Solution
Solving a Polynomial
x4  6x2  8x + 24  0.
Now we can solve the original equation as follows.
x4  6x2  8x + 24  0 This is the given equation.
(x – 2)(x – 2)(x2  4x  6)  0
x–20
x2
or
x–20
x2
or
This was obtained from the
second synthetic division.
x2  4x  6  0
Set each factor equal to
zero.
x2  4x  6  0
Solve.
Ise Quadratic Formula
The solution set of the original equation is {2, 2 i i,
2, 2
i 2 i
}.
f (x)  x 4  3x 3  6x 2  2x  60
(Given that 1 + 3i is a zero of f)
f (x)  x 3  7x 2  x  87
(Given that 5 + 2i is a zero of f)
f (x)  x  x  2x 12x  8
5
3
2
f (x)  3x  4 x  8x  8
3
2
FIND ALL RATIONAL ROOTS:
List all possible rational zeros of f (x)  15x3  14x2  3x – 2.
Solution
The constant term is –2 and the leading coefficient is 15.
Factors of the constant term,  2
Factors of the leading coefficient, 15
1,  2

1,  3,  5,  15
Possible rational zeros 
 1,  2,
Divide 1
and 2
by 1.
 13 ,  32 ,
Divide 1
and 2
by 3.
 15 ,  52 ,
Divide 1
and 2
by 5.
1 ,  2
 15
15
Divide 1
and 2
by 15.
There are 16 possible rational zeros. The actual solution set to f (x) 
15x3  14x2  3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16
possible solutions.
FIND ALL POSSIBLE RATIONAL ROOTS
Solve:
x4  6x2  8x + 24  0.
Solution Recall that we refer to the zeros of a polynomial function
and the roots of a polynomial equation. Because we are given an
equation, we will use the word "roots," rather than "zeros," in the
solution process. We begin by listing all possible rational roots.
Factors of the constant term, 24
Factors of the leading coefficient, 1
1,  2  3,  4,  6,  8,  12,  24

1
 1,  2  3,  4,  6,  8,  12,  24
Possible rational zeros 
Arrange the terms of the polynomial P(x) in
descending degree:
• The number of times the coefficients of the terms
of P(x) change sign = the number of Positive Real
Roots (or less by any even number)
• The number of times the coefficients of the terms
of P(-x) change sign = the number of Negative
Real Roots (or less by any even number)
In the examples that follow, use Descartes’ Rule of Signs to predict the number
of + and - Real Roots!
Determine the possible number of positive and negative real zeros of
f (x)  x3  2x2  5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign
changes in the equation for f (x). Because all the terms are positive,
there are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of
sign changes in the equation for f (x). We obtain this equation by
replacing x with x in the given function.
f (x)  x3  2x2  5x + 4 This is the given polynomial function.
Replace x with x.
f (x)  (x)3  2(x)2  5x  4
 x3  2x2  5x + 4
3.4: Zeros of Polynomial Functions
Determine the possible number of positive and negative real zeros of
f (x)  x3  2x2  5x + 4.
Solution
Now count the sign changes.
f (x)  x3  2x2  5x + 4
1
2
3
There are 0 positive roots, and 3 negative roots
For higher degree polynomials, finding the complex
roots (real and imaginary) is easier if we know one
of the roots.
Descartes’ Rule of Signs can help get you started.
Complete the table below:
Polynomial
y = x 4 + 2x 2 + 3
y = x 3 - 7x 2 + 17x - 15
y = 3x 4 + x 3 - 3x 2 - x + 1
# + Real # - Real
Roots
Roots
For the polynomial:
All possible values of:
f (x) = x -3x + 5x -15
3
2
p: ± 1, ± 3, ± 5
q: ± 1
All possible Rational Roots of the form p/q:
p
: ± 1, ± 3, ± 5
q
For the polynomial:
Descartes’ Rule:
f (x) = x -3x + 5x -15
# + Real Roots = 3 or 1
# - Real Roots = 0
# Imag. Roots = 2 or 0
3
2
All possible Rational Roots of the form p/q:
p
: 1, 3, 5
q
For the polynomial:
f (x) = x -3x + 5x -15
3
2
Substitute each of our possible rational roots into f(x).
If a value, a, is a root, then f(a) = 0. (Roots are
solutions to an equation set equal to zero!)
f (1) = 1- 3 + 5 - 15 = -12
f (3) = 27 - 27 + 15 - 15 = 0
f (5) = 125 - 75 + 25 - 15 = 60
*
Descartes’s Rule of Signs
EXAMPLES: describe the possible real zeros
f (x)  3x 3  5x 2  6x  4
f (x)  3x 3  2x 2  x  3
f  x   2x3  x 2  13x  6
Find the zeros of
Hint: 2 is a zero
2
2
2
X
1
-13
6
4
10
-6
5
-3
0
 x  2  2x 2  5x  3   0
 x  2 2x  1 x  3   0
1
x  2, ,  3
2
Find the zeros of
f  x   x3  11x  20
Hint: 4 is a zero
4
1
1
X
0
-11
-20
4
16
20
4
5
0
 x  4   x 2  4x  5   0
4  16  4 1 5 
x  4,  2  i,  2  i
2
4  4
2
2  i,  2  i
f  x   x3  2x 2  3x  6
Find the zeros of
Hint: 2 is a zero
2
1
1
X
-2
-3
6
2
0
-6
0
-3
0
 x  2  x2  3   0
x2  3
x  3,  3
x  2, 3,  3
f  x   x3  5x 2  2x  24
Find the zeros of
Hint: 2 is a zero
2
1
1
X
5
-2
-24
2
14
24
7
12
0
 x  2   x 2  7x  12   0
 x  2  x  4  x  3   0
x  2,  4,  3
No Calculator
f  x   x3  2x 2  5x  6,
Given –2 is a zero of
find ALL the zeros of the function.
-2
1
1
-2
-5
6
-2
8
-6
-4
3
0
 x  2  x 2  4x  3   0
 x  2 x  1 x  3   0
x  2, 1, 3
No Calculator
f  x   x 4  5x3  4x 2  20x,
Given 5 is a zero of
find ALL the zeros of the function.

f  x   x x 3  5x 2  4x  20
No constant, so 0 is a zero:
5
1
1
-5
-4
20
5
0
-20
0
-4

0

x  x  5 x2  4  0
x  x  5  x  2  x  2   0
x  0, 5, 2,  2

No Calculator
f  x   x 4  9x3  23x 2  3x  36,
Given -1 and 3 are zeros of
find ALL the zeros of the function.
-1
3
1
1
1
-9
23
-3
-36
-1
10
-33
36
-10
33
-36
0
3
-21
36
-7
12
0
 x  1 x  3   x 2  7x  12   0
 x  1 x  3  x  4  x  3   0
x  1, 3  d.r. , 4
No Calculator
Given
3
2
f  x   2x3  9x 2  13x  6,
is a zero of
find ALL the zeros of the function.
3 /2
-9
13
-6
3
-9
6
2
-6
4
0
1
-3
2
2


2
2x

3
x

  3x  2  0
 2x  3  x  1 x  2   0
3
x  , 1, 2
2
No Calculator
Given
2
is a zero of
3
f  x   3x3  8x 2  5x  6,
find ALL the zeros of the function.
2 /3
-8
-5
6
2
-4
-6
3
-6
-9
0
1
-2
-3
3


2
3x

2
x

  2x  3  0
 3x  2 x  3  x  1  0
2
x  , 3,  1
3
No Calculator
f  x   x3  6x 2  13x  10,
Given 2 is a zero of
find ALL the zeros of the function.
2
1
1
-6
13
-10
2
-8
10
-4
5
0
 x  2   x 2  4x  5   0
4  16  4 1 5 
2
4  4
2
2  i, 2  i
x  2, 2  i, 2  i
No Calculator
f  x   x3  3x 2  x  3,
Given –3 is a zero of
find ALL the zeros of the function.
-3
1
1
3
1
3
-3
0
-3
0
1
0
 x  3   x 2  1  0
x2  1
x  i,  i
x  3, i,  i
No Calculator
Find a polynomial function with real coefficients which has
zeros of 1, -2, and 3.
f  x    x  1 x  2  x  3 

f  x    x  1 x 2  x  6


 
 6x    x

 x  6
f  x   x x2  x  6  1 x2  x  6

f  x   x3  x2
f  x   x3  2x 2  5x  6
2
No Calculator
Find a polynomial function with real coefficients which has
zeros of 0, 2, -2, and 5.
f  x   x  x  2  x  2  x  5 


f  x   x  x  x  3x  10   2  x  3x  10 
f  x   x  x  3x  10x    2x  6x  20 
f  x   x  x  2  x 2  3 x  10
2
3
2
2
2
f  x   x  x 3  5x 2  4x  20 
f  x   x 4  5x3  4x 2  20x
No Calculator
Find a polynomial function with real coefficients which has
zeros of 3/2, 2, and 1.
f  x    2x  3  x  2  x  1

f  x    2x  3  x 2  3 x  2


 
 4x    3x
f  x   2x x 2  3 x  2  3 x 2  3x  2

f  x   2x 3  6x 2
f  x   2x3  9x 2  13x  6
2

 9x  6

No Calculator
Find a polynomial function with real coefficients which has
zeros of 2 and i.
If i is a root, then –i is a root as well
f  x    x  2  x  i  x  i 


f  x    x  2 x2  1

 
f  x    x  x    2x

 2
f  x   x x2  1  2 x2  1
3
2
f  x   x3  2x 2  x  2
No Calculator
Find a polynomial function with real coefficients which has
zeros of 1 and 2 + i.
If 2 + i is a root, then 2 – i is a root as well
f  x    x  1   x  2  i    x  2   i 

f  x    x  1  x  2   i2
2



f  x    x  1  x  4 x  5 
f  x   x  x  4x  5   1 x  4 x  5 
f  x    x  4x  5x     x  4x  5 
f  x    x  1 x 2  4x  4  1
2
2
3
2
2
f  x   x3  5x 2  9x  5
2