# Ch 3 Polynomial Functions

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```					        Chapter 3: Polynomial Functions

3.1   Complex Numbers
3.4   Further Applications of Quadratic Functions and Models
3.5   Higher Degree Polynomial Functions and Graphs
3.6   Topics in the Theory of Polynomial Functions (I)
3.7   Topics in the Theory of Polynomial Functions (II)
3.8   Polynomial Equations and Inequalities; Further
Applications and Models

Slide 3-1

An equation that can be written in the form
ax2 + bx + c = 0
where a, b, and c are real numbers with a  0, is a
• Consider the
– zeros of P( x)  2 x 2  4 x  16
– x-intercepts of P( x)  2 x 2  4 x  16
– solution set of 2 x  4 x  16  0.
2

• Each is solved by finding the numbers that make
2 x  4 x  16  0 true.
2

Slide 3-2
3.3 Zero-Product Property

• If a and b are complex numbers and ab = 0, then a = 0 or
b = 0 or both equal zero.
Example Solve 2 x 2  4 x  16  0.
Solution   2 x 2  4 x  16  0
x2  2x  8  0
( x  4)( x  2)  0   x40       or    x2  0
x  4     or      x2

The solution set of the equation is {– 4,2}.
Note that the zeros of P(x) = 2x2 + 4x – 16
are –4 and 2, and the x-intercepts are (– 4,0)
and (2,0).
Slide 3-3
3.3 Example of a Quadratic with One
Distinct Zero
Solve x  6 x  9  0.
2
Example

Solution      x2  6x  9  0         ( x  3) 2  0
x  3  0 or       x 3  0
x  3 or          x3
There is one distinct zero, 3. It is sometimes called a double
zero, or double solution (root) of the equation.

Graphing Calculator Solution

Slide 3-4
3.3 Solving x2 = k

x2  k
x2  k  0
( x  k )( x  k )  0  x  k  0          or x  k  0
x k           or     x k
Square Root Property
The solution of x 2  k is
(a) { k} if k  0 (b) {0} if k  0 (c) {i k } if k  0.

Figure 26 pg 3-38

Slide 3-5
3.3 Examples Using the Square Root
Property
Solve each equation.
(a) 2 x 2  10         (b) 2( x  1) 2  14
x 2  5               ( x  1) 2  7
x  i 5                  x 1   7  x  1 7

Graphing Calculator Solution
(a)                    (b)

This solution is an approximation.

Slide 3-6

• Complete the square on P( x)  ax 2  bx  c and rewrite P in the form
P( x)  a( x  h) 2  k . To find the zeros of P, use the square root property
to solve for x.
ax 2  bx  c  0
b      c
x2  x   0
a      a
b        c
x  x
2

a        a
b      b2      b2 c
x  x 2  2 
2

a      4a     4a a
b 2  4ac
2
      b 
x  
      2a        4a 2
b         b 2  4ac
x       
2a            4a 2
b       b 2  4ac  b  b 2  4ac
x                     
2a          2a           2a
Slide 3-7
3.3 The Quadratic Formula and the
Discriminant
The solutions of the equation ax 2  bx  c  0, where a  0 ,
are
b  b 2  4 ac
x                   .
2a

• The expression under the radical, b 2  4ac, is called the
discriminant.
• The discriminant determines whether the quadratic
equation has
– two real solutions if b  4ac  0,
2

– one real solution if b  4ac  0,
2

– no real solutions if b  4ac  0.
2

Slide 3-8
3.3 Solving Equations with the Quadratic
Formula
Solve x( x  2)  2 x  2.
Analytic Solution
x ( x  2)  2 x  2
x2  2x  2x  2                       Rewrite in the form
x2  4x  2  0                            ax2 + bx + c = 0

 (4)  (4) 2  4(1)(2)
x                              Substitute into the
2(1)             quadratic formula a = 1,
b = – 4, and c = 2, and
4  16  8
x                              simplify.
2
4 8

2
42 2
           2 2
2
Slide 3-9
3.3 Solving Equations with the Quadratic
Formula
Graphical Solution

With the calculator, we get an approximation for
2  2 and 2  2.

Slide 3-10
3.3 Solving a Quadratic Equation with
No Solution
Example    Solve 2 x 2  x  4  0.
 (1)  (1) 2  4(2)(4)
Analytic Solution       x
2(2)
1  1  32

4
1   31 1  i 31 1       31
                      i
4          4      4  4
Graphing Calculator Solution

Figure 30 pg 3-43

Slide 3-11
Function Graphs (a > 0)
P(x) a>0        Solution Set of   Is

P(x)=0            {a,b}
x-axis   P(x)<0            interval (a,b)
(a,0)           (b,0)              P(x)>0            ( , a )  ( b ,  )

P(x) a>0        Solution Set of   Is

P(x)=0            {a}
x-axis   P(x)<0            
P(x)>0            ( , a )  ( a ,  )
(a,0)

P(x) a>0        Solution Set of   Is

P(x)=0            
x-axis   P(x)<0            
P(x)>0            ( ,  )
a

• Similar for a quadratic function P(x) with a < 0.
Slide 3-12
Analytically
Solve x 2  x  12  0.
x 2  x  12  0
( x  3)( x  4)  0          x  3 or        x4
The function is 0 when x = – 3 or x = 4. It will be greater than 0 or
less than zero for any other value of x. So we use a sign graph to
determine the sign of the product (x + 3)(x – 4).

For the interval (–3,4), one factor is positive and the other is
negative, giving a negative product. The product is positive
elsewhere since (+)(+) and (–)(–) is positive.
Slide 3-13
Graphically
Verify the analytic solution to x  x  12  0 : Y1 < 0 in the
2

interval (–3,4).

1.   Solve the corresponding quadratic equation.
2.   Identify the intervals determined by its solutions.
3.   Use a sign graph to determine the solution interval(s).
4.   Decide whether or not the endpoints are included.
Slide 3-14

 d2
Example Solve (a) A               for d (b) rt 2  st  k (r  0) for t.
4
Solution 2
d
(a) A      4A   d 2
4        4A
d2 

4A      A
d        2
           

(b) rt  st  k  rt  st  k  0
2                 2

 b  b 2  4ac
t                       where a  r , b   s, c  k
2a
 ( s )  ( s ) 2  4(r )(k ) s  s 2  4rk
t                                  
2r                          2r
Slide 3-15
3.3 Modeling Length of Life

The survival rate after age 65 is approximated by S ( x)  1  .058 x  .076 x 2 ,
where x is measured in decades. This function gives the probability that an
individual who reaches age 65 will live at least x decades (10x years)
longer. Find the age for which the survival rate is .5 for people who reach
the age of 65. Interpret the result.
Solution Let Y1 = S(x), Y2 = .5, and find the positive x-value of an
intersection point.

Discard x = – 3. For x  2.2, half the people who reach age
65 will live about 2.2 decades or 22 years longer to age 87.

Slide 3-16

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