Chabot Mathematics
§5.7 PolyNomial
Eqns & Apps
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Review § 5.6
MTH 55
 Any QUESTIONS About
• §5.6 → Factoring Strategies
 Any QUESTIONS About HomeWork
• §5.6 → HW-20
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
§5.7 Solving PolyNomial Eqns
 The Principle of Zero Products
 Factoring to Solve Equations
 Algebraic-Graphical Connection
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Quadratic Equations
 Second degree equations such as
9t2 – 4 = 0 and x2 + 6x + 9 = 0 are
called quadratic equations
 A quadratic equation is an equation
equivalent to one of the form
ax2 + bx + c = 0
• where a, b, and c are constants, with a ≠ 0
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Principle of Zero Products
 An equation AB = 0 is true if and
only if A = 0 or B = 0, or both = 0.
 That is, a product is 0 if and only if
at LEAST ONE factor in the
multiplication-chain is 0
• i.e.; Need a Zero-FACTOR to create
a Zero-PRODUCT
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Solve (x + 4)(x − 3) = 0
 In order for a product to be 0, at least
one factor must be 0. Therefore, either
x + 4 = 0 or x − 3 = 0
 Solving each equation:
x+4=0
or
x−3=0
x = −4 or
x=3
 Both −4 and 3 should be checked in the
original equation.
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Check for (x + 4)(x − 3) = 0
For x = −4:
For x = 3:
(x + 4)(x − 3) = 0
(x + 4)(x − 3) = 0
(−4 + 4)(−4 − 3)
(3 + 4)(3 − 3)
0(−7)
7(0)
0=0
0=0
True
True
The solutions are −4 and 3.
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Solve  4(3x + 1)(x − 4) = 0
 Since the factor 4 is constant, the only
way for 4(3x + 1)(x − 4) to be 0 is for
one of the other factors to be 0. That is,
3x + 1 = 0
or
x−4=0
3x = −1
or
x=4
x = −1/3
 So the solutions to the Equation are
x = −1/3
and
x=4
{−1/3 , 4}
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Check 4(3x + 1)(x – 4) = 0
 For −1/3:
4(3x + 1)(x − 4) = 0
4(3•[−1/3] + 1)([−1/3] − 4) = 0
4(−1 + 1)(−1/3 − 12/3) = 0
4(0)(−13/3 ) = 0
0=0
 For 4:
4(3x + 1)(x − 4) = 0
4((3(4) + 1)(4 − 4) = 0
4(13)(0) = 0
0=0
 The solutions are −1/3 and 4
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Solve  3y(y − 7) = 0
 SOLUTION
3  y (y − 7) = 0
y=0
or
y=0
or
y−7=0
y =7
 The solutions are 0 and 7
• The Check is Left to the Student
– Should be easily “EyeBalled”
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Factoring to Solve Equations
 By factoring and using the principle of
zero products, we can now solve a
variety of quadratic equations.
 Example: Solve x2 + 9x + 14 = 0
 SOLUTION: This equation requires us
to FIRST factor the polynomial since
there are no like terms to combine and
there is a squared term. THEN we use
the principle of zero products
Chabot College Mathematics
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Solve  x2 + 9x + 14 = 0
 Factor the Left Hand Side (LHS) by
Educated Guessing (FOIL Factoring):
x2 + 9x + 14 = 0
(x + 7)(x + 2) = 0
x+7=0
or x + 2 = 0
x = −7
or
x = −2.
 The Tentative Solutions are −7 and −2
• Let’s Check
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Check  x = −7
For −7:
&
x = −2
For −2:
x2 + 9x + 14 = 0
(−7)2 + 9(−7) + 14
49 − 63 + 14
−14 + 14
0=0
x2 + 9x + 14 = 0
(–2)2 + 9(–2) + 14
4 − 18 + 14
−14 + 14
0=0
True
True
 Thus −7 and −2 are VERIFIED as
Solutions to x2 + 9x + 14 = 0
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Solve x2 + 9x = 0
 SOLUTION: Although there is no
constant term, because of the x2-term,
the equation is still quadratic. Try
factoring:
x2 + 9x = 0 → see GCF = x
x(x + 9) = 0
x=0
or
x+9=0
x=0
or
x = −9
 The solutions are 0 and −9.
Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Caveat Mathematicus
 CAUTION: We must have 0 on one
side of the equation before the
principle of zero products can be used.
 Get all nonzero terms on one side of the
equation and 0 on the other
 Example: Solve: x2 − 12x = −36
Chabot College Mathematics
15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Solve  x2 − 12x = −36


SOLUTION: We first add 36 to BOTH
Sides to get 0 on one side:
x2 − 12x = −36
x2 − 12x + 36 = −36 + 36 = 0
(x − 6)(x − 6) = 0
x − 6 = 0 or
x−6=0
x = 6 or
x=6
There is only one solution, 6.
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Standard Form
 To solve a quadratic equation using
the principle of zero products, we
first write it in standard form: with
0 on one side of the equation and
the leading coefficient POSITIVE.
 We then factor and determine when
each factor is 0.
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Functional Eval
 Given f(x) = x2 + 10x + 26.
Find a such that f(a) = 1.
 SOLUTION
f (a) = a2 + 10a + 26 = 1
Set f (a) = 1
a2 + 10a + 25 = 0
(a + 5)(a + 5) = 0
a + 5 = 0 so a = −5
The check is left for you & I to do Later
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Solve 9x2 = 49
 SOLUTION: 9x2 = 49
9x2 − 49 = 0
► Diff of Sqs: (3x)2 & 72
(3x − 7)(3x + 7) = 0
3x − 7 = 0 or 3x + 7 = 0
3x = 7 or 3x = −7
x = 7/3 or x = −7/3
 The solutions are 7/3 & −7/3
Chabot College Mathematics
19
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Solve: 14x2 + 9x + 2 = 10x + 6
 SOLUTION: Be careful with an equation like this!
Since we need 0 on one side, we subtract
10x and 6 from Both Sides to get the RHS = 0
14x2 + 9x + 2 = 10x + 6
14x2 + 9x − 10x + 2 − 6 = 0
14x2 − x − 4 = 0
(7x − 4)(2x + 1) = 0
7x − 4 = 0 or 2x + 1 = 0
7x = 4 or
2x = −1
x = 4/7 or
x = −1/2
 The solutions are 4/7 and −1/2.
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Solve Eqns by Zero Products
1. Write an equivalent equation with
0 on one side, using the addition
principle.
2. Factor the nonzero side of the
equation
3. Set each factor that is not a
constant equal to 0
4. Solve the resulting equations
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Parabola intercepts
 Find the x-intercepts for
the graph of the
equation shown.
Parabola
 The x-intercepts
occur where the plot
crosses y = 0
 Thus at the x-intercepts
y = x2 + 2x  8
y = 0 = x2 + 2x − 8
 So Can use the principle of zero products
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Parabola intercepts cont.
 SOLUTION: To find the
intercepts, let y = 0 and
solve for x.
Parabola
0 = x2 + 2x − 8
0 = (x + 4)(x − 2)
x+4=0
x = −4
or x − 2 = 0
or x = 2
y = x2 + 2x  8
 The x-intercepts are (−4, 0) and (2, 0).
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
−x2 − x + 6 = 0  Solve with Graph
8
 Solve
y
7
0  x  x  6
2
y   x  x  6  Recall from
2
6
5
Graphing that
the x-axis is the
Location where
y=0
4
3
2
1
x
0
-5
-4
-3
-2
-1
0
-1
-2
-3
M65_§71_Graphs_0607.xls5_Graphs_0607.xls
Chabot College Mathematics
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1
2
3
4
5
 Thus on Graph
find x for y = 0
 Solns: x = −3
and x = 2
-4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Find x-Intercepts
y
-5
 Find the xintercepts for
x 5 graph (at Left) of
Equation
y  2 x  3x  9
2
 At intercepts
y = 0; so Use
ZERO Products:
y  0  2 x  3x  9
2
M65_§71_Graphs_0607.xls5_Graphs_0607.xls
Chabot College Mathematics
25
-12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
y
x
-5
Example  Find x-Intercepts
 At Intercepts y = 0, so
y = 0 = 2x2 + 3x − 9
 FOIL-factor the Quadratic expression
0 = 2x2 + 3x − 9 = (2x − 3)(x + 3) = 0
 By ZERO PRODUCTS
(2x − 3) = 0 or (x + 3) = 0
 Solving for x (the intercept values):
x = 3/2 or x = −3
M65_§71_Graphs_0607.xls5_Graphs_0607.xls
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
-12
5
y
 3,0
3 
 ,0 
2 
-5
M65_§71_Graphs_0607.xls5_Graphs_0607.xls
Chabot College Mathematics
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x
5
Example 
x-Intercepts
-12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
PolyNomial Fcns and Graphs
 Consider, for
example the eqn,
x2 − 2x = 8.
One way to begin to
solve this equation
is to graph the
function
f (x) = x2 − 2x. Then
look for any x-value
that is paired with 8,
as shown at Right
Chabot College Mathematics
28
f ( x)  x 2  2 x
y
y=8
8
7
6
5
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2 3 4 5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
x
PolyNomial Fcns and Graphs
2
 Equivalently, we
1
Root-1
0
could graph the
-6
-5
-4
-3
-2
-1
0
-1
function given by
-2
g(x) = x2 − 2x − 8
-3
and look for the
-4
values of x for which
-5
g(x) = 0. These
-6
values are what we
g x   x 2  2 x  8 -7
call the roots, or
-8
zeros, of a
-9
polynomial function
file =XY_Plot_0211.xls
Chabot College Mathematics
29
y
x
Root-2
1
2
3
-10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
4
5
6
Problem Solving
 Some problems can be translated
to quadratic equations, which we
can now solve.
 The problem-solving process is the
same as for other kinds of
problems.
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
The Pythagorean Theorem
 Recall Pythagorus’ Great Discovery:
 In any right triangle, if a and b are the
lengths of the legs and c is the length of
the hypotenuse, then
a2 + b2 = c2
c
a
b
Chabot College Mathematics
31
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Screen Diagonal
 A computer screen has the dimensions
(in inches) shown below.
 Find the length of the diagonal of the
screen.
x6
x
x3
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Screen Diagonal
 Familiarize. A right triangle is formed
using the diagonal and sides of the
screen. x + 6 is the hypotenuse with x
and x + 3 as legs.
 Translate. Applying the Pythagorean
Theorem, Use the Diagram to translate
as follows:
a2 + b2 = c2
x2 + (x + 3)2 = (x + 6)2
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Screen Diagonal
 Carry out. Solve the equation by:
x2 + (x2 + 6x + 9) = x2 + 12x + 36
2x2 + 6x + 9 = x2 + 12x + 36
x 2 – 6x – 27 = 0
(x + 3)(x – 9) = 0
x + 3 = 0 or x – 9 = 0
x = –3 or x = 9
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Screen Diagonal
 Check. The integer −3 cannot be the
length of a side because it is negative.
For x = 9, we have x + 3 = 12 and
x + 6 = 15. Since 81 + 144 = 225,
the lengths determine a right triangle.
Thus 9, 12,and 15 check.
 State. The length of the diagonal of the
screen is 15 inches
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Area Allotment
 A LandScape Architect designs a flower
bed of uniform width around a small
reflecting pool. The pool is 6ft by 10ft.
The plans call for 36 ft2 of plant coverage.
How WIDE should the Border be?
 Familiarize
with
Diagram
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Area Allotment
 Now LET x ≡ the Border Width
 Translate using Diagram
 The OverAll
• Width is 6ft + 2x
• Length is
10ft + 2x
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Area Allotment
Chabot College Mathematics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Area Allotment
 Carry Out
x  9 or x  1
Zero Products
 Since a Length can Never be Negative,
Discard −9 as solution
Chabot College Mathematics
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Area Allotment
 State: The reflecting pool border
width should be 1 ft
 Check: 2·(12ft·1ft) + 2·(6ft·1ft)
= 24ft2 + 12ft2
12 ft
= 36ft2 
8 ft
Chabot College Mathematics
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Rocket Ballistics
 A Model Rocket is fired
UpWards from the Ground.
The Height, h, in feet of the
rocket can be found from
this equation:
ht   80t  16t
2
– Where t is time in seconds
 Find the time that it takes
for the Rocket to reach a
height of 48 feet
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Rocket Ballistics
 Familiarize: We must find
t such that h(t) = 48.
• Thus Substitute 48 for h(t) in
the ballistics equation
48  80t  16t
2
 Carry Out:
• Subtract 48 from both sides
0  80t  16t  48
2
Chabot College Mathematics
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Rocket Ballistics
• Divide Both Sides by −16
0  5t  t  48
2
• Write in Standard form
0  t  5t  3
2
Prime 
Not Factorable
• Use QUADRATIC Formula
with a = 1, b = −5, and c = 3
Chabot College Mathematics
43
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Rocket Ballistics
• Approximate the Sq-Roots
 Since “What goes UP must
come DOWN” The Rocket
will reach 48ft TWICE;
Once while blasting-UP
and again while
Free-Falling DOWN
Chabot College Mathematics
44
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Example  Rocket Ballistics
 State: The rocket will climb
to 48ft in about 0.7
seconds, continue its
climb, and then it will
descend (fall) to a height of
48ft after a total flite time of
4.3 seconds as it continues
its FreeFall to the Ground
Chabot College Mathematics
45
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
WhiteBoard Work
 Problems From §5.7 Exercise Set
• 74, 80, 82

Model
Rockets
Chabot College Mathematics
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
All Done for Today
Blast
Off!
Chabot College Mathematics
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
49
5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
4
5
6
y
5
2
y
1
x
4
0
-6
-5
-4
-3
-2
-1
3
0
1
2
3
4
-1
-2
2
-3
1
-4
x
-5
5
-6
0
-3
-2
-1
0
1
2
3
4
-1
-7
-8
-2
-9
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
50
file =XY_Plot_0211.xls
-10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt
5
6
Quadratic Formula
Chabot College Mathematics
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt