### Chapter Five 5.7

```§ 5.7
Polynomial Equations and Their Applications
Solving Polynomial Equations
We have spent much time on learning how to factor
polynomials.
Now we will look at one important use of factoring.
In this section, we will use factoring to solve equations of
degree 2 and higher. Up to this point, we have only looked at
solving equations of degree one.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 5.7
Solving Polynomial Equations,
pp 368-369
A quadratic equation in x is an equation that can be
written in the standard form
ax2  bx  c  0
where a, b, and c are real numbers, with a  0 . A
quadratic equation in x is also called a second-degree
polynomial equation in x.
The Zero-Product Rule
If the product of two algebraic expressions is zero, then at
least one of the factors is equal to zero.
If AB = 0, then A = 0 or B = 0.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 5.7
pp 368-369
Solving Polynomial Equations,
p 369
Solving a Quadratic Equation by Factoring
1)If necessary, rewrite the equation in the standard form
ax2  bx  c  0 , moving all terms to one side, thereby
obtaining zero on the other side.
2) Factor completely.
3) Apply the zero-product principle, setting each factor
containing a variable equal to zero.
4) Solve the equations in step 3.
5) Check the solutions in the original equation.
pp 369
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 5.7
Solving Polynomial Equations
EXAMPLE (similar to example 1 on page 369)
Solve: x 2  4 x  45.
SOLUTION
1) Move all terms to one side and obtain zero on the other
side. Subtract 45 from both sides and write the equation in
standard form.
x 2  4 x  45  45  45
x 2  4 x  45  0
Subtract 45 from both sides
Simplify
2) Factor.
x  9x  5  0
Factor
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x9  0
x9
or
x5  0
x  5
5) Check the solutions in the original equation.
Check 9:
Check -5:
x 2  4 x  45
x 2  4 x  45
92  49 ? 45
81 49 ? 45
 52  4 5 ? 45
?
25  4 5 
45
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 5.7
Solving Polynomial Equations
CONTINUED
Check 9:
Check -5:
81  36 ? 45
25  20 ? 45
45  45 true
45  45 true
The solutions are 9 and -5. The solution set is {9,-5}.
80
The graph of y  x 2  4 x  45
60
40
lies to the right.
20
0
-10
-5
0
-20
-40
-60
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 5.7
5
10
15
Solving Polynomial Equations
Check Point 1
2x  9x  5
Solve
2
2x  9x  5  0
2
2x  1x  5  0
2x  1  0 x  5  0
1
x 2
x5
 1 
 ,5
 2 
P 370
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 5.2
Solving Polynomial Equations
EXAMPLE (similar to example 2a on page 371
Solve: x 2  4 x.
SOLUTION
1) Move all terms to one side and obtain zero on the other
side. Subtract 4x from both sides and write the equation in
standard form. NOTE: DO NOT DIVIDE BOTH SIDES BY x.
WE WOULD LOSE A POTENTIAL SOLUTION!!!
x2  4x  4x  4x
Subtract 4x from both sides
x2  4x  0
Simplify
2) Factor.
xx  4  0
Factor
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x0
or
x40
x4
5) Check the solutions in the original equation.
Check 0:
Check 4:
x2  4x
x2  4x
02 ? 40
42 ? 44
00
16  16
true
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 5.7
true
Solving Polynomial Equations
CONTINUED
The solutions are 0 and 4. The solution set is {0,4}.
80
The graph of y  x 2  4x
70
60
50
lies to the right.
40
30
20
10
0
-10
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 5.7
-5
-10 0
5
10
15
Solving Polynomial Equations
EXAMPLE (similar to 2c on page 371
Solve: x  1x  4  14.
SOLUTION
Be careful! Although the left side of the original equation is
factored, we cannot use the zero-product principle since the right
side of the equation is NOT ZERO!!
1) Move all terms to one side and obtain zero on the other
side. Subtract 14 from both sides and write the equation in
standard form.
x 1x  4 14  14 14
x 1x  4 14  0
Subtract 14 from both sides
Simplify
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 5.7
Solving Polynomial Equations
CONTINUED
2) Factor. Before we can factor the equation, we must simplify it
first.
x 1x  4 14  0
x 2  4 x  x  4  14  0
x 2  3x  18  0
FOIL
Simplify
Now we can factor the polynomial equation.
x  3x  6  0
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x3  0
x3
or
x60
x  6
5) Check the solutions in the original equation.
Check 3:
x 1x  4  14
Check -6:
x 1x  4  14
3 13  4 ? 14
 6 1 6  4 ? 14
27 ? 14
 7 2 ? 14
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 5.7
Solving Polynomial Equations
CONTINUED
Check 3:
14  14 true
Check -6:
14  14 true
The solutions are 3 and -6. The solution set is {3,-6}.
120
100
80
The graph of y  x 2  3x 18
60
40
lies to the right.
20
0
-15
-10
-5
-20
-40
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 5.7
0
5
10
15
Solving Polynomial Equations
EXAMPLE (similar to 2b on page 371)
Solve by factoring: x3  2 x 2  x  2  0.
SOLUTION
1) Move all terms to one side and obtain zero on the other
2) Factor. Use factoring by grouping. Group terms that have a
common factor.
x 3  2x 2 +  x  2  0
Common
2
factor is x
Common
factor is -1.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 5.7
Solving Polynomial Equations
CONTINUED
x 2 x  2  x  2  0
x  2x2 1  0
x  2x  1x 1  0
Factor x 2 from the first two
terms and -1 from the last
two terms
Factor out the common
binomial, x – 2, from each
term
Factor completely by
factoring x 2  1 as the
difference of two squares
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
or
or
x20
x 1  0
x 1  0
x2
x  1
x 1
5) Check the solutions in the original equation. Check the
three solutions 2, -1, and 1, by substituting them into the original
equation. Can you verify that the solutions are 2, -1, and 1?
150
100
The graph of y  x  2 x  x  2
3
2
50
0
lies to the right.
-6
-4
-2
0
-50
-100
-150
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 5.7
2
4
6
Solving Polynomial Equations
Check Point 2a
Solve
3x 2  2 x
 2
0, 
 3
Check Point 2b
Solve
x 2  7  10x  18
Check Point 2c
Solve
x  2x  3  6
5
 4,3
P 372
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 5.7
Solving Polynomial Equations
Check Point 3
2 x  3x  8x  12
3
Solve
2x
3
2

 3x 2   8x 12  0
x 2x  3  42 x  3  0
2
2x  3x2  4  0
2x  3x  2x  2  0
3 

 2, ,2
2 

Blitzer, Intermediate Algebra, 5e – Slide #20 Section 5.7
P 370
Modeling Motion Application
EXAMPLE 4 See book p 373-4 (figure on p 374)
finds when the ball hits the ground
building height is 384
similar to problem 65 on p. 378
similar to 98 on p. 388
Solve
st   16t 2  32t  384
0  16t 2  32t  384
Time to this
height,
substitute 0
for s(t)
thrust
Beginning
height
0  16(t 2  2t  24)
0  16t  4t  6
Factor out -16
factor
t  4  0 t  6  0
t  4
 4,6
t 6
The ball hits the ground after 6
seconds.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 5.7
Modeling Motion Application
Check Point 4 on page 374
finds when the ball height is at 336 feet
Solve
st   16t 2  32t  384
336  16t 2  32t  384
Time to this
height,
substitute
336 for s(t)
0  16t 2  32t  48
Subtract 336
0  16(t 2  2t  3)
0  16t  1t  3
Factor out -16
factor
t  1  0 t  3  0
t  1
 1,3
t3
The ball passes 336 after 3 seconds.
(3,336) is the point on the graph
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 5.7
Modeling Motion Application
Flight of a baseball
The height in feet of a
baseball after t seconds is
given by the following.
At what values of t is the
height of the baseball 75
feet?
s (t )  16t 2  88t  3
Beginning
height
Time to this
height,
substitute 75 for
s(t)
thrust
75  16t 2  88t  3
Time to this
height,
substitute 75
for h
Modeling Motion Application
Flight of a baseball, continued
75  16t 2  88t  3
Multiply both sides by -1
 75  16t 2  88t  3
0  16t 2  88t  3  75
16t 2  88t  72  0
Combine like terms
8(2t 2 11t  9)  0
8t  12t  9  0
Answer is 1 second going up
and 4.5 seconds going down
t  1  0
2t  9  0
t 1
t  9 2  4.5
Both
valid
Modeling Motion Application
Model Rocket #98 on
page 388
A model rocket is launched
from the top of a cliff 144
feet above sea level.
The rocket misses the cliff
and falls in the ocean. How
long will it take for the rocket
to hit the water?
s (t )  16t 2  128t  144
Beginning
height
thrust
0  16t 2  128t  144
Time to this height,
substitute 0 for s(t)
Modeling Motion Application
0  16t 2  128t  144
Model rocket #98 on page 388, continued
0  16(t 2  8t  9)
 16t  9 t  1  0
t  9  0
t  1  0
t 9
t  1
Factor out -16
Modeling Motion Application
Model Rocket
A model rocket is launched
from the ground.
s (t )  16t 2  48t
How long will it take for the
rocket to hit the ground?
0  16t 2  48t
Beginning
height is zero
0  16t (t  3)
Time to this height,
substitute 0 for s(t)
The rocket strikes the ground after 3
seconds.
-16t = 0
t= 0
t-3 = 0
t= 3
Modeling Motion Application
Car breaking, #99 on 388 (from extra credit)
How much distance do you
need to bring your car to a
complete stop?
d(x) is the stopping distance,
in feet, for a care traveling at
x miles per hour.
It takes 40 feet to come
to a complete stop,
substitute 40 for d(x)
X+40=0
X-20=0
t = -40
t = 20
x2
d ( x) 
x
20
x2
40 
x
20
800  x 2  20 x
0  x 2  20 x  800
0  ( x  40)( x  20)
The car was traveling 20 miles per hour.
Solving Polynomial Equations-Area
Do Check Point 5 on page 375 – see word problem diagram
(also see Example 5 on page 374.)
Solve
12  2x16  2x  320
192 24x  32x  4 x 2  320
4 x 2  56x  192  320
4 x 2  56x  128  0
4x  16x  2  0
factor
16,2
The path should be 2 feet wide.
P 375
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 5.7
The Pythagorean Theorem,
p 375
The Pythagorean Theorem
The sum of the squares of the lengths of the legs of a right
triangle equals the square of the length of the hypotenuse.
If the legs have lengths a and b, and the hypotenuse has
length c, then a 2  b 2  c 2 .
B
Hypotenuse has length c
The two legs have lengths of a and b
c
A
b
a
C
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 5.7
The Pythagorean Theorem
EXAMPLE
A tree is supported by a wire anchored in the ground 5 feet from its base.
The wire is 1 foot longer than the height that it reaches on the tree. Find the
length of the wire.
SOLUTION
Since the wire is 1 foot longer than the height that it reaches on the tree, if
we call the length of the wire (the quantity we wish to determine) x, then the
height of the tree would be x -1.
Tree
x
This is the equation arising from
the Pythagorean Theorem
Tree
x-1
a 2  b2  c2 .
x 12  52  x2
5 feet
5 feet
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 5.7
The Pythagorean Theorem
CONTINUED
We can now use the Pythagorean Theorem to solve for x, the
length of the wire.
x 12  52  x2
x 2  2 x  1  25  x 2
x 2  2 x  26  x 2
 2 x  26  0
26  2 x
13  x
This is the equation arising from
the Pythagorean Theorem
Square x – 1 and 5
Subtract x 2 from both sides
Divide both sides by 2
Therefore, the solution is x = 13 feet.
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 5.7
The Pythagorean Theorem
#82 on page 380
See picture on page 380 to solve for x, how far up the pole.
x  42  x2  202
x 2  8x  16  x 2  400
2 x 2  8x  16  400
2 x 2  8x  384  0
2( x2  4x 192)  0
2( x  12)(x  16)
x  12  0
x  12
x  16  0
x  16
This is the equation arising from
the Pythagorean Theorem
Square x + 4 and 20
Combine terms
Subtract 400 from both sides
GCF
Factor
Therefore, the solution is x = 12 feet.
Blitzer, Intermediate Algebra, 5e – Slide #33 Section 5.7
Solving Polynomial Equations
Important to Note:
A polynomial equation is the result of setting two polynomials equal to each
other.
The equation is in standard form if one side is 0 and the polynomial on the
other side is in standard form, that is in descending powers of the variable.
A polynomial equation of degree one is a linear equation and of degree two
Some polynomial equations can be solved by writing the equation in standard
form, factoring, and then using the zero-product principle: If a product is 0,
then at least one of its factors is equal to 0.
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 5.7
DONE
Modeling Motion Application
h  16t 2  88t  3
Time to this
height,
substitute 75
for h
thrust
Beginning
height
75  16t 2  88t  3
Time to this height,
substitute 75 for h
h  16t 2  132t  0
thrust
Time to this height,
substitute 0 for h
Beginning
height
(height of
building)
Polynomial Equations in Application
EXAMPLE
A gymnast dismounts the uneven parallel bars at a height of 8
feet with an initial upward velocity of 8 feet per second. The
function st   16t 2  8t  8 describes the height of the
gymnast’s feet above the ground, s (t), in feet, t seconds after
dismounting. The graph of the function is shown below.
10
Height (feet)
8
6
4
2
0
0
0.5
1
1.5
Time (seconds)
Blitzer, Intermediate Algebra, 5e – Slide #37 Section 5.7
2
Polynomial Equations in Application
CONTINUED
When will the gymnast be 8 feet above the ground? Identify the
solution(s) as one or more points on the graph.
SOLUTION
We note that the graph of the equation passes through the line y
= 8 twice. Once when x = 0 and once when x = 0.5. This can be
verified by determining when y = s (t) = 8. That is,
st   16t 2  8t  8
8  16t 2  8t  8
0  16t 2  8t
0  8t 2t  1
Original equation
Replace s (t) with 8
Subtract 8 from both sides
Factor
Blitzer, Intermediate Algebra, 5e – Slide #38 Section 5.7
Polynomial Equations in Application
CONTINUED
Now we set each factor equal to zero.
 8t  0
t0
or
2t  1  0
2t  1
1
t
2
We have just verified the information we deduced from the graph.
That is, the gymnast will indeed be 8 feet off the ground at t = 0
seconds and at t = 0.5 seconds. These solutions are identified by
the dots on the graph on the next page.
Blitzer, Intermediate Algebra, 5e – Slide #39 Section 5.7
Polynomial Equations in Application
CONTINUED
10
Height (feet)
8


0
0.5
6
4
2
0
1
1.5
Time (seconds)
Blitzer, Intermediate Algebra, 5e – Slide #40 Section 5.7
2
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