§ 5.7 Polynomial Equations and Their Applications Solving Polynomial Equations We have spent much time on learning how to factor polynomials. Now we will.
Download ReportTranscript § 5.7 Polynomial Equations and Their Applications Solving Polynomial Equations We have spent much time on learning how to factor polynomials. Now we will.
Slide 1
§ 5.7
Polynomial Equations and Their Applications
Slide 2
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
Slide 3
Solving Polynomial Equations
Definition of a Quadratic Equation
A quadratic equation in x is an equation that can be
written in the standard form
ax bx c 0
2
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
Slide 4
Solving Polynomial Equations
Solving a Quadratic Equation by Factoring
1)If necessary, rewrite the equation in the standard form
2
ax 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.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 5.7
Slide 5
Solving Polynomial Equations
EXAMPLE
Solve:
x 4 x 45 .
2
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 4 x 45 45 45
2
x 4 x 45 0
2
Subtract 45 from both sides
Simplify
2) Factor.
x 9 x 5 0
Factor
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 5.7
Slide 6
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x9 0
or
x50
x9
x 5
5) Check the solutions in the original equation.
Check 9:
Check -5:
x 4 x 45
x 4 x 45
9 2 4 9 ? 45
5 2 4 5 ? 45
81 4 9 ? 45
25 4 5 ? 45
2
2
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 5.7
Slide 7
Solving Polynomial Equations
CONTINUED
Check 9:
Check -5:
81 36 ? 45
25 20 ? 45
45 45
45 45
true
true
The solutions are 9 and -5. The solution set is {9,-5}.
80
The graph of
y x 4 x 45
2
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
Slide 8
Solving Polynomial Equations
EXAMPLE
Solve:
x 4 x.
2
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!!!
x 4x 4x 4x
Subtract 4x from both sides
x 4x 0
Simplify
2
2
2) Factor.
xx 4 0
Factor
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 5.7
Slide 9
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x0
or
x40
x 4
5) Check the solutions in the original equation.
Check 0:
x 4x
x 4x
2
0 2
2
?
4 0
0 0
Check 4:
4 2
true
?
4 4
16 16
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 5.7
true
Slide 10
Solving Polynomial Equations
CONTINUED
The solutions are 0 and 4. The solution set is {0,4}.
80
The graph of
70
y x 4x
2
60
50
lies to the right.
40
30
20
10
0
-10
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 5.7
-5
-10 0
5
10
15
Slide 11
Solving Polynomial Equations
EXAMPLE
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 #11 Section 5.7
Slide 12
Solving Polynomial Equations
CONTINUED
2) Factor. Before we can factor the equation, we must simplify it
first.
x 1 x 4 14 0
2
FOIL
x 4 x x 4 14 0
2
Simplify
x 3 x 18 0
Now we can factor the polynomial equation.
x 3 x 6 0
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 5.7
Slide 13
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x3 0
x 3
or
x60
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 #13 Section 5.7
Slide 14
Solving Polynomial Equations
CONTINUED
Check 3:
14 14
Check -6:
14 14 true
true
The solutions are 3 and -6. The solution set is {3,-6}.
120
100
80
The graph of
y x 3 x 18
2
60
40
lies to the right.
20
0
-15
-10
-5
-20
-40
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 5.7
0
5
10
15
Slide 15
Solving Polynomial Equations
EXAMPLE
Solve by factoring:
x 2 x x 2 0.
3
2
SOLUTION
1) Move all terms to one side and obtain zero on the other
side. This is already done.
2) Factor. Use factoring by grouping. Group terms that have a
common factor.
x 2x
3
Common
2
factor is x
2
+ x2 0
Common
factor is -1.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 5.7
Slide 16
Solving Polynomial Equations
CONTINUED
x
2
x 2 x 2 0
x 2 x 2 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 #16 Section 5.7
Slide 17
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
or
or
x20
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 #17 Section 5.7
2
4
6
Slide 18
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 16 t 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 #18 Section 5.7
2
Slide 19
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 16 t 8 t 8
2
8 16 t 8 t 8
2
0 16 t 8 t
2
0 8 t 2 t 1
Original equation
Replace s (t) with 8
Subtract 8 from both sides
Factor
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 5.7
Slide 20
Polynomial Equations in Application
CONTINUED
Now we set each factor equal to zero.
8t 0
t 0
or
2t 1 0
2t 1
t
1
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 #20 Section 5.7
Slide 21
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 #21 Section 5.7
2
Slide 22
The Pythagorean Theorem
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 #22 Section 5.7
Slide 23
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
Let’s begin with a picture.
Tree
5 feet
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 5.7
Slide 24
The Pythagorean Theorem
CONTINUED
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.
x
Tree
x-1
5 feet
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 5.7
Slide 25
The Pythagorean Theorem
CONTINUED
We can now use the Pythagorean Theorem to solve for x, the
length of the wire.
x 12 5 2
x
2
x 2 x 1 25 x
2
x 2 x 26 x
2 x 26 0
2
2
2
26 2 x
13 x
This is the equation arising from
the Pythagorean Theorem
Square x – 1 and 5
Add 1 and 25
Subtract x 2 from both sides
Add 2x to both sides
Divide both sides by 2
Therefore, the solution is x = 13 feet.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 5.7
Slide 26
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
is a quadratic equation.
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 #26 Section 5.7
§ 5.7
Polynomial Equations and Their Applications
Slide 2
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
Slide 3
Solving Polynomial Equations
Definition of a Quadratic Equation
A quadratic equation in x is an equation that can be
written in the standard form
ax bx c 0
2
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
Slide 4
Solving Polynomial Equations
Solving a Quadratic Equation by Factoring
1)If necessary, rewrite the equation in the standard form
2
ax 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.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 5.7
Slide 5
Solving Polynomial Equations
EXAMPLE
Solve:
x 4 x 45 .
2
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 4 x 45 45 45
2
x 4 x 45 0
2
Subtract 45 from both sides
Simplify
2) Factor.
x 9 x 5 0
Factor
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 5.7
Slide 6
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x9 0
or
x50
x9
x 5
5) Check the solutions in the original equation.
Check 9:
Check -5:
x 4 x 45
x 4 x 45
9 2 4 9 ? 45
5 2 4 5 ? 45
81 4 9 ? 45
25 4 5 ? 45
2
2
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 5.7
Slide 7
Solving Polynomial Equations
CONTINUED
Check 9:
Check -5:
81 36 ? 45
25 20 ? 45
45 45
45 45
true
true
The solutions are 9 and -5. The solution set is {9,-5}.
80
The graph of
y x 4 x 45
2
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
Slide 8
Solving Polynomial Equations
EXAMPLE
Solve:
x 4 x.
2
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!!!
x 4x 4x 4x
Subtract 4x from both sides
x 4x 0
Simplify
2
2
2) Factor.
xx 4 0
Factor
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 5.7
Slide 9
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x0
or
x40
x 4
5) Check the solutions in the original equation.
Check 0:
x 4x
x 4x
2
0 2
2
?
4 0
0 0
Check 4:
4 2
true
?
4 4
16 16
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 5.7
true
Slide 10
Solving Polynomial Equations
CONTINUED
The solutions are 0 and 4. The solution set is {0,4}.
80
The graph of
70
y x 4x
2
60
50
lies to the right.
40
30
20
10
0
-10
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 5.7
-5
-10 0
5
10
15
Slide 11
Solving Polynomial Equations
EXAMPLE
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 #11 Section 5.7
Slide 12
Solving Polynomial Equations
CONTINUED
2) Factor. Before we can factor the equation, we must simplify it
first.
x 1 x 4 14 0
2
FOIL
x 4 x x 4 14 0
2
Simplify
x 3 x 18 0
Now we can factor the polynomial equation.
x 3 x 6 0
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 5.7
Slide 13
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
x3 0
x 3
or
x60
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 #13 Section 5.7
Slide 14
Solving Polynomial Equations
CONTINUED
Check 3:
14 14
Check -6:
14 14 true
true
The solutions are 3 and -6. The solution set is {3,-6}.
120
100
80
The graph of
y x 3 x 18
2
60
40
lies to the right.
20
0
-15
-10
-5
-20
-40
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 5.7
0
5
10
15
Slide 15
Solving Polynomial Equations
EXAMPLE
Solve by factoring:
x 2 x x 2 0.
3
2
SOLUTION
1) Move all terms to one side and obtain zero on the other
side. This is already done.
2) Factor. Use factoring by grouping. Group terms that have a
common factor.
x 2x
3
Common
2
factor is x
2
+ x2 0
Common
factor is -1.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 5.7
Slide 16
Solving Polynomial Equations
CONTINUED
x
2
x 2 x 2 0
x 2 x 2 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 #16 Section 5.7
Slide 17
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting
equations.
or
or
x20
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 #17 Section 5.7
2
4
6
Slide 18
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 16 t 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 #18 Section 5.7
2
Slide 19
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 16 t 8 t 8
2
8 16 t 8 t 8
2
0 16 t 8 t
2
0 8 t 2 t 1
Original equation
Replace s (t) with 8
Subtract 8 from both sides
Factor
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 5.7
Slide 20
Polynomial Equations in Application
CONTINUED
Now we set each factor equal to zero.
8t 0
t 0
or
2t 1 0
2t 1
t
1
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 #20 Section 5.7
Slide 21
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 #21 Section 5.7
2
Slide 22
The Pythagorean Theorem
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 #22 Section 5.7
Slide 23
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
Let’s begin with a picture.
Tree
5 feet
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 5.7
Slide 24
The Pythagorean Theorem
CONTINUED
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.
x
Tree
x-1
5 feet
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 5.7
Slide 25
The Pythagorean Theorem
CONTINUED
We can now use the Pythagorean Theorem to solve for x, the
length of the wire.
x 12 5 2
x
2
x 2 x 1 25 x
2
x 2 x 26 x
2 x 26 0
2
2
2
26 2 x
13 x
This is the equation arising from
the Pythagorean Theorem
Square x – 1 and 5
Add 1 and 25
Subtract x 2 from both sides
Add 2x to both sides
Divide both sides by 2
Therefore, the solution is x = 13 feet.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 5.7
Slide 26
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
is a quadratic equation.
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 #26 Section 5.7