Transcript Algebra for College Students, 6e

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, Algebra for College Students, 6e – Slide #2 Section 5.7
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 2  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, Algebra for College Students, 6e – Slide #3 Section 5.7
Solving Polynomial Equations
Solving a Quadratic Equation by Factoring
1)If necessary, rewrite the equation in the standard form
ax 2  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, Algebra for College Students, 6e – Slide #4 Section 5.7
Solving Polynomial Equations
EXAMPLE
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, Algebra for College Students, 6e – 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, Algebra for College Students, 6e – 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, Algebra for College Students, 6e – Slide #7 Section 5.7
5
10
15
Solving Polynomial Equations
EXAMPLE
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, Algebra for College Students, 6e – Slide #8 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
true
16  16 true
Blitzer, Algebra for College Students, 6e – Slide #9 Section 5.7
Solving Polynomial Equations
CONTINUED
The solutions are 0 and 4. The solution set is {0,4}.
80
The graph of y  x 2  4 x
70
60
50
lies to the right.
40
30
20
10
0
-10
-5
-10 0
Blitzer, Algebra for College Students, 6e – Slide #10 Section 5.7
5
10
15
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, Algebra for College Students, 6e – Slide #11 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
FOIL
x 2  4 x  x  4  14  0
Simplify
x 2  3x  18  0
Now we can factor the polynomial equation.
x  3x  6  0
Blitzer, Algebra for College Students, 6e – Slide #12 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
27 ? 14
 6 1 6  4 ? 14
 7 2 ? 14
Blitzer, Algebra for College Students, 6e – Slide #13 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, Algebra for College Students, 6e – Slide #14 Section 5.7
0
5
10
15
Solving Polynomial Equations
EXAMPLE
Solve by factoring: x 3  2 x 2  x  2  0.
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 3  2x 2 +  x  2  0
Common
2
factor is x
Common
factor is -1.
Blitzer, Algebra for College Students, 6e – Slide #15 Section 5.7
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, Algebra for College Students, 6e – Slide #16 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, Algebra for College Students, 6e – Slide #17 Section 5.7
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4
6
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
2
Time (seconds)
Blitzer, Algebra for College Students, 6e – Slide #18 Section 5.7
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, Algebra for College Students, 6e – Slide #19 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, Algebra for College Students, 6e – Slide #20 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, Algebra for College Students, 6e – Slide #21 Section 5.7
2
5.7
Assignment
p. 376 (2-30 even)