5.7 polynomial equations and their applications. blitzer, algebra for college students, 6e – slide...

5.7

Polynomial Equations and Their Applications

Blitzer, Algebra for College Students, 6e – Slide #2 Section 5.7

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.



Definition of a Quadratic EquationA quadratic equation in x is an equation that can be written in the standard form

where a, b, and c are real numbers, with . A quadratic equation in x is also called a second-degree polynomial equation in x.

02 cbxax

0a

The Zero-Product RuleIf 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.



Solving a Quadratic Equation by Factoring1)If necessary, rewrite the equation in the standard form

, 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.

02 cbxax



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.xx 4542 Solve:

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.

45454542 xx

04542 xx

Subtract 45 from both sides

Simplify

2) Factor.

059 xx Factor



4542 xx

3) & 4) Set each factor equal to zero and solve the resulting equations.

09 x or

CONTINUECONTINUEDD

05 x

9x 5x

5) Check the solutions in the original equation.Check 9: Check -5:

4542 xx

45949 2 45545 2

459481 455425 ?

?

?

?



CONTINUECONTINUEDD Check 9: Check -5:

453681 452025

4545 4545

The solutions are 9 and -5. The solution set is {9,-5}.

? ?

-60

-40

-20

0

20

40

60

80

-10 -5 0 5 10 15

The graph of

lies to the right.

4542 xxy

truetrue



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

x.x 42 Solve:

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!

xxxx 4442

042 xx

Subtract 4x from both sides

Simplify

2) Factor.

04 xx Factor




0x or

CONTINUECONTINUEDD

04 x

4x

5) Check the solutions in the original equation.Check 0: Check 4:

040 2 444 2

00 1616

? ?

xx 42 xx 42

truetrue



CONTINUECONTINUEDD

The solutions are 0 and 4. The solution set is {0,4}.

The graph of

lies to the right.

xxy 42

-10

0

10

20

30

40

50

60

70

80

-10 -5 0 5 10 15



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.xx 1441 Solve:

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.

Simplify

14141441 xx

01441 xx




CONTINUECONTINUEDD2) Factor. Before we can factor the equation, we must simplify it first.

01441 xx

014442 xxx

01832 xx

FOIL

Simplify

Now we can factor the polynomial equation.

063 xx



CONTINUECONTINUEDD3) & 4) Set each factor equal to zero and solve the resulting equations.

03 x or 06 x

6x3x

5) Check the solutions in the original equation.Check 3: Check -6:

? ?

1441 xx 1441 xx

144313 144616

1472 1427 ? ?



CONTINUECONTINUEDD

Check 3: Check -6:

true1414 true1414

The solutions are 3 and -6. The solution set is {3,-6}.

The graph of

lies to the right.

1832 xxy

-40

-20

0

20

40

60

80

100

120

-15 -10 -5 0 5 10 15



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

.xxx 022 23 Solve by factoring:

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.

23 2xx 2 x 0

Common factor is

Common factor is -1.2x



0222 xxx

CONTINUECONTINUEDD

012 2 xx

0112 xxx

Factor from the first two terms and -1 from the last two terms

2x

Factor out the common binomial, x – 2, from each term

Factor completely by factoring as the difference of two squares

12 x




CONTINUECONTINUEDD

02 x 01x 01xor or2x 1x 1x

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?

The graph of

lies to the right.

22 23 xxxy

-150

-100

-50

0

50

100

150

-6 -4 -2 0 2 4 6


Polynomial Equations in Application

EXAMPLEEXAMPLE

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 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.

8816 2 ttts

0

2

4

6

8

10

0 0.5 1 1.5 2

Time (seconds)

Hei

gh

t (f

eet)



SOLUTIONSOLUTION

When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph.

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,

CONTINUECONTINUEDD

8816 2 ttts

88168 2 tt

tt 8160 2

1280 tt

Original equation

Replace s (t) with 8


Factor



Now we set each factor equal to zero.

CONTINUECONTINUEDD

08 t

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.

012 tor0t 12 t

2

1t



CONTINUECONTINUEDD

0

2

4

6

8

10

0 0.5 1 1.5 2

Time (seconds)

Hei

gh

t (f

eet)

5.7 Assignment

p. 376 (2-30 even)

5.7 polynomial equations and their applications. blitzer, algebra for college students, 6e – slide...

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