Tải bản đầy đủ - 0 (trang)
3: Factoring Trinomials of the Form x2 + bx + c

3: Factoring Trinomials of the Form x2 + bx + c

Tải bản đầy đủ - 0trang

6.3 • Factoring Trinomials of the Form x 2 ϩ bx ϩ c



243



The third line contains the numbers that we want. Thus

x2 Ϫ 11x ϩ 24 ϭ (x Ϫ 3) (x Ϫ 8)

Classroom Example

Factor x2 ϩ 2x Ϫ 24.



EXAMPLE 3



Factor x2 ϩ 3x Ϫ 10.



Solution

To factor x2 ϩ 3x Ϫ10, we want to find two numbers whose product is Ϫ10 and whose sum is 3.

Product



1(Ϫ10)

Ϫ1(10)

2(Ϫ5)

Ϫ 2(5)



ϭ Ϫ10

ϭ Ϫ 10

ϭ Ϫ 10

ϭ Ϫ 10



Sum



1 ϩ (Ϫ10) ϭ Ϫ9

Ϫ1 ϩ 10 ϭ 9

2 ϩ (Ϫ5) ϭ Ϫ 3

Ϫ2 ϩ 5 ϭ 3



The bottom line is the key line. Thus

x2 ϩ 3x Ϫ 10 ϭ (x ϩ 5)(x Ϫ 2)

Classroom Example

Factor n2 Ϫ 4n Ϫ 32.



EXAMPLE 4



Factor x2 Ϫ 2x Ϫ 8.



Solution

We are looking for two numbers whose product is Ϫ8 and whose sum is Ϫ2.

Product



1(Ϫ8)

Ϫ1(8)

2(Ϫ4)

Ϫ 2(4)



ϭ Ϫ8

ϭ Ϫ8

ϭ Ϫ8

ϭ Ϫ8



Sum



1 ϩ (Ϫ8) ϭ Ϫ7

Ϫ1 ϩ 8 ϭ 7

2 ϩ (Ϫ4) ϭ Ϫ 2

Ϫ2 ϩ 4 ϭ 2



The third line has the information we want.

x2 Ϫ 2x Ϫ 8 ϭ (x Ϫ 4)(x ϩ 2)

The tables in the last four examples illustrate one way of organizing your thoughts for such

problems. We show complete tables; that is, for Example 4, we include the bottom line even

though the desired numbers are obtained in the third line. If you use such tables, keep in mind

that as soon as you get the desired numbers, the table need not be completed any further.

Furthermore, you may be able to find the numbers without using a table. The key ideas are the

product and sum relationships.

Classroom Example

Factor x2 ϩ 13x Ϫ 14.



EXAMPLE 5



Factor x2 Ϫ 13x ϩ 12.



Solution

Product



(Ϫ1)(Ϫ12) ϭ 12



Sum



(Ϫ1) ϩ (Ϫ12) ϭ Ϫ13



We need not complete the table.

x2 Ϫ 13x ϩ 12 ϭ (x Ϫ 1)(x Ϫ 12)



244



Chapter 6 • Factoring, Solving Equations, and Problem Solving



In the next example, we refer to the concept of absolute value. Recall that the absolute

value of any nonzero real number is positive. For example,

0 40 ϭ 4  and  0Ϫ40 ϭ 4



Classroom Example

Factor x2 Ϫ 6x Ϫ 7.



EXAMPLE 6



Factor x2 Ϫ x Ϫ 56.



Solution

Notice that the coefficient of the middle term is Ϫ1. Therefore, we are looking for two

numbers whose product is Ϫ56; because their sum is Ϫ1, the absolute value of the negative

number must be one larger than the absolute value of the positive number. The numbers are

Ϫ8 and 7, and we have

x2 Ϫ x Ϫ 56 ϭ (x Ϫ 8)(x ϩ 7)



Classroom Example

Factor m2 ϩ m ϩ 3.



EXAMPLE 7



Factor x2 ϩ 10x ϩ 12.



Solution

Product



Sum



1(12) ϭ 12    1 ϩ 12 ϭ 13

2(6) ϭ 12

2ϩ6ϭ8

3(4) ϭ 12

3ϩ4ϭ7

Since the table is complete and no two factors of 12 produce a sum of 10, we conclude that

x2 ϩ 10x ϩ 12

is not factorable using integers.



In a problem such as Example 7, we need to be sure that we have tried all possibilities before

we conclude that the trinomial is not factorable.



Back to Solving Equations

The property ab ϭ 0 if and only if a ϭ 0 or b ϭ 0 continues to play an important role as

we solve equations that involve the factoring ideas of this section. Consider the following

examples.



Classroom Example

Solve x2 ϩ 15x ϩ 26 ϭ 0.



EXAMPLE 8



Solve x2 ϩ 8x ϩ 15 ϭ 0.



Solution

x2 ϩ 8x ϩ 15 ϭ 0

Factor the left side

(x ϩ 3)(x ϩ 5) ϭ 0

x ϩ 3 ϭ 0   or  x ϩ 5 ϭ 0

Use ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

x ϭ Ϫ3  or  

x ϭ Ϫ5

The solution set is {Ϫ5, Ϫ3}.



6.3 • Factoring Trinomials of the Form x 2 ϩ bx ϩ c



Classroom Example

Solve x2 Ϫ 8x Ϫ 9 ϭ 0.



EXAMPLE 9



245



Solve x2 ϩ 5x Ϫ 6 ϭ 0.



Solution

x2 ϩ 5x Ϫ 6 ϭ 0

(x ϩ 6)(x Ϫ 1) ϭ 0

x ϩ 6 ϭ 0   or  x Ϫ 1 ϭ 0

x ϭ Ϫ6  or  

xϭ1

The solution set is {Ϫ6, 1}.

Classroom Example

Solve m2 Ϫ 8m ϭ 33.



EXAMPLE 10



Solve y2 Ϫ 4y ϭ 45.



Solution

y2 Ϫ 4y ϭ 45

y2 Ϫ 4y Ϫ 45 ϭ 0

(y Ϫ 9)(y ϩ 5) ϭ 0

y Ϫ 9 ϭ 0  or  y ϩ 5 ϭ 0

y ϭ 9  or   

y ϭ Ϫ5

The solution set is {Ϫ5, 9}.

Don’t forget that we can always check to be absolutely sure of our solutions. Let’s check

the solutions for Example 10. If y ϭ 9, then y2 Ϫ 4y ϭ 45 becomes

92 Ϫ 4(9) ՘ 45

81 Ϫ 36 ՘ 45

45 ϭ 45

If y ϭ Ϫ5, then y2 Ϫ 4y ϭ 45 becomes

(Ϫ5) 2 Ϫ 4(Ϫ5) ՘ 45

25 ϩ 20 ՘ 45

45 ϭ 45



Back to Problem Solving

The more we know about factoring and solving equations, the more easily we can solve word

problems.

Classroom Example

Find two consecutive odd integers

whose product is 35.



EXAMPLE 11



Find two consecutive integers whose product is 72.



Solution

Let n represent one integer. Then n ϩ 1 represents the next integer.

n(n ϩ 1) ϭ 72 The product of the two integers is 72

n2 ϩ n ϭ 72

n2 ϩ n Ϫ 72 ϭ 0

(n ϩ 9)(n Ϫ 8) ϭ 0

n ϩ 9 ϭ 0   or  n Ϫ 8 ϭ 0

n ϭ Ϫ9  or   n ϭ 8

If n ϭ Ϫ9, then n ϩ 1 ϭ Ϫ9 ϩ 1 ϭ Ϫ8. If n ϭ 8, then n ϩ 1 ϭ 8 ϩ 1 ϭ 9. Thus the

consecutive integers are Ϫ9 and Ϫ8 or 8 and 9.



246



Chapter 6 • Factoring, Solving Equations, and Problem Solving



Classroom Example

A triangular lot has a height that is 8

yards longer than the base. The area

of the lot is 24 square yards. Find

the base and height of the lot.



EXAMPLE 12

A rectangular plot is 6 meters longer than it is wide. The area of the plot is 16 square meters.

Find the length and width of the plot.



w



w+6

Figure 6.3



Solution

We let w represent the width of the plot, and then w ϩ 6 represents the length (see Figure 6.3).

Using the area formula A ϭ lw, we obtain

w(w ϩ 6) ϭ 16

w2 ϩ 6w ϭ 16

w2 ϩ 6w Ϫ 16 ϭ 0

(w ϩ 8)(w Ϫ 2) ϭ 0

w ϩ 8 ϭ 0   or  w Ϫ 2 ϭ 0

w ϭ Ϫ8  or   

wϭ2

The solution of Ϫ8 is not possible for the width of a rectangle, so the plot is 2 meters wide

and its length (w ϩ 6) is 8 meters.



The Pythagorean theorem, an important theorem pertaining to right triangles, can also

serve as a guideline for solving certain types of problems. The Pythagorean theorem states

that in any right triangle, the square of the longest side (called the hypotenuse) is equal to

the sum of the squares of the other two sides (called legs); see Figure 6.4. We can use this

theorem to help solve problems.



c



b



a2 + b2 = c2



a

Figure 6.4



Classroom Example

Suppose that the lengths of the three

sides of a right triangle are consecutive even integers. Find the lengths

of the three sides.



EXAMPLE 13

Suppose that the lengths of the three sides of a right triangle are consecutive whole numbers.

Find the lengths of the three sides.



Solution

Let s represent the length of the shortest leg. Then s ϩ 1 represents the length of the other leg,

and s ϩ 2 represents the length of the hypotenuse. Using the Pythagorean theorem as a guideline, we obtain the following equation:



6.3 • Factoring Trinomials of the Form x 2 ϩ bx ϩ c



247



Sum of squares of two legs ϭ Square of hypotenuse

6447448

64748



s2 ϩ (s ϩ 1) 2



ϭ



(s ϩ 2) 2



Solving this equation yields

s2 ϩ s2 ϩ 2s ϩ 1 ϭ s2 ϩ 4s ϩ 4

2s2 ϩ 2s ϩ 1 ϭ s2 ϩ 4s ϩ 4

s2 ϩ 2s ϩ 1 ϭ 4s ϩ 4

s2 Ϫ 2s ϩ 1 ϭ 4

s2 Ϫ 2s Ϫ 3 ϭ 0

(s Ϫ 3)(s ϩ 1) ϭ 0

s Ϫ 3 ϭ 0  or  s ϩ 1 ϭ 0

s ϭ 3  or   

s ϭ Ϫ1



Remember (a ϩ b)2 ϭ a2 ϩ 2ab ϩ b2

Add Ϫs2 to both sides



The solution of Ϫ1 is not possible for the length of a side, so the shortest side is of length 3.

The other two sides (s ϩ 1 and s ϩ 2) have lengths of 4 and 5.



Concept Quiz 6.3

For Problems 1–10, answer true or false.

1. Any trinomial of the form x2 ϩ bx ϩ c can be factored (using integers) into the product

of two binomials.

2. To factor x2 Ϫ 4x Ϫ 60 we look for two numbers whose product is Ϫ60 and whose

sum is Ϫ4.

3. A trinomial of the form x2 ϩ bx ϩ c will never have a common factor other than 1.

4. If n represents an odd integer, then n ϩ 1 represents the next consecutive odd integer.

5. The Pythagorean theorem only applies to right triangles.

6. In a right triangle the longest side is called the hypotenuse.

7. The polynomial x2 ϩ 25x ϩ 72 is not factorable.

8. The polynomial x2 ϩ 27x ϩ 72 is not factorable.

9. The solution set of the equation x2 ϩ 2x Ϫ 63 ϭ 0 is {Ϫ9, 7}.

10. The solution set of the equation x2 Ϫ 5x Ϫ 66 ϭ 0 is {Ϫ11, Ϫ6}.



Problem Set 6.3

For Problems 1–30, factor each trinomial completely. Indicate any that are not factorable using integers. (Objective 1)



19. y2 Ϫ y Ϫ 72



20. y2 Ϫ y Ϫ 30



21. x2 ϩ 21x ϩ 80



22. x2 ϩ 21x ϩ 90



23. x2 ϩ 6x Ϫ 72



24. x2 Ϫ 8x Ϫ 36



25. x2 Ϫ 10x Ϫ 48



26. x2 Ϫ 12x Ϫ 64



27. x2 ϩ 3xy Ϫ 10y2



28. x2 Ϫ 4xy Ϫ 12y2



29. a2 Ϫ 4ab Ϫ 32b2



30. a2 ϩ 3ab Ϫ 54b2



1. x ϩ 10x ϩ 24



2. x ϩ 9x ϩ 14



3. x ϩ 13x ϩ 40



4. x ϩ 11x ϩ 24



5. x Ϫ 11x ϩ 18



6. x Ϫ 5x ϩ 4



7. n Ϫ 11n ϩ 28



8. n Ϫ 7n ϩ 10



9. n ϩ 6n Ϫ 27



10. n ϩ 3n Ϫ 18



11. n2 Ϫ 6n Ϫ 40



12. n2 Ϫ 4n Ϫ 45



For Problems 31– 50, solve each equation. (Objective 2)



13. t ϩ 12t ϩ 24



14. t ϩ 20t ϩ 96



31. x2 ϩ 10x ϩ 21 ϭ 0



32. x2 ϩ 9x ϩ 20 ϭ 0



15. x2 Ϫ 18x ϩ 72



16. x2 Ϫ 14x ϩ 32



33. x2 Ϫ 9x ϩ 18 ϭ 0



34. x2 Ϫ 9x ϩ 8 ϭ 0



17. x2 ϩ 5x Ϫ 66



18. x2 ϩ 11x Ϫ 42



35. x2 Ϫ 3x Ϫ 10 ϭ 0



36. x2 Ϫ x Ϫ 12 ϭ 0



2

2

2



2

2



2



2

2

2



2

2



2



248



Chapter 6 • Factoring, Solving Equations, and Problem Solving



37. n2 ϩ 5n Ϫ 36 ϭ 0

38. n2 ϩ 3n Ϫ 18 ϭ 0

39. n2 Ϫ 6n Ϫ 40 ϭ 0

40. n2 Ϫ 8n Ϫ 48 ϭ 0

41. t2 ϩ t Ϫ 56 ϭ 0

42. t2 ϩ t Ϫ 72 ϭ 0

43. x2 Ϫ 16x ϩ 28 ϭ 0

44. x2 Ϫ 18x ϩ 45 ϭ 0

45. x2 ϩ 11x ϭ 12

46. x2 ϩ 8x ϭ 20

47. x(x Ϫ 10) ϭ Ϫ16

48. x(x Ϫ 12) ϭ Ϫ35

49. Ϫx2 Ϫ 2x ϩ 24 ϭ 0

50. Ϫx2 ϩ 6x ϩ 16 ϭ 0

For Problems 51– 68, set up an equation and solve each

problem. (Objectives 3 and 4)

51. Find two consecutive integers whose product is 56.

52. Find two consecutive odd whole numbers whose product is 63.

53. Find two consecutive even whole numbers whose product is 168.

54. One number is 2 larger than another number. The sum

of their squares is 100. Find the numbers.

55. Find four consecutive integers such that the product of

the two larger integers is 22 less than twice the product

of the two smaller integers.

56. Find three consecutive integers such that the product of

the two smaller integers is 2 more than ten times the

largest integer.

57. One number is 3 smaller than another number. The square

of the larger number is 9 larger than ten times the smaller

number. Find the numbers.



59. Suppose that the width of a certain rectangle is

3 inches less than its length. The area is numerically

6 less than twice the perimeter. Find the length and

width of the rectangle.

60. The sum of the areas of a square and a rectangle is

64 square centimeters. The length of the rectangle is 4

centimeters more than a side of the square, and the

width of the rectangle is 2 centimeters more than a side

of the square. Find the dimensions of the square and the

rectangle.

61. The perimeter of a rectangle is 30 centimeters, and the

area is 54 square centimeters. Find the length and width

of the rectangle. [Hint: Let w represent the width; then

15 Ϫ w represents the length.]

62. The perimeter of a rectangle is 44 inches, and its area

is 120 square inches. Find the length and width of the

rectangle.

63. An apple orchard contains 84 trees. The number of trees

per row is five more than the number of rows. Find the

number of rows.

64. A room contains 54 chairs. The number of rows is

3 less than the number of chairs per row. Find the number of rows.

65. Suppose that one leg of a right triangle is 7 feet shorter

than the other leg. The hypotenuse is 2 feet longer than

the longer leg. Find the lengths of all three sides of the

right triangle.

66. Suppose that one leg of a right triangle is 7 meters longer

than the other leg. The hypotenuse is 1 meter longer than

the longer leg. Find the lengths of all three sides of the

right triangle.

67. Suppose that the length of one leg of a right triangle is

2 inches less than the length of the other leg. If the

length of the hypotenuse is 10 inches, find the length of

each leg.

68. The length of one leg of a right triangle is 3 centimeters more than the length of the other leg. The length of

the hypotenuse is 15 centimeters. Find the lengths of the

two legs.



58. The area of the floor of a rectangular room is 84 square

feet. The length of the room is 5 feet more than its

width. Find the length and width of the room.



Thoughts Into Words

69. What does the expression “not factorable using integers”

mean to you?

70. Discuss the role that factoring plays in solving

equations.



71. Explain how you would solve the equation

(x Ϫ 3)(x ϩ 4) ϭ 0

and also how you would solve

(x Ϫ 3)(x ϩ 4) ϭ 8.



6.4 • Factoring Trinomials of the Form ax 2 ϩ bx ϩ c



249



Further Investigations

For Problems 72 – 75, factor each trinomial and assume

that all variables appearing as exponents represent positive integers.

72. x2a ϩ 10xa ϩ 24



73. x2a ϩ 13xa ϩ 40



74. x2a Ϫ 2xa Ϫ 8



75. x2a ϩ 6xa Ϫ 27



76. Suppose that we want to factor n2 ϩ 26n ϩ 168 so that

we can solve the equation n2 ϩ 26n ϩ 168 ϭ 0. We

need to find two positive integers whose product is 168

and whose sum is 26. Since the constant term, 168, is

rather large, let’s look at it in prime factored form:

168 ϭ 2



n2 ϩ 26n ϩ 168 ϭ 0

(n ϩ 12)(n ϩ 14) ϭ 0

n ϩ 12 ϭ 0    or    n ϩ 14 ϭ 0

n ϭ Ϫ12   or   

n ϭ Ϫ14

The solution set is {Ϫ14, Ϫ12}.

Solve each of the following equations.

(a) n2 ϩ 30n ϩ 216 ϭ 0

(b) n2 ϩ 35n ϩ 294 ϭ 0



и2и2и3и7



Now we can mentally form two numbers by using all of

these factors in different combinations. Using two 2s and

the 3 in one number and the other 2 and the 7 in another

number produces 2 и 2 и 3 ϭ 12 and 2 и 7 ϭ 14.

Answers to the Concept Quiz

1. False

2. True

3. True

4. False

9. True

10. False



6.4



Therefore, we can solve the given equation as follows:



5. True



(c) n2 Ϫ 40n ϩ 384 ϭ 0

(d) n2 Ϫ 40n ϩ 375 ϭ 0

(e) n2 ϩ 6n Ϫ 432 ϭ 0

(f) n2 Ϫ 16n Ϫ 512 ϭ 0



6. True



7. True



8. False



Factoring Trinomials of the Form ax 2 ؉ bx ؉ c



OBJECTIVES



1



Factor trinomials where the leading coefficient is not 1



2



Solve equations that involve factoring



Now let’s consider factoring trinomials where the coefficient of the squared term is not 1. We

present here an informal trial and error technique that works quite well for certain types of

trinomials. This technique simply relies on our knowledge of multiplication of binomials.

Classroom Example

Factor 3x2 ϩ 10x ϩ 8.



EXAMPLE 1



Factor 2x2 ϩ 7x ϩ 3.



Solution

By looking at the first term, 2x2, and the positive signs of the other two terms, we know that

the binomials are of the form

(2x ϩ _____)(x ϩ _____)

Since the factors of the constant term, 3, are 1 and 3, we have only two possibilities to try:

(2x ϩ 3)(x ϩ 1)



or



(2x ϩ 1)(x ϩ 3)



By checking the middle term of both of these products, we find that the second one yields the

correct middle term of 7x. Therefore,

2x2 ϩ 7x ϩ 3 ϭ (2x ϩ 1)(x ϩ 3)



250



Chapter 6 • Factoring, Solving Equations, and Problem Solving



Classroom Example

Factor 15y2 Ϫ 13y ϩ 2.



EXAMPLE 2



Factor 6x2 Ϫ 17x ϩ 5.



Solution

First, we note that 6x2 can be written as 2x и 3x or 6x и x. Second, since the middle term of the

trinomial is negative, and the last term is positive, we know that the binomials are of the form

(2x Ϫ _____)(3x Ϫ _____)



or



(6x Ϫ _____)(x Ϫ _____)



Since the factors of the constant term, 5, are 1 and 5, we have the following possibilities:

(2x Ϫ 5)(3x Ϫ 1)      (2x Ϫ 1)(3x Ϫ 5)

(6x Ϫ 5)(x Ϫ 1)

(6x Ϫ 1)(x Ϫ 5)

By checking the middle term for each of these products, we find that the product (2x Ϫ 5)(3x Ϫ 1)

produces the desired term of Ϫ17x. Therefore,

6x2 Ϫ 17x ϩ 5 ϭ (2x Ϫ 5)(3x Ϫ 1)



Classroom Example

Factor 18n2 ϩ 12n Ϫ 16.



EXAMPLE 3



Factor 8x2 Ϫ 8x Ϫ 30.



Solution

First, we note that the polynomial 8x 2 Ϫ 8x Ϫ 30 has a common factor of 2. Factoring out the

common factor gives us 2(4x 2 Ϫ 4x Ϫ 15). Now we need to factor 4x 2 Ϫ 4x Ϫ 15.

We note that 4x2 can be written as 4x и x or 2x и 2x. The last term, Ϫ15, can be written

as (1)(Ϫ15), (Ϫ1)(15), (3)(Ϫ5), or (Ϫ3)(5). Thus we can generate the possibilities for the

binomial factors as follows:







Using 1 and ؊15



Using ؊1 and 15



(4x Ϫ 15)(x ϩ 1)

(4x ϩ 1)(x Ϫ 15)

(2x ϩ 1)(2x Ϫ 15)



(4x Ϫ 1)(x ϩ 15)

(4x ϩ 15)(x Ϫ 1)

(2x Ϫ 1)(2x ϩ 15)



Using 3 and ؊5



Using ؊3 and 5



(4x ϩ 3)(x Ϫ 5)

(4x Ϫ 5)(x ϩ 3)

(2x Ϫ 5)(2x ϩ 3)



(4x Ϫ 3)(x ϩ 5)

(4x ϩ 5)(x Ϫ 3)

(2x ϩ 5)(2x Ϫ 3)



By checking the middle term of each of these products, we find that the product indicated with

a check mark produces the desired middle term of Ϫ4x. Therefore,

8x2 Ϫ 8x Ϫ 30 ϭ 2(4x2 Ϫ 4x Ϫ 15) ϭ 2(2x Ϫ 5)(2x ϩ 3)

Let’s pause for a moment and look back over Examples 1, 2, and 3. Obviously, Example 3

created the most difficulty because we had to consider so many possibilities. We have suggested one possible format for considering the possibilities, but as you practice such problems, you

may develop a format that works better for you. Regardless of the format that you use, the key

idea is to organize your work so that you consider all possibilities. Let’s look at another example.

Classroom Example

Factor 9x2 ϩ 10x ϩ 4.



EXAMPLE 4



Factor 4x2 ϩ 6x ϩ 9.



Solution

First, we note that 4x2 can be written as 4x и x or 2x и 2x. Second, since the middle term is

positive and the last term is positive, we know that the binomials are of the form

(4x ϩ _____)(x ϩ _____)



or



(2x ϩ _____)(2x ϩ _____)



Tài liệu bạn tìm kiếm đã sẵn sàng tải về

3: Factoring Trinomials of the Form x2 + bx + c

Tải bản đầy đủ ngay(0 tr)

×