Domains of Rational Expressions

In this section, we will factor the type of trinomials that result from multiplying two different binomials.We will do this only for trinomials in which the coefficient of x2, the leading coefficient, is 1. Factoring trinomials with a leading coefficient not equal to 1 will be done in Section 5.4.

To find the product of the binomials x + m and x + n, where x is the variable and m and n are constants, we use the distributive property as follows:

Notice that in the trinomial the coefficient of x is the sum m + n and the constant term is the product mn. This observation is the key to factoring the trinomial ax2 + bx + c with a = 1. We first find two numbers that have a product of c (the constant term) and a sum of b (the coefficient of x). Then reverse the steps that we used in finding the product (x + m)(x + n). We summarize these ideas with the following strategy.

To factor x2 + bx + c:

1. Find two integers that have a product of c and a sum equal to b.

2. Replace bx by the sum of two terms whose coefficients are the two numbers found in (1).

3. Factor the resulting four-term polynomial by grouping.

Factoring ax2+bx+c with a=1

Factor.

a)	To factor x2 + 5x + 6, we need two integers that have a product of 6 and a sum of 5. If the product is positive and the sum is positive, then both integers must be positive. We can list all of the possibilities: Product Sum 6 = 1 ⋅ 6 1 + 6 = 7 6 = 2 ⋅ 3 2 + 3 = 5   The only integers that have a product of 6 and a sum of 5 are 2 and 3. Now replace 5x with 2x + 3x and factor by grouping: Check by FOIL: (x + 3)(x + 2) = x2 + 5x + 6.
b)	To factor x2 + 8x + 12, we need two integers that have a product of 12 and a sum of 8. Since the product and sum are both positive, both integers are positive. Product Sum 12 = 1 ⋅ 12 1 + 12 = 13 12 = 2 ⋅ 6 2 + 6 = 8 12 = 3 ⋅ 4 3 + 4 = 7 The only integers that have a product of 12 and a sum of 8 are 2 and 6. Now replace 8x by 2x + 6x and factor by grouping: Check by FOIL: (x + 6)(x + 2) = x2 + 8x + 12.
c)	To factor a2 − 9a + 20, we need two integers that have a product of 20 and a sum of −9. Since the product is positive and the sum is negative, both integers must be negative. Product Sum 20 = (−1)(−20) −1 + (−20) = −21 20 = (−2)(−10) −2 + (−10) = −12 20 = (−4)(−5) −4 + (−5) = −9 Only −4 and −5 have a product of 20 and a sum of −9. Now replace −9a by −4a + (−5a) or −4a − 5a and factor by grouping: Check by FOIL: (a − 5)(a − 4) = a2 − 9a + 20.

Now do Exercises 1–14

We usually do not write out all of the steps shown in Example 1. We saw prior to Example 1 that

So once you know m and n, you can simply write the factors, as shown in Example 2.

Factor.

a)	To factor x2 + 5x + 4 we need two integers with a product of 4 and a sum of 5. The only possibilities for a product of 4 are Only 1 and 4 have a sum of 5. So, Check by using FOIL on (x + 1)(x + 4) to get x2 + 5x + 4.
b)	To factor y2 + 6y − 16 we need two integers with a product of −16 and a sum of 6. The only possibilities for a product of −16 are Only −2 and 8 have a sum of 6. So, Check by using FOIL on (y + 8)(y−2) to get y2 + 6y − 16.
c)	To factor w2 − 5w − 24 we need two integers with a product of −24 and a sum of −5. The only possibilities for a product of −24 are Only −8 and 3 have a sum of −5. So, Check by using FOIL on (w−8)(w + 3) to get w2 − 5w − 24.

Now do Exercises 15–22

Polynomials are easiest to factor when they are in the form ax2 + bx + c. So if a polynomial can be rewritten into that form, rewrite it before attempting to factor it. In Example 3, we factor polynomials that need to be rewritten.

Factor.

a)	Before factoring, write the trinomial as x2 + 2x − 8. Now, to get a product of −8 and a sum of 2, use −2 and 4:
b)	Before factoring, write the trinomial as t2 − 9t − 36. Now, to get a product of −36 and a sum of −9, use −12 and 3:

Now do Exercises 23–24

To factor x2 + bx + c, we search through all pairs of integers that have a product of c until we find a pair that has a sum of b. If there is no such pair of integers, then the polynomial cannot be factored and it is a prime polynomial. Before you can conclude that a polynomial is prime, be sure that you have tried all possibilities.

Factor.

Now do Exercises 25–52

Don't confuse a2 + b2 with the difference of two squares a2 − b2 which is not a prime polynomial:

The prime polynomial x2 + 9 in Example 4(b) is a sum of two squares. There are many other sums of squares that are prime. For example,

are prime. However, not every sum of two squares is prime. For example, 4x2 + 16 is a sum of two squares that is not prime because 4x2 + 16 = 4(x2 + 4).

The sum of two squares a2 + b2 is prime, but not every sum of two squares is prime.

In Example 5, we factor polynomials that have two variables using the same technique that we used for one variable.

Factoring with Two Variables

Factor.

a)	To factor x2 + 2xy − 8y2 we need two integers with a product of −8 and a sum of 2. The only possibilities for a product of −8 are Only −2 and 4 have a sum of 2. Since (−2y)(4y) = −8y2, we have Check by using FOIL on (x − 2y)(x + 4y) to get x2 + 2xy − 8y2.
b)	To factor a2 − 7ab + 10b2 we need two integers with a product of 10 and a sum of −7. The only possibilities for a product of 10 are Only −2 and −5 have a sum of −7. Since (−2b)(−5b) = 10b2, we have Check by using FOIL on (a − 2b)(a − 5b) to get a2 − 7ab + 10b2.
c)	As in part (a), we need two integers with a product of −8 and a sum of −2. The integers are −4 and 2. Since 1 factors as 1 ⋅ 1 and −8x2y2 = (−4xy)(2xy), we have Check by using FOIL.

Now do Exercises 53–64

In Section 5.2 you learned that binomials such as 3x − 5 (with no common factor) are prime polynomials. In Example 4 of this section we saw a trinomial that is a prime polynomial. There are infinitely many prime trinomials. When factoring a polynomial completely, we could have a factor that is a prime trinomial.

Factoring Completely

Factor each polynomial completely.

a)
b)	First factor out 4x, the greatest common factor: To factor x2 + x + 1, we would need two integers with a product of 1 and a sum of 1. Because there are no such integers, x2 + x + 1 is prime, and the factorization is complete.

Now do Exercises 65–106

1. If there are no two integers that have a of c and a of b, then x2 + bx + c is prime.

2. We can check all factoring by the factors.

3. The sum of two squares a2 + b2 is .

4. Always factor out the first.

1. x2 − 6x + 9 = (x − 3)2

2. x2 + 6x + 9 = (x + 3)2

3. x2 + 10x + 9 = (x − 9)(x − 1)

4. x2 + 8x − 9 = (x − 1)(x + 9)

5. x2 − 10xy + 9y2 = (x − y)(x − 9y)

6. x2 + 1 = (x + 1)(x + 1)

7. x2 + x + 1 = (x + 1)(x + 1)

• Put important facts on note cards.Work on memorizing the note cards when you have a few spare minutes.

• Post some note cards on your refrigerator door. Make this course a part of your life.

Factor each trinomial. Write out all of the steps as shown in Example 1. See the Strategy for Factoring x2 + bx + c by Grouping on page 339.

Factor each polynomial. If the polynomial is prime, say so. See Examples 2–4.

1. y2 + 7y + 10

2. x2 + 8x + 15

3. a2−6a + 8

4. b2−8b + 15

   Exercise 19 - Factoring ax2+bx+c with a=1: Factors Out

5. m2−10m + 16

6. m2−17m + 16

7. w2 + 9w−10

8. m2 + 6m−16

9. w2−8 −2w

10. −16 + m2−6m

11. a2−2a−12

    Exercise 26 - Factoring ax2+bx+c with a=1: Prime

12. x2 + 3x + 3

13. 15m−16 + m2

14. 3y + y2 − 10

15. a2−4a + 12

16. y2−6y−8

17. z2−25

18. p2−1

19. h2 + 49

20. q2 + 4

21. m2 + 12m + 20

22. m2 + 21m + 20

23. t2−3t + 10

24. x2−5x−3

    Exercise 39 - Factoring ax2+bx+c with a=1: Rearrange then factor

25. m2−18−17m

26. h2−36 + 5h

27. m2−23m + 24

28. m2 + 23m + 24

29. 5t−24 + t2

30. t2−24−10t

31. t2−2t−24

32. t2 + 14t + 24

33. t2−10t−200

34. t2 + 30t + 200

35. x2−5x−150

36. x2−25x + 150

37. 13y + 30 + y2

38. 18z + 45 + z2

Factor each polynomial. See Example 5.

1. x2 + 5ax + 6a2

2. a2 + 7ab + 10b2

3. x2−4xy−12y2

4. y2 + yt−12t2

5. x2−13xy + 12y2

6. h2−9hs + 9s2

7. x2 + 4xz−33z2

8. x2−5xs−24s2

9. 1 + 3ab − 28a2b2

10. 1 − xy − 20x2y2

11. 15a2b2 + 8ab + 1

12. 12m2n2 − 8mn + 1

Factor each polynomial completely. Use the methods discussed in Sections 5.1 through 5.3. If the polynomial is prime say so. See Example 6.

1. 5x3 + 5x

2. b3 + 49b

3. w2−8w

4. x4−x3

5. 2w2−162

6. 6w4−54w2

7. −2b2−98

8. −a3−100a

9. x3−2x2−9x + 18

10. x3 + 7x2−x−7

11. 4r2 + 9

12. t2 + 4z2

13. x2w2 + 9x2

14. a4b + a2b3

15. w2−18w + 81

16. w2 + 30w + 81

17. 6w2−12w−18

18. 9w−w3

• 3y2 + 75

• 5x2 + 500

• ax + ay + cx + cy

• y3 + y2−4y−4

• −2x2−10x−12

• −a3−2a2−a

  Exercise 89 - Factoring Completely

• 32x2−2x4

• 20w2 + 100w + 40

• 3w2 + 27w + 54

• w3−3w2−18w

• 18w2 + w3 + 36w

• 18a2 + 3a3 + 36a

• 9y2 + 1 + 6y

• 2a2 + 1 + 3a

• 8vw2 + 32vw + 32v

• Exercise 98 - Factoring Completely: GCF has a Variable and Further Factors to Perfect Square Trinomial

• 6x3y + 30x2y2 + 36xy3

• 3x3y2−3x2y2 + 3xy2

• 5 + 8w + 3w2

• −3 + 2y + 21y2

• −3y3 + 6y2−3y

• −4w3−16w2 + 20w

• a3 + ab + 3b + 3a2

• ac + xc + aw2 + xw2

Use factoring to solve each problem.

1. Area of a deck. The area in square feet for a rectangular deck is given by A(x) = x2 + 6x + 8.

   a)	Find A(6).
   b)	If the width of the deck is x + 2 feet, then what is the length?

   Figure for Exercise 107

2. Area of a sail. The area in square meters for a triangular sail is given by A(x) = x2 + 5x + 6.

   a)	Find A(5).
   b)	If the height of the sail is x + 3 meters, then what is the length of the base of the sail?

   Figure for Exercise 108

3. Volume of a cube. Hector designed a cubic box with volume x3 cubic feet. After increasing the dimensions of the bottom, the box has a volume of x3 + 8x2 + 15x cubic feet. If each of the dimensions of the bottom was increased by a whole number of feet, then how much was each increase?

4. Volume of a container. A cubic shipping container had a volume of a3 cubic meters. The height was decreased by a whole number of meters and the width was increased by a whole number of meters so that the volume of the container is now a3 + 2a2 − 3a cubic meters. By how many meters were the height and width changed?

1. Discussion

   Which of the following products is not equivalent to the others? Explain your answer.

   a)	(2x − 4)(x + 3)
   b)	(x − 2)(2x + 6)
   c)	2(x − 2)(x + 3)
   d)	(2x − 4)(2x + 6)

2. Discussion

   When asked to factor completely a certain polyno mial, four students gave the following answers. Only one student gave the correct answer. Which one must it be? Explain your answer.

   a)	3(x2 − 2x − 15)
   b)	(3x − 5)(5x − 15)
   c)	3(x − 5)(x − 3)
   d)	(3x − 15)(x − 3)

Find the prime factorization of each integer.

1. 48

2. 140

Find the greatest common factor for each group of integers.

1. 36, 45

2. 60, 144, 240

Factor each expression by factoring out the greatest common factor.

1. 8w − 6y

2. 12x3 − 30x2

3. 15ab3 − 25a2b2 + 35a3b

Factor each expression.

1. (x + 3)x − (x + 3)5

2. m(m − 9) − 6(m − 9)

Factor completely.

1. 4y2 − 9w2

2. 4h2 + 12h + 9

3. w2 − 16w + 64

4. 10x3 − 250x

5. −6x2 − 36x − 54

6. aw − 3w + 6a − 18

7. bx − 5b − 6x + 30

8. ax2 − a + x2 − 1

9. x3 − 5x − 4x2

10. 2x3 + 18x

11. a2 − 12as + 32s2