1. Copyright © Cengage Learning. All rights reserved.7
Transitioning to
Intermediate Algebra
Copyright © Cengage Learning. All rights reserved.

2. Review of Factoring and Solving Quadratic Equations 7.4
Section
7.4
Copyright © Cengage Learning. All rights reserved.

3. Objectives
Factor a polynomial by factoring out the greatest common factor (GCF).
Factor a polynomial with four terms or six terms by grouping.
Factor a difference of two squares.
Factor a trinomial by trial and error and by grouping (ac method). 1 2 3 4

4. Objectives 5 6 Factor a sum and difference of two cubes.
Solve a quadratic equation by factoring. 5 6

5. Factor a polynomial by factoring out the greatest common factor (GCF)1.

6. Example 1
Factor out the greatest common factor (GCF): 3xy2z3 + 6xz2 – 9xyz4. Solution: We begin by factoring each monomial: 3xy2z3 = 3  x  y  y  z  z  z 6xz2 = 3  2  x  z  z –9xyz4 = –3  3  x  y  z  z  z  z Since each term has one factor of 3, one factor of x, and two factors of z and there are no other common factors, 3xz2 is the GCF of the three terms.

7. Example 1 – Solution
cont’d
We can use the distributive property to factor it out. 3xy2z3 + 6xz2 – 9xyz4 = 3xz2  y2z + 3xz2  2 + 3xz2 (–3yz2) = 3xz2(y2z + 2 – 3yz2)

8. Factor a polynomial with four terms or six terms by grouping2.

9. Factor a polynomial with four terms or six terms by grouping
Although there is no factor common to all four terms of ac + ad + bc + bd, there is a factor of a in the first two terms and a factor of b in the last two terms. We can factor out these common factors. ac + ad + bc + bd = a(c + d ) + b(c + d ) We can now factor out c + d on the right side. ac + ad + bc + bd = (c + d )(a + b) The grouping in this type of problem is not always unique.

10. Factor a polynomial with four terms or six terms by grouping
For example, if we write the expression ac + ad + bc + bd in the form ac + bc + ad + bd and factor c from the first two terms and d from the last two terms, we obtain the result with the factors in reverse order. ac + bc + ad + bd = c(a + b) + d(a + b) = (a + b)(c + d )

11. Example 4
Factor: 3ax2 + 3bx2 + a + 5bx + 5ax + b Solution: Although there is no factor common to all six terms, 3x2 can be factored out of the first two terms, and 5x can be factored out of the fourth and fifth terms to obtain 3ax2 + 3bx2 + a + 5bx + 5ax + b = 3x2(a + b) + a + 5x(b + a) + b

12. Example 4 – Solution
cont’d
This result can be written in the form 3ax2 + 3bx2 + a + 5bx + 5ax + b = 3x2(a + b) + 5x(a + b) + 1(a + b) Since a + b is common to all three terms, it can be factored out. 3ax2 + 3bx2 + a + 5bx + 5ax + b = (a + b)(3x2 + 5x + 1)

13. Factor a difference of two squares3.

14. Factor a difference of two squares
We know the formula for factoring the difference of two squares. Factoring the Difference of Two Squares x2 – y2 = (x + y)(x – y) If we think of the difference of two squares as the square of a First quantity minus the square of a Last quantity, we have the formula F2 – L2 = (F + L)(F – L)

15. Factor a difference of two squares
In words, we say, to factor the square of a First quantity minus the square of a Last quantity, we multiply the First plus the Last by the First minus the Last. To factor 49x2 – 16, for example, we write 49x2 – 16 in the form (7x)2 – (4)2 and use the formula for factoring the difference of two squares: 49x2 – 16 = (7x)2 – (4)2 = (7x + 4)(7x – 4) We can verify this result by multiplying 7x + 4 and 7x – 4 and observing that the result is 49x2 – 16.

16. Example 6
Factor: (x + y)4 – z4 Solution: This expression is the difference of two squares that can be factored: (x + y)4 – z4 = [(x + y)2]2 – (z2)2 = [(x + y)2 + z2][(x + y)2 – z2] The factor (x + y)2 + z2 is the sum of two squares and is prime.

17. Example 6 – Solution
cont’d
The factor (x + y)2 – z2 is the difference of two squares and can be factored as (x + y + z)(x + y – z). Thus, (x + y)4 – z4 = [(x + y)2 + z2][(x + y)2 – z2] = [(x + y)2 + z2](x + y + z)(x + y – z)

18. Factor a trinomial by trial and error and by grouping (ac method)4.

19. Factor a trinomial by trial and error and by grouping (ac method)
To factor a trinomial of the form x2 + bx + c by trial and error, we follow these steps: Factoring Trinomials 1. Write the trinomial in descending powers of one variable. 2. Factor out any GCF, including –1. 3. List the factorizations of the third term of the trinomial. 4. Select the factorization in which the sum of the factors is the coefficient of the middle term.

20. Example 8
Factor: a. x2 – 6x + 8 b. 30x – 4xy – 2xy2 Solution: a. Since this trinomial is already written in descending powers of x and there are no common factors, we can move to Step 3 and list the possible factorizations of the third term, which is 8. 8(1) 4(2) –8(–1) –4(–2) In this trinomial, the coefficient of the middle term is –6, and the only factorization where the sum of the factors is –6 is –4(–2).
The one to choose

21. Example 8 – Solution
cont’d
Thus, x2 – 6x + 8 = (x – 4)(x – 2) Because of the commutative property of multiplication, the order of the factors is not important. We can verify this result by multiplication. b. We begin by writing the trinomial in descending powers of y: –2xy2 – 4xy + 30x = – 2xy2 – 4xy + 30x

22. Example 8 – Solution
cont’d
Since each term in the trinomial has a common factor of –2x, it can be factored out as a GCF. – 2xy2 – 4xy + 30x = –2x(y2 + 2y – 15) To factor y2 + 2y – 15, we list the factors of –15 and find the pair whose sum is 2. 15(–1) 5(–3) 1(–15) 3(–5) The only factorization where the sum of the factors is 2 (the coefficient of the middle term of y2 + 2y – 15) is 5(–3).
The one to choose

23. Example 8 – Solution
cont’d
Thus, 30x – 4xy – 2xy2 = –2x(y2 + 2y – 15) = –2x(y + 5)(y – 3) Verify this result by multiplication.

24. Factor a trinomial by trial and error and by grouping (ac method)
There are usually more combinations of factors to consider when factoring trinomials with leading coefficients other than 1. For example, to factor 5x2 + 7x + 2, we must find two binomials of the form ax + b and cx + d such that 5x2 + 7x + 2 = (ax + b)(cx + d) Since the first term of the trinomial 5x2 + 7x + 2 is 5x2, the first terms of the binomial factors must be 5x and x. 5x2 + 7x + 2 = (5x + b)(x + d)
5x2

25. Factor a trinomial by trial and error and by grouping (ac method)
The product of the last terms must be +2, and the sum of the products of the outer and inner terms must be +7x. We must find two numbers whose product is +2 that will give a middle term of +7x. However, the factors of 2 must be multiplied by the factors of 5x2 to produce this middle term of +7x. 5x2 + 7x + 2 = (5x + b)(x + d) 2
O + I = 7x

26. Factor a trinomial by trial and error and by grouping (ac method)
Since 2(1) and (–2)(–1) both give a product of 2, there are four combinations to consider:
(5x + 2)(x + 1)
(5x – 2)(x – 1)
(5x + 1)(x + 2)
(5x – 1)(x – 2)
Of these combinations, only the first gives the correct middle term of 7x.
5x2 + 7x + 2 = (5x + 2)(x + 1)
We can verify this result by multiplication.

27. Factor a trinomial by trial and error and by grouping (ac method)
If a trinomial has the form ax2 + bx + c, with integer coefficients and a  0, we can test to see if it is factorable. If the value of b2 – 4ac is a perfect square, the trinomial can be factored using only integers. If the value is not a perfect square, the trinomial is prime and cannot be factored using only integers. For example, 5x2 + 7x + 2 is a trinomial in the form ax2 + bx + c with a = 5, b = 7, and c = 2

28. Factor a trinomial by trial and error and by grouping (ac method)
For this trinomial, the value of b2 – 4ac is b2 – 4ac = 72 – 4(5)(2) = 49 – 40 = 9 Since 9 is a perfect square, the trinomial is factorable. Its factorization is shown in Equation 1. Test for Factorability A trinomial of the form ax2 + bx + c, with integer coefficients and a  0, will factor into two binomials with integer coefficients if the value of b2 – 4ac is a perfect square. If b2 – 4ac = 0, the factors will be the same. If b2 – 4ac is not a perfect square, the trinomial is prime.

29. Factor a trinomial by trial and error and by grouping (ac method)
We know the following steps for factoring trinomials by trial and error. Factoring a General Trinomial 1. Write the trinomial in descending powers of one variable. 2. Factor out any GCF (including –1 if that is necessary to make the coefficient of the first term positive). 3. Test the trinomial for factorability.

30. Factor a trinomial by trial and error and by grouping (ac method)
4. Factor the remaining trinomial. a. If the sign of the third term is +, the signs between the terms of the binomial factors are the same as the sign of the middle term. If the sign of the third term is –, the signs between the terms of the binomial factors are opposite. b. Try combinations of the factors of the first term and last term until you find one where the sum of the product of the outer and inner terms is the middle term of the trinomial. If no combination works, the trinomial is prime. 5. Check the factorization by multiplication, including any GCF found in Step 2.

31. Factor a trinomial by trial and error and by grouping (ac method)
Factoring by grouping, also called the ac method, can be used to factor trinomials of the form ax2 + bx + c. For example, to factor the trinomial 6x2 + 7x – 3 where a = 6, b = 7, and c = –3, we proceed as follows: 1. Find the product ac: 6(–3) = –18. This number is called the key number. 2. Find the two factors of the key number –18 whose sum is b = 7. These are 9 and –2. 9(–2) = –18 and 9 + (–2) = 7

32. Factor a trinomial by trial and error and by grouping (ac method)
3. Use the factors 9 and –2 as coefficients of terms, the sum of which is +7x. We replace +7x with 9x and –2x: 6x2 + 7x – 3 = 6x2 + 9x – 2x – 3 4. Factor by grouping: 6x2 + 9x – 2x – 3 = 3x(2x + 3) – (2x + 3) = (2x + 3)(3x – 1) We can verify this factorization by multiplication.
Factor out 2x + 3.

33. Factor a sum and difference of two cubes5.

34. Factor a sum and difference of two cubes
We know the following formula for factoring the sum of two cubes. Factoring the Sum of Two Cubes x3 + y3 = (x + y)(x2 – xy + y2) If we think of the sum of two cubes as the cube of a First quantity plus the cube of a Last quantity, we have the formula F3 + L3 = (F + L)(F2 – FL + L2)

35. Factor a sum and difference of two cubes
In words, we say, to factor the cube of a First quantity plus the cube of a Last quantity, we multiply the First plus the Last by • the First squared • minus the First times the Last • plus the Last squared.

36. Example 13
Factor: a. x3 + 8 b. 8b3 + 27c3 Solution: a. The binomial x3 + 8 is the sum of two cubes, because x3 + 8 = x Thus, x3 + 8 factors as (x + 2) times the trinomial x2 – 2x F3 + L3 = (F + L)(F2 – F L + L2) x = (x + 2)(x2 – x  ) = (x + 2)(x2 – 2x + 4)

37. Example 13 – Solution
cont’d
b. The binomial 8b3 + 27c3 is the sum of two cubes, because 8b3 + 27c3 = (2b)3 + (3c)3 Thus, 8b3 + 27c3 factors as (2b + 3c) times the trinomial (2b)2 – (2b)(3c) + (3c)2. F3 + L3 = (F + L) (F2 – F L + L2) (2b)3 + (3c)3 = (2b + 3c)[(2b)2 – (2b)(3c) + (3c)2] = (2b + 3c)(4b2 – 6bc + 9c2)

38. Factor a sum and difference of two cubes
We know the following formula for factoring the difference of two cubes. Factoring the Difference of Two Cubes x3 – y3 = (x – y)(x2 + xy + y2) If we think of the difference of two cubes as the cube of a First quantity minus the cube of a Last quantity, we have the formula F3 – L3 = (F – L)(F2 + FL + L2)

39. Factor a sum and difference of two cubes
In words, we say, to factor the cube of a First quantity minus the cube of a Last quantity, we multiply the First minus the Last by • the First squared • plus the First times the Last • plus the Last squared.

40. Solve a quadratic equation by factoring6.

41. Solve a quadratic equation by factoring
We know that an equation such as 3x2 + 4x – 7 = 0 or –5y2 + 3y + 8 = 0 is called a quadratic (or second-degree) equation. Quadratic Equations A quadratic equation is any equation that can be written in the form ax2 + bx + c = 0 where a, b, and c are real numbers and a  0.

42. Solve a quadratic equation by factoring
Many quadratic equations can be solved by factoring and then applying the zero-factor property. Zero-Factor Property Assume a and b are real numbers. If ab = 0, then a = 0 or b = 0. The zero-factor property and its extensions state that if the product of two or more factors is 0, then at least one of the factors must be 0.

43. Solve a quadratic equation by factoring
Many equations that do not appear to be quadratic can be put into quadratic form (ax2 + bx + c = 0) and then solved by factoring.

44. Example 16
Solve: Solution: We write the equation in quadratic form and then solve by factoring. 5x = 6 – 6x2 6x2 + 5x – 6 = 0
Multiply both sides by 5 to clear fractions.
Add 6x2 to both sides and subtract 6
from both sides.

45. Example 16 – Solution
cont’d
(3x – 2) (2x + 3) = 0 3x – 2 = 0 or 2x + 3 = 0 3x = 2 2x = –3 Verify that both solutions check.
Factor the trinomial.
Set each factor equal to 0.

46. Solve a quadratic equation by factoring
Comment To solve a quadratic equation by factoring, be sure to set the equation equal to 0 before applying the zero-factor property.