1. College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

2. Prerequisites P

3. Factoring P.7

4. Factoring We use the Distributive Property to expand algebraic expressions. We sometimes need to reverse this process (again using the Distributive Property) by: Factoring an expression as a product of simpler ones.

5. Factoring For example, we can write: We say that x – 2 and x + 2 are factors of x 2 – 4.

6. Common Factors

7. Factoring The easiest type of factoring occurs when: The terms have a common factor.

8. E.g. 1—Factoring Out Common Factors Factor each expression. (a)3x 2 – 6x (b)8x 4 y 2 + 6x 3 y 3 – 2xy 4

9. E.g. 1—Common Factors The greatest common factor of the terms 3x 2 and –6x is 3x. So, we have: 3x 2 – 6x = 3x(x – 2) Example (a)

10. E.g. 1—Common Factors We note that: 8, 6, and –2 have the greatest common factor 2. x 4, x 3, and x have the greatest common factor x. y 2, y 3, and y 4 have the greatest common factor y 2. Example (b)

11. E.g. 1—Common Factors So, the greatest common factor of the three terms in the polynomial is 2xy 2. Thus, we have: 8x 4 y 2 + 6x 3 y 3 – 2xy 4 = (2xy 2 )(4x 3 ) + (2xy 2 )(3x 2 y) + (2xy 2 )(–y 2 ) = 2xy 2 (4x 3 + 3x 2 y – y 2 ) Example (b)

12. E.g. 2—Factoring Out a Common Factor Factor : (2x + 4)(x – 3) – 5(x – 3)

13. E.g. 2—Factoring Out a Common Factor The two terms have the common factor x – 3. (2x + 4)(x – 3) – 5(x – 3) = [(2x + 4) – 5](x – 3) (Distributive Property) = (2x – 1)(x – 3) (Simplify)

14. Factoring Trinomials

15. Factoring x 2 + bx + c In order to factor a trinomial of the form x 2 + bx + c, we note that: (x + r)(x + s) = x 2 + (r + s)x + rs So, we need to choose numbers r and s so that r + s = b and rs = c.

16. E.g. 3—Factoring x 2 + bx + c by Trial and Error Factor: x 2 + 7x + 12 We need to find two integers whose product is 12 and whose sum is 7. By trial and error, we find that they are 3 and 4. Thus, the factorization is: x 2 + 7x + 12 = (x + 3)(x + 4)

17. Factoring ax 2 + bx + c To factor a trinomial of the form ax 2 + bx + c with a ≠ 1, we look for factors of the form px + r and qx + s: ax 2 + bx + c = (px + r)(qx + s) = pqx 2 + (ps + qr)x + rs

18. Factoring ax 2 + bx + c Thus, we try to find numbers p, q, r, and s such that: pq = a rs = c ps + qr = b If these numbers are all integers, then we will have a limited number of possibilities to try for p, q, r, and s.

19. E.g. 4—Factoring ax 2 + bx + c by Trial and Error Factor: 6x 2 + 7x – 5 We can factor: 6 as 6 ∙ 1 or 3 ∙ 2 –5 as –25 ∙ 1 or –5 ∙ (–1) By trying these possibilities, we arrive at the factorization 6x 2 + 7x – 5 = (3x + 5)(2x – 1)

20. E.g. 5—Recognizing the Form of an Expression Factor each expression. (a) x 2 – 2x – 3 (b) (5a + 1) 2 – 2(5a + 1) – 3

21. E.g. 5—Recognizing the Form x 2 – 2x – 3 = (x – 3)(x + 1) (Trial and error) Example (a)

22. E.g. 5—Recognizing the Form This expression is of the form __ 2 – 2__ – 3 where __ represents 5a + 1. Example (b)

23. E.g. 5—Recognizing the Form This is the same form as the expression in part (a). So, it will factor as ( __ – 3)( __ + 1). (5a + 1) 2 – 2(5a + 1) – 3 = [(5a + 1) – 3][(5a + 1) + 1] = (5a – 2)(5a + 2) Example (b)

24. Special Factoring Formulas

25. Some special algebraic expressions can be factored using the following formulas. The first three are simply Special Product Formulas written backward.

26. E.g. 6—Factoring Differences of Squares Factor each polynomial. (a) 4x 2 – 25 (b) (x + y) 2 – z 2

27. E.g. 6—Differences of Squares Using the Difference of Squares Formula with A = 2x and B = 5, we have: 4x 2 – 25 = (2x) 2 – 5 2 = (2x – 5)(2x + 5) Example (a)

28. E.g. 6—Differences of Squares We use the Difference of Squares Formula with A = x + y and B = z. (x + y) 2 – z 2 = (x + y – z)(x + y + z) Example (b)

29. E.g. 7—Factoring Differences and Sums of Cubes Factor each polynomial. (a) 27x 3 – 1 (b) x 6 + 8

30. E.g. 7—Difference of Cubes Using the Difference of Cubes Formula with A = 3x and B = 1, we get: 27x 3 – 1 = (3x) 3 – 1 3 = (3x – 1)[ (3x) 2 + (3x)(1) + 1 2 ] = (3x – 1)(9x 2 + 3x + 1) Example (a)

31. E.g. 7—Sum of Cubes Using the Sum of Cubes Formula with A = x 2 and B = 2, we have: x 6 + 8 = (x 2 ) 3 + 2 3 = (x 2 + 2)(x 4 – 2x 2 + 4) Example (b)

32. Perfect Square A trinomial is a perfect square if it is of the form A 2 + 2AB + B 2 or A 2 – 2AB + B 2 So, we recognize a perfect square if the middle term (2AB or –2AB) is plus or minus twice the product of the square roots of the outer two terms.

33. E.g. 8—Recognizing Perfect Squares Factor each trinomial. (a) x 2 + 6x + 9 (b) 4x 2 – 4xy + y 2

34. E.g. 8—Perfect Squares Here, A = x and B = 3. So, 2AB = 2. x. 3 = 6x. Since the middle term is 6x, the trinomial is a perfect square. By the Perfect Square Formula, we have: x 2 + 6x + 9 = (x + 3) 2 Example (a)

35. E.g. 8—Perfect Squares Here, A = 2x and B = y. So, 2AB = 2. 2x. y = 4xy. Since the middle term is –4xy, the trinomial is a perfect square. By the Perfect Square Formula, we have: 4x 2 – 4xy + y 2 = (2x – y) 2 Example (b)

36. Factoring an Expression Completely

37. When we factor an expression, the result can sometimes be factored further. In general, We first factor out common factors. Then, we inspect the result to see if it can be factored by any of the other methods of this section. We repeat this process until we have factored the expression completely.

38. E.g. 9—Factoring an Expression Completely Factor each expression completely. (a) 2x 4 – 8x 2 (b) x 5 y 2 – xy 6

39. E.g. 9—Factoring Completely We first factor out the power of x with the smallest exponent. 2x 4 – 8x 2 = 2x 2 (x 2 – 4) (Common factor is 2x 2 ) = 2x 2 (x – 2)(x + 2) (Factor x 2 – 4 as a difference of squares) Example (a)

40. E.g. 9—Factoring Completely We first factor out the powers of x and y with the smallest exponents. x 5 y 2 – xy 6 = xy 2 (x 4 – y 4 ) (Common factor is xy 2 ) = xy 2 (x 2 + y 2 )(x 2 – y 2 ) (Factor x 4 – y 4 as a difference of squares) = xy 2 (x 2 + y 2 )(x + y)(x – y) (Factor x 2 – y 2 as a difference of squares) Example (b)

41. Factoring Expressions with Fractional Exponents In the next example, we factor out variables with fractional exponents. This type of factoring occurs in calculus.

42. E.g. 10—Factoring with Fractional Exponents Factor each expression. (a)3x 3/2 – 9x 1/2 + 6x –1/2 (b)(2 + x) –2/3 x + (2 + x) 1/3

43. E.g. 10—Fractional Exponents Factor out the power of x with the smallest exponent—that is, x –1/2. 3x 3/2 – 9x 1/2 + 6x –1/2 = 3x –1/2 (x 2 – 3x + 2) (Factor out 3x –1/2 ) = 3x –1/2 (x – 1)(x – 2) (Factor the quadratic x 2 – 3x + 2) Example (a)

44. E.g. 10—Fractional Exponents Factor out the power of 2 + x with the smallest exponent—that is, (2 + x) –2/3. (2 + x) –2/3 x + (2 + x) 1/3 = (2 + x) –2/3 [x + (2 + x)] (Factor out (2 + x) –2/3 ) = (2 + x) –2/3 (2 + 2x) (Simplify) = 2(2 + x) –2/3 (1 + x) (Factor out 2) Example (b)

45. Factoring by Grouping Terms

46. Factoring by Grouping Polynomials with at least four terms can sometimes be factored by grouping terms. The following example illustrates the idea.

47. E.g. 11—Factoring by Grouping Factor each polynomial. (a) x 3 + x 2 +4x + 4 (b) x 3 – 2x 2 – 3x + 6

48. E.g. 11—Factoring by Grouping x 3 + x 2 +4x + 4 = (x 3 + x 2 ) + (4x + 4) (Group terms) = x 2 (x + 1) + 4(x + 1) (Factor out common factors) = (x 2 + 4)(x + 1) (Factor out x + 1 from each term) Example (a)

49. E.g. 11—Factoring by Grouping x 3 – 2x 2 – 3x + 6 = (x 3 – 2x 2 ) – (3x – 6) (Group terms) = x 2 (x – 2) – 3(x – 2) (Factor out common factors) = (x 2 – 3)(x – 2) (Factor out x – 2 from each term) Example (b)