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College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson. Prerequisites. P. Factoring. P.7. Factoring. We use the Distributive Property to expand algebraic expressions. We sometimes need to reverse this process (again using the Distributive Property) by:

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College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

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  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: • Factoringan expression as a product of simpler ones.

  5. Factoring • For example, we can write: • We say that x – 2 and x + 2 are factorsof x2 – 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. • 3x2 – 6x • 8x4y2 + 6x3y3 – 2xy4

  9. Example (a) E.g. 1—Common Factors • The greatest common factor of the terms 3x2 and –6x is 3x. • So,we have: 3x2 – 6x = 3x(x – 2)

  10. Example (b) E.g. 1—Common Factors • We note that: • 8, 6, and –2 have the greatest common factor 2. • x4, x3, and x have the greatest common factor x. • y2, y3, and y4 have the greatest common factor y2.

  11. Example (b) E.g. 1—Common Factors • So, the greatest common factor of the three terms in the polynomial is 2xy2. • Thus, we have: 8x4y2 + 6x3y3 – 2xy4 • = (2xy2)(4x3) + (2xy2)(3x2y) + (2xy2)(–y2) • = 2xy2(4x3 + 3x2y – y2)

  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 x2 + bx + c • In order to factor a trinomial of the form x2 + bx + c, we note that: (x + r)(x + s) = x2 + (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 x2 + bx + c by Trial and Error • Factor: x2 + 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: x2 + 7x + 12 = (x + 3)(x + 4)

  17. Factoring ax2 + bx + c • To factor a trinomial of the form ax2 + bx + c with a≠ 1, we look for factors of the form px + r and qx + s: • ax2 + bx + c = (px + r)(qx + s) • = pqx2 + (ps + qr)x + rs

  18. Factoring ax2 + 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 ax2 + bx + c by Trial and Error • Factor: 6x2 + 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 6x2 + 7x – 5 = (3x + 5)(2x – 1)

  20. E.g. 5—Recognizing the Form of an Expression • Factor each expression. • x2 – 2x – 3 • (5a + 1)2 – 2(5a + 1) – 3

  21. Example (a) E.g. 5—Recognizing the Form • x2 – 2x – 3 = (x – 3)(x + 1) (Trial and error)

  22. Example (b) E.g. 5—Recognizing the Form • This expression is of the form __2 – 2__ – 3 where __ represents 5a + 1.

  23. Example (b) 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)

  24. Special Factoring Formulas

  25. Special Factoring Formulas • 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. • 4x2 – 25 • (x + y)2 – z2

  27. Example (a) E.g. 6—Differences of Squares • Using the Difference of Squares Formula with A =2x and B = 5, we have: • 4x2 – 25 = (2x)2 – 52 = (2x – 5)(2x + 5)

  28. Example (b) E.g. 6—Differences of Squares • We use the Difference of Squares Formula with A = x + y and B = z. (x + y)2 – z2 = (x + y – z)(x + y + z)

  29. E.g. 7—Factoring Differences and Sums of Cubes • Factor each polynomial. • 27x3 – 1 • x6 + 8

  30. Example (a) E.g. 7—Difference of Cubes • Using the Difference of Cubes Formula with A = 3x and B = 1, we get: • 27x3 – 1 = (3x)3 – 13 • = (3x – 1)[ (3x)2 + (3x)(1) + 12] • = (3x – 1)(9x2 + 3x + 1)

  31. Example (b) E.g. 7—Sum of Cubes • Using the Sum of Cubes Formula with A = x2 and B = 2, we have: x6 + 8 = (x2)3 + 23 = (x2 + 2)(x4 – 2x2 + 4)

  32. Perfect Square • A trinomial is a perfect square if it is of the form A2 + 2AB + B2 or A2 – 2AB + B2 • So, we recognize a perfect squareif 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. • x2 + 6x + 9 • 4x2 – 4xy + y2

  34. Example (a) 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:x2 + 6x + 9 = (x + 3)2

  35. Example (b) 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: 4x2 – 4xy + y2 = (2x – y)2

  36. Factoring an Expression Completely

  37. Factoring an Expression Completely • 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. • 2x4 – 8x2 • x5y2 – xy6

  39. Example (a) E.g. 9—Factoring Completely • We first factor out the power of x with the smallest exponent. • 2x4 – 8x2= 2x2(x2 – 4) (Common factor is 2x2) • = 2x2(x – 2)(x + 2) (Factor x2 – 4 as a difference of squares)

  40. Example (b) E.g. 9—Factoring Completely • We first factor out the powers of x and y with the smallest exponents. x5y2 – xy6= xy2(x4 – y4) (Common factor is xy2) • = xy2(x2 + y2)(x2 – y2) (Factor x4 – y4 as a difference of squares) • = xy2(x2 + y2)(x + y)(x – y) (Factor x2 – y2 as a difference of squares)

  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. • 3x3/2 – 9x1/2 + 6x–1/2 • (2 + x)–2/3x + (2 + x)1/3

  43. Example (a) E.g. 10—Fractional Exponents • Factor out the power of x with the smallest exponent—that is, x–1/2. • 3x3/2 – 9x1/2 + 6x–1/2 • = 3x–1/2(x2 – 3x + 2) (Factor out 3x–1/2) • = 3x–1/2(x – 1)(x – 2) (Factor the quadratic x2 – 3x + 2)

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

  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. • x3 + x2 +4x + 4 • x3 – 2x2 – 3x + 6

  48. Example (a) E.g. 11—Factoring by Grouping • x3 + x2 +4x + 4 • = (x3 + x2) + (4x + 4) (Group terms) • = x2(x + 1) + 4(x + 1) (Factor out common factors) • = (x2 + 4)(x + 1) (Factor out x + 1 from each term)

  49. Example (b) E.g. 11—Factoring by Grouping • x3 – 2x2 – 3x + 6 • = (x3 – 2x2) – (3x – 6) (Group terms) • = x2(x – 2) – 3(x – 2) (Factor out common factors) • = (x2 – 3)(x – 2) (Factor out x – 2 from each term)

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