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MTH 10905 Algebra

MTH 10905 Algebra. Factoring Trinomials of the form ax 2 + bx + c a ≠ 1 Chapter 5 Section 4. Factoring Trinomials ax 2 + bx + c, a ≠ 1. The squared term has a numerical coefficient not equal to 1. There are two methods Trial and Error Factor by Grouping.

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MTH 10905 Algebra

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  1. MTH 10905Algebra Factoring Trinomials of the form ax2 + bx + c a ≠ 1 Chapter 5 Section 4

  2. Factoring Trinomialsax2 + bx + c, a ≠ 1 • The squared term has a numerical coefficient not equal to 1. • There are two methods • Trial and Error • Factor by Grouping. • Remember factoring is the reverse of multiplication.

  3. Factoring Trinomialsax2 + bx + c, a ≠ 1by Trial and Error • Determine whether there is a factor common to all three terms. If yes, factor it out. (we do not factor out 1 or -1) • Write all pairs of factors of the coefficient of the squared term, a. • Write all pairs of factors of the constant term, c. • Try various combination of these factors until the correct middle term, bx, is found. NOTE: You may want to select a style when deciding the position of the factors of the coefficient of the squared term (Exp: put the larger value in the first binomial)

  4. Trial and Error Example: 2x2 + 11x + 12 (2x + 3)(x + 4) (2)(1) Factors of 2 Since the constant is positive and the x-term is positive both factors are positive. You can always check with the FOIL method

  5. Trial and Error Factors of 7 Example: 7x2 – 11x – 6 (7x + 3)(x – 2) (7)(1) Since the last term is negative, one factors is positive and one is negative. When we change the sign of the constant in the binomial , x-term in trinomial changes)

  6. Trial and Error Example: 6x2 + 31x + 5 (6x + 1)(x + 5) (6)(1) (2)(3) Factors of 6 Since the constant is positive and the middle-term is positive both factors are positive. You can always check with the FOIL method

  7. Trial and Error Factors of 16 Example: 16x2 - 8x + 1 (16)(1) (4x - 1)(4x - 1) (2)(8) (4x - 1)2 (4)(4) Since the constant is positive and the middle-term is negative both factors are negative. You can always check with the FOIL method

  8. Trial and Error Example: 2x2 + 11x + 7 PRIME (2)(1) Factors of 2 Since the constant is positive and the middle-term is positive both factors are positive. You can always check with the FOIL method

  9. Trial and Error Example: 6a2 + 11ab + 5b2 (6a + 5b)(a + b) (6)(1) (2)(3) Factors of 6 Since the last term is positive and the middle-termis positive both factors are positive. You can always check with the FOIL method

  10. Trial and Error Factors of 6 Example: 6x2 – 19xy – 7y2 (6)(1) (3x + y)(2x – 7y) (2)(3) Since the last term is negativeone factors is positive and one is negative.

  11. Trial and Error Factors of 3 Example: 9x3 + 15x2 + 6x GCF (3)(1) 3x(3x2 + 5x + 2x) 3x 3x(3x + 2)(x + 1) Since the constant is positive and the middle-term is positive both factors are positive.

  12. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping • Determine whether there is a factor common to all three terms. If yes, factor it out. • Find two numbers whose product is equal to the product of a times c, and whose sum is equal to b. • Rewrite the middle term, bx, as the sum or difference of two terms using the numbers found in step 2. • Factor by grouping.

  13. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 3x2 + 14x + 15 3x2 + 9x + 5x + 15 3x2 + 9x + 5x + 15 3x(x + 3) + 5(x + 3) (3x + 5)(x + 3) a = 3, b = 14, c = 15 a • c = 3 • 15 = 45 Since a • c is positive and the middle-term is positive both factors are positive.

  14. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 3x2 - 7x - 6 3x2 - 9x + 2x - 6 3x2 - 9x + 2x - 6 3x(x - 3) + 2(x - 3) (3x + 2)(x - 3) a = 3, b = -7, c = -6 a • c = 3 • -6 = -18 Since a • c is negativeone factors is positive and one is negative.

  15. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 6x2 + 31x + 5 6x2 + x + 30x + 5 6x2 + x + 30x + 5 x(6x + 1) + 5(6x + 1) (6x + 1)(x + 5) a = 6, b = 31, c = 5 a • c = 6 • 5 = 30 Since a • c is positive and the middle-term is positive both factors are positive.

  16. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 64p2 – 16p + 1 64p2 - 8p - 8p + 1 64p2 - 8p - 8p + 1 8p(8p - 1) - 1(8p - 1) (8p - 1)(8p - 1) (8p – 1)2 a = 64, b = -16, c = 1 a • c = 64 • 1 = 64 Since a • c is positive and the middle-term is negative both factors are negative.

  17. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 3x2 + 20x + 5 3x2 + ____ + 5 PRIME a = 3, b = 20, c = 5 a • c = =3 • 5 = 15 Since a • c is positive and the middle-term is positive both factors would have been positive.

  18. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 5a2 + 9ab + 4b2 5a2 + 4ab + 5ab + 4b2 5a2 + 5ab + 4ab + 4b2 5a(a + b) + 4b(a + b) (a + b)(5a + 4b) a = 5, b = 9, c = 4 a • c = 5 • 4 = 20 Since a • c is positive and the middle-term is positive both factors are positive.

  19. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 6x2 – 25xy – 9y2 6x2 + 2xy – 27xy – 9y2 6x2 + 2xy – 27xy + 9y2 2x(3x + y) – 9y(3x + y) (2x – 9y)(3x + y) a = 6, b = -25, c = -9 a • c = 6 • -9 = -54 Since a • c is negativeone factors is positive and one is negative.

  20. Factoring Trinomialsax2 + bx + c, a ≠ 1by Grouping Example: 6x3 + 3x2 – 45x GCF = 3x 3x(2x2 + x - 15) 3x(2x2 + 6x - 5x - 15) 3x(2x2 + 6x - 5x - 15) 3x(2x)(x + 3) - 5(x + 3) 3x(2x – 5)(x + 3) a = 2, b = 1, c = -15 a • c = 2 • -15 = -30 Since a • c is negativeone factors is positive and one is negative.

  21. REMEMBER • Always put the polynomial in standard form before attempting to factor. • For Trial and Error you my want to setup a table listing all the possible factors of a and c, and then use these to form all possible binomial factor pairs for the polynomial. • If the original expression has no common factor, then the two factors will not have any common factors. • Check your results by multiplying. • I prefer Grouping. Which method do you prefer?

  22. HOMEWORK 5.4 Page 323: #5, 7, 15, 21, 23, 45, 55, 57

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