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**Factoring Trinomials of the form**

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

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**Factoring Trinomials ax2 + 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.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by 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)

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**Trial and Error Factors of 2 Example: 2x2 + 11x + 12**

Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (1)(12) (2x + 1)(x + 12) (2x)(12) + (1)(x) = 24x + x = 25x (2)(6) (2x + 2)(x + 6) (2x)(6) + (2)(x) = 12x + 2x = 14x (3)(4) (2x + 3)(x + 4) (2x)(4) + (3)(x) = 8x + 3x = 11x (12)(1) (2x + 12)(x + 1) (2x)(1) + (12)(x) = 2x + 12x = 14x (6)(2) (2x + 6)(x + 2) (2x)(2) + (6)(x) = 4x + 6x = 10x (4)(3) (2x + 4)(x + 3) (2x)(3) + (4)(x) = 6x + 4x = 10x Since the constant is positive and the x-term is positive both factors are positive. You can always check with the FOIL method

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**Trial and Error Factors of 7 Example: 7x2 – 11x – 6**

Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (-1)(6) (7x - 1)(x + 6) (7x)(6) + (-1)(x) = 42x + -x = 41x (-2)(3) (7x - 2)(x + 3) (7x)(3) + (-2)(x) = 21x + -2x = 19x (-3)(2) (7x - 3)(x + 2) (7x)(2) + (-3)(x) = 14x + -3x = 11x (-6)(1) (7x - 6)(x + 1) (7x)(1) + (-6)(x) = 7x + -6x = 1x (1)(-6) (7x + 1)(x - 6) (7x)(-6) + (1)(x) = -42x + 1x = -41x (2)(-3) (7x + 2)(x - 3) (7x)(-3) + (2)(x) = -21x + 2x = -19x (3)(-2) (7x + 3)(x - 2) (7x)(-2) + (3)(x) = -14x + 3x = -11x (6)(-1) (7x + 6)(x - 1) (7x)(-1) + (6)(x) = -7x + 6x = -1x 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)

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**Trial and Error Factors of 6 Example: 6x2 + 31x + 5**

(2)(3) Factors of 5 Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (1)(5) (6x + 1)(x + 5) (6x)(5) + (1)(x) = 30x + 1x = 31x (5)(1) (6x + 5)(x + 1) (6x)(1) + (5)(x) = 6x + 5x = 11x (2x + 1)(3x + 5) (2x)(5) + (1)(3x) = 10x + 3x = 13x (2x + 5)(3x + 1) (2x)(1) + (5)(3x) = 2x + 15x = 17x Since the constant is positive and the middle-term is positive both factors are positive. You can always check with the FOIL method

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**Trial and Error Example: 16x2 - 8x + 1 (16)(1) (4x - 1)(4x - 1) (2)(8)**

Factors of 16 Example: x2 - 8x (16)(1) (4x - 1)(4x - 1) (2)(8) (4x - 1)2 (4)(4) Factors of 1 Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (-1)(-1) (1x - 1)(x - 1) (1x)(-1) + (-1)(x) = -1x + -1x = -2x (2X - 1)(8X - 1) (2X)(-1) + (-1)(8X) = -2X + -8X = -10X (4X - 1)(4X - 1) (4X)(-1) + (-1)(4X) = -4X + -4X = -8X Since the constant is positive and the middle-term is negative both factors are negative. You can always check with the FOIL method

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**Trial and Error Factors of 2 Example: 2x2 + 11x + 7 PRIME (2)(1)**

Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (1)(7) (2x + 1)(x + 7) (2x)(7) + (1)(x) = 14x + 1x = 15x (7)(1) (2x + 7)(x + 1) (2x)(1) + (7)(x) = 2x + 7x = 9x Since the constant is positive and the middle-term is positive both factors are positive. You can always check with the FOIL method

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**Trial and Error Factors of 6 Example: 6a2 + 11ab + 5b2**

(6a + 5b)(a + b) (6)(1) (2)(3) Factors of 5 Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (1)(5) (6a + 1)(a + 5) (6a)(5) + (1)(a) = 30a + 1a = 31a (5)(1) (6a + 5)(a + 1) (6a)(1) + (5)(a) = 6a + 5a = 11a (2a + 1)(3a + 5) (2a)(5) + (1)(3a) = 10a + 3a = 13a (2a + 5)(3a + 1) (2a)(1) + (5)(3a) = 2a + 15a = 17a Since the last term is positive and the middle-term is positive both factors are positive. You can always check with the FOIL method

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Trial and Error Factors of 6 Example: 6x2 – 19xy – 7y2 (6)(1) (3x + y)(2x – 7y) (2)(3) Factors of 7 Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (-1)(7) (6x - 1)(x + 7) (6x)(7) + (-1)(x) = 42x + -1x = 41x (-7)(1) (6x - 7)(x + 1) (6x)(1) + (-7)(x) = 6x + -7x = -1x (1)(-7) (2x + 1)(3x - 7) (2x)(-7) + (1)(3x) = -14x + 3x = -11x (7)(-1) (2x + 7)(3x - 1) (2x)(-1) + (7)(3x) = -2x + 21x = 19x (2x - 1)(3x + 7) (2x)(7) + (-1)(3x) = 14x - 3x = 11x (2x - 7)(3x + 1) (2x)(1) + (-7)(3x) = 2x - 21x = -19x Since the last term is negative one factors is positive and one is negative.

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Trial and Error Factors of 3 Example: 9x3 + 15x2 + 6x GCF (3)(1) x(3x2 + 5x + 2x) 3x 3x(3x + 2)(x + 1) Factors of 2 Possible factors of trinomial Products of the Outer and Inner Terms Sums to the b term in the Trinomial (1)(2) (3x + 1)(x + 2) (3x)(2) + (1)(x) = 6x + 1x = 7x (2)(1) (3x + 2)(x + 1) (3x)(1) + (2)(x) = 3x + 2x = 5x Since the constant is positive and the middle-term is positive both factors are positive.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by 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.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping**

Example: 3x2 + 14x + 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 Factors of 45 Sum 14 (1)(45) =46 (3)(15) = 18 (5)(9) 5 + 9 = 14 Since a • c is positive and the middle-term is positive both factors are positive.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping**

Example: 3x2 - 7x - 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 Factors of -18 Sum -7 (-1)(18) = 17 (-2)(9) = 7 (-3)(6) = 3 (1)(-18) = -17 (2)(-9) = -7 (3)(-6) = -3 Since a • c is negative one factors is positive and one is negative.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping**

Example: 6x2 + 31x + 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 Factors of 30 Sum 31 (1)(30) = 31 Since a • c is positive and the middle-term is positive both factors are positive.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping**

Example: 64p2 – 16p + 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 Factors of 64 Sum -16 (-1)(-64) = -65 (-2)(-32) = -34 (-4)(-16) = -20 (-8)(-8) = -16 Since a • c is positive and the middle-term is negative both factors are negative.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping**

Example: 3x2 + 20x + 5 3x2 + ____ + 5 PRIME a = 3, b = 20, c = 5 a • c = =3 • 5 = 15 Factors of 15 Sum 5 (1)(15) = 16 (3)(5) 3 + 5 = 8 Since a • c is positive and the middle-term is positive both factors would have been positive.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by 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 Factors of 20 Sum 9 (1)(20) = 21 (2)(10) = 12 (4)(5) 4 + 5 = 9 Since a • c is positive and the middle-term is positive both factors are positive.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by 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 Factors of -54 Sum -25 (-1)(54) = 53 (-2)(27) = 25 (1)(-54) = -53 (2)(-27) = -25 Since a • c is negative one factors is positive and one is negative.

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**Factoring Trinomials ax2 + bx + c, a ≠ 1 by Grouping**

Example: 6x3 + 3x2 – 45x GCF = 3x 3x(2x2 + x - 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 Factors of -30 Sum 1 (-1)(30) = 29 (-2)(15) = 13 (-3)(10) = 7 (-5)(6) = 1 Since a • c is negative one factors is positive and one is negative.

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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?

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

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