MTH 10905 Algebra Factoring Trinomials of the form ax 2 + bx + c a ≠ 1 Chapter 5 Section 4.

MTH 10905Algebra

Factoring Trinomials of the form

ax2 + bx + c a ≠ 1

Chapter 5 Section 4

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.

1. Determine whether there is a factor common to all three terms. If yes, factor it out. (we do not factor out 1 or -1)

2. Write all pairs of factors of the coefficient of the squared term, a.

3. Write all pairs of factors of the constant term, c.

4. 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)

Factoring Trinomialsax2 + bx + c, a ≠ 1by Trial and Error

Trial and Error

Example: 2x2 + 11x + 12(2x + 3)(x + 4) (2)(1)

You can always check with the FOIL method

Factors of 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

Factors of 2

Since the constant is positive and the x-term is positive both factors are positive.

Trial and Error

Example: 7x2 – 11x – 6 (7x + 3)(x – 2) (7)(1)

Factors of -6



(-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

Factors of 7

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)

Trial and Error

Example: 6x2 + 31x + 5 (6x + 1)(x + 5) (6)(1)

(2)(3)


Factors of5



(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

(1)(5) (2x + 1)(3x + 5) (2x)(5) + (1)(3x) = 10x + 3x = 13x

(5)(1) (2x + 5)(3x + 1) (2x)(1) + (5)(3x) = 2x + 15x = 17x

Factors of 6

Since the constant is positive and the middle-term is positive both factors are positive.

Trial and Error

Example: 16x2 - 8x + 1 (16)(1)(4x - 1)(4x - 1) (2)(8)(4x - 1)2 (4)(4)


Factors of1



(-1)(-1) (1x - 1)(x - 1) (1x)(-1) + (-1)(x) = -1x + -1x = -2x

(-1)(-1) (2X - 1)(8X - 1) (2X)(-1) + (-1)(8X) = -2X + -8X = -10X

(-1)(-1) (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.

Factors of 16

Trial and Error

Example: 2x2 + 11x + 7 PRIME (2)(1)


Factors of7



(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


Factors of 2

Trial and Error

Example: 6a2 + 11ab + 5b2 (6a + 5b)(a + b) (6)(1)

(2)(3)


Factors of5



(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

(1)(5) (2a + 1)(3a + 5) (2a)(5) + (1)(3a) = 10a + 3a = 13a

(5)(1) (2a + 5)(3a + 1) (2a)(1) + (5)(3a) = 2a + 15a = 17a

Since the last term is positive and the middle-termis positive both factors are positive.

Factors of 6

Trial and Error

Example: 6x2 – 19xy – 7y2 (6)(1)(3x + y)(2x – 7y) (2)(3)

Factors of7



(-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

(-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

Since the last term is negativeone factors is positive and one is negative.

Factors of 6

Trial and Error

Example: 9x3 + 15x2 + 6x GCF (3)(1) 3x(3x2 + 5x + 2x) 3x

3x(3x + 2)(x + 1)

Factors of2



(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

Factors of 3


1. Determine whether there is a factor common to all three terms. If yes, factor it out.

2. Find two numbers whose product is equal to the product of a times c, and whose sum is equal to b.

3. Rewrite the middle term, bx, as the sum or difference of two terms using the numbers found in step 2.

4. Factor by grouping.


by Grouping


by Grouping

Example: 3x2 + 14x + 15

3x2 + 9x + 5x + 153x2 + 9x + 5x + 15 3x(x + 3) + 5(x + 3)(3x + 5)(x + 3)

Factors of 45 Sum 14

(1)(45) 1 + 45 =46

(3)(15) 3 + 15 = 18

(5)(9) 5 + 9 = 14

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.


by Grouping

Example: 3x2 - 7x - 6

3x2 - 9x + 2x - 63x2 - 9x + 2x - 6 3x(x - 3) + 2(x - 3)(3x + 2)(x - 3)

Factors of -18 Sum -7

(-1)(18) -1 + 18 = 17

(-2)(9) -2 + 9 = 7

(-3)(6) -3 + 6 = 3

(1)(-18) 1 + -18 = -17

(2)(-9) 2 + -9 = -7

(3)(-6) 3 + -6 = -3

a = 3, b = -7, c = -6

a • c = 3 • -6 = -18

Since a • c is negativeone factors is positive and one is negative.


by Grouping

Example: 6x2 + 31x + 5

6x2 + x + 30x + 56x2 + x + 30x + 5 x(6x + 1) + 5(6x + 1)(6x + 1)(x + 5)

Factors of 30 Sum 31

(1)(30) 1 + 30 = 31

a = 6, b = 31, c = 5

a • c = 6 • 5 = 30



by Grouping

Example: 64p2 – 16p + 1

64p2 - 8p - 8p + 164p2 - 8p - 8p + 1 8p(8p - 1) - 1(8p - 1)(8p - 1)(8p - 1)(8p – 1)2

Factors of 64 Sum -16

(-1)(-64) -1 + -64 = -65

(-2)(-32) -2 + -32 = -34

(-4)(-16) -4 + -16 = -20

(-8)(-8) -8 + -8 = -16

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.


by Grouping

Example: 3x2 + 20x + 5

3x2 + ____ + 5

PRIMEFactors of 15 Sum 5

(1)(15) 1 + 15 = 16

(3)(5) 3 + 5 = 8

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.


by Grouping

Example: 5a2 + 9ab + 4b2

5a2 + 4ab + 5ab + 4b2

5a2 + 5ab + 4ab + 4b2 5a(a + b) + 4b(a + b)(a + b)(5a + 4b)

Factors of 20 Sum 9

(1)(20) 1 + 20 = 21

(2)(10) 2 + 10 = 12

(4)(5) 4 + 5 = 9

a = 5, b = 9, c = 4

a • c = 5 • 4 = 20



by Grouping

Example: 6x2 – 25xy – 9y2

6x2 + 2xy – 27xy – 9y2

6x2 + 2xy – 27xy + 9y2 2x(3x + y) – 9y(3x + y)(2x – 9y)(3x + y)

Factors of -54 Sum -25

(-1)(54) -1 + 54 = 53

(-2)(27) -2 + 27 = 25

(1)(-54) 1 + -54 = -53

(2)(-27) 2 + -27 = -25

a = 6, b = -25, c = -9

a • c = 6 • -9 = -54



by Grouping

Example: 6x3 + 3x2 – 45xGCF = 3x3x(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)

Factors of -30 Sum 1

(-1)(30) -1 + 30 = 29

(-2)(15) -2 + 15 = 13

(-3)(10) -3 + 10 = 7

(-5)(6) -5 + 6 = 1

a = 2, b = 1, c = -15

a • c = 2 • -15 = -30


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?

HOMEWORK 5.4

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MTH 10905 Algebra Factoring Trinomials of the form ax 2 + bx + c a ≠ 1 Chapter 5 Section 4.

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