Lesson 8-6 Warm-Up

ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx + c” (8-6)

How do you factor a trinomial in which the second degree variable has a coefficient?

To factor a trinomial of the form ax 2 + bx + c, you must find two factor pairs of

a

and

c

whose sum is

b

. There are two methods to factoring a trinomial in the form of

a

x 2 +

b

x +

c

.

Method 1: Create a Three Column Table:

Title one column

“Factors of First Term”

, the second column “

Sum of the Factors

”, and the last column “

Factors of Third Term

”. Then, fill in the table and choose the products whose sum is the middle number, b.

Example: Factor 6 n 2 +

23

n +

7

Factors of 1 st Term 1

and

6

= 6

1 2

and

6

and

3

= 6 = 6

Sum of the Factors 1

•

1

+

6

•

7

 23

1

•

7

+

6

•

1

 23

2

•

1

+

3

•

7

= 23 

Factors of 3 rd Term 1

and

7

= 7

1 1

and

7

and

7

= 7 = 7

ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx +c” (8-6)

So, use the factors 2 and 3 for the first terms of the binomials and use 1 and 7 for the second terms of the binomials. Now, we need to figure out what order to put those combinations in.

(2 n + 1 )(3 n + 7 )

Check (using FOIL): ( 2 n + 1 )( 3 n + 7 ) = 6 n 2 + 14 n + 3 n + 7 = 6 n 2 + 17 n + 7 ≠ 6 n 2 + 23 n + 7 Since this combination doesn’t work, switch the first of last terms. Below, the last terms are switched.

(2 n + 7 )(3 n + 1 )

Check (using FOIL): ( 2 n + 7 )( 3 n + 1 ) = 6 n 2 + 21 n + 2 n + 7 =

6

n 2 +

23

n +

7



ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx +c” (8-6)

Method 2: Use the FOIL Method in Reverse:

Look for binomials that have the following characteristics for

a

x 2 +

b

x +

c

.

Example: Factor 6 n 2 +

23

n +

7

ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx +c” (8-6)

Method 3: Use an Area Model in Reverse:

Arrange the Algebra Tiles that model the trinomial into a rectangle

.

The sides of the rectangle (

length and width

) are the factors of the trinomial.

Tip: Think about how to end with the number of “1” tiles.

Example: Factor 6 n 2 +

23

n +

7

2n + 7 Algebra Tiles Key n 2 n 2 n n n n n n n n n 1 1 1 1 n 2 n 2 n n n n n n n n 2 n n n n 2 n 2 n n n n n n n n 1 n n 2n + 7 1 1 1 1 1 1 1 ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx +c” (8-6)

Method 4: Use an Area Model in Reverse (“X-Box Method”):

1. Find

two numbers whose product is

a c

and

sum is

b

. These numbers will be the coefficients of the

x

terms. 2. Then,

create a box divided into two columns and two rows

. The

top-left box will be the a term

, the

bottom right box will be the c term

, and the

middle two boxes will be the b terms

.

3. Finally,

find common factors of each column and row

. The dimensions

(length and width)

of the box are factors

(binomial times binomial)

of the trinomials.

Example: Factor 6 n 2 +

23

n +

7

2n 7

42n 2

3n 6

n 2 21 n 21 n 2 n

23 1

2 n

7

Answer: ( 2n + 7 )( 3n + 1 )

ALGEBRA 1

Factoring Trinomials of the Type ax 2 + bx + c LESSON 8-6 Additional Examples Factor 20x 2 + 17x + 3.

Method 1: Table Factors of 1 st Term 20

and

1

= 20

20

and

1

= 20

10

and

2

= 20

10

and

2

= 20

5

and

4

= 20

5

and

4

= 20

Sum of the Factors 20

•

1

+

1

•

3

 17

20

•

3

+

1

•

1

 17

10

•

1

+

2

•

3

 17

10

•

3

+

2

•

1

 17

5

•

1

+

4

•

3

= 17 

5

•

3

+

4

•

1

 17

Factors of 3 rd Term 1

and

3

= 3

1

and

3

= 3

1

and

3

= 3

1

and

3

= 3

1

and

3

= 3

1

and

3

= 3

ALGEBRA 1

Factoring Trinomials of the Type ax 2 + bx + c LESSON 8-6 Additional Examples

So, use the factors 5 and 4 for the first terms of the binomials and use 1 and 3 for the second terms of the binomials. Now, we need to figure out what order to put those combinations.

(5 x + 1 )(4 x + 3 )

Check (using FOIL): (5x + 1 )(4x + 3 ≠ 20 x 2 3 ) = 20 x 2 + 17 x + 3 + 15 x + 4 x + 3 = 20 x 2 + 19 x + Since this combination doesn’t work, switch the first of last terms. Below, the last terms are switched.

(5x + 3 )(4x + 1 )

Check (FOIL): ( 5 x + 1 )( 4 x + 3 ) = 20 x 2 + 5 x + 12 x + 3 = 20 x 2 +

17

x + 3



ALGEBRA 1

Factoring Trinomials of the Type ax 2 + bx + c LESSON 8-6 Additional Examples Factor 20x 2 + 17x + 3.

Method 2: FOIL 20

x

factors of

a

F

1 • 20 1 • 20 2 • 10 2 • 10 4 • 5 +

17

x

O I

1 1 • 3 • 1 + 1 + 3 • 20 • 20 = 23 = 61 2 2 • 3 • 1 + 1 + 3 • 10 • 10 = 16 = 32 4 • 3 + 1 • 5 =

17

+ 3

L

1 • 3 3 • 1 1 • 3 3 • 1 1 • 3 factors of

c

20

x

2 + 17

x

+ 3 =

( 4x + 1 )( 5 x + 3 ) ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx +c” (8-6)

Factor 20x 2 + 17x + 3.

Method 3: “X-Box Method” Factor 20

x

2 + 17 x + 3 5x 3

60x 2

4x 20

x 2 12 x 5 x 12 x

17 1

5 x

3 Answer: ( 5x + 3 )( 4x + 1 ) ALGEBRA 1

Factoring Trinomials of the Type ax 2 + bx + c LESSON 8-6 Additional Examples

( ( ( 3 ( 1 )( 3 ) 1 1 1

n

F

)( )( )( 2 3 3 3 ) ) )

Factor 3n 2 – 7n – 6.

–7

n

O I

( 1 )( –6 ) + ( 3 )( 1 ) = –3 ( 1 )( 1 ) + ( 3 )( -6 ) = –17 ( 1 )( -3 ) + ( 3 )( 2 ) = 3 ( 1 )( 2 ) + ( 3 )( -3 ) =

–7

–6

L

(1)( –6) (-6)(1) (2)( –3) ( –3)(2) 3

n

2 – 7

n

– 6 =

( 1

n

– 3 )( 3 n + 2 ) ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx +c” (8-6)

Note:

Some polynomials have terms with a common factor. If this is the case, “factor out “ that monomial factor using the Distributive Property in reverse before factoring the trinomial.

Example: Factor 20 x 2 +

80

x +

35

Step 1: 20x 2 + 80x + 35 =

5

(4x 2 + 16x + 7) Step 2: Factor 4x 2 + 16x + 7.

Factors of 1 st Term 1

and

4

= 4

Sum of the Factors 1

•

1

+

4

•

7

 16

5 is a common factor of all 3 terms, so factor it out.

Factors of 3 rd Term 1

and

7

= 7

1 2

and

4

and

2

= 4 = 4

1

•

7

+

4

•

1

 16

2

•

1

+

2

•

7

= 16 

1 1

and

7

and

7

= 7 = 7 Step 3: Find the correct combination of the factor pairs 2, 2 and 1,7 to equal 4x 2 + 16x + 7. (2x + 1)(2x + 7)

Check (FOIL): (2x + 1)(2x + 7) = 4x 2 + 14 x + 2 x + 7 = 4x 2 + 16 x + 7



ALGEBRA 1

“Factoring Trinomials of the Type ax

2

+ bx +c” (8-6)

Don’t forget to multiply the binomials by the factors you pulled out (i

n other words, put the common factor in front of the answer

).

5

(2x + 1)(2x + 7) So, the 20 x 2 + 80 x + 35 completely factored is

5(2x + 1)(2x + 7).

ALGEBRA 1

Factoring Trinomials of the Type ax 2 + bx + c LESSON 8-6 Additional Examples Factor 18x 2 + 33x – 30 completely.

18

x

2 + 33

x

– 30 =

3

( 6

x

2 + 11

x

– 10 ) Factor out the common factor.

( ( ( ( ( Factor 6

x

2

F

6

x

2 + 11

x

– 10.

O I

+ 11

x

2 2 2 )( )( )( 3 3 3 ) ) ) ( 2 )( –10 ) + ( 3 )( 1 ) = –17 ( 2 )( 1 ) + ( 3 )( -10 ) = –28 ( 2 )( –5 ) + ( 3 )( 2 ) = –4 2 2 )( )( 3 3 ) ) ( 2 )( 2 ) + ( 3 )( -5 ) = –11 ( 2 )( –2 ) + ( 3 )( 5 ) =

11

( ( ( –10 ( 1 )( –10 ) ( –10 )( 1 ) 2 5 )( –5 )(

L

–5 )( 2 –2 ) ) ) 6

x

2 + 11

x

– 10 = ( 2

x

+ 5 )( 3

x

– 2 ) 18

x

2 + 33

x

– 30 =

3 (2x + 5)(3x – 2)

Include the common factor in your final answer.

ALGEBRA 1

Factoring Trinomials of the Type ax 2 + bx + c LESSON 8-6 Lesson Quiz

Factor each expression.

1.

3

x

2 – 14

x

+ 11 (

x

– 1)(3

x

– 11)

2.

6

t

2 + 13

t

– 63 (2

t

+ 9)(3

t

– 7)

3.

9

y

2 – 48

y

– 36 3(3

y

+ 2)(

y

– 6)

ALGEBRA 1