algebra 1 lesson 8-6 warm-up algebra 1 “factoring trinomials of the type ax 2 + bx + c” (8-6)...

ALGEBRA 1

Lesson 8-6 Warm-Up

ALGEBRA 1

“Factoring Trinomials of the Type ax2 + 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 ax2 + 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 ax2 + bx + 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 6n2 + 23n + 7

S

Factors of 3rd Term

1 and 7 = 7

1 and 7 = 7

1 and 7 = 7

Sum of the Factors

1•1 + 6•7 23

1•7 + 6•1 23

2•1 + 3•7 = 23

Factors of 1st Term

1 and 6 = 6

1 and 6 = 6

2 and 3 = 6

ALGEBRA 1

“Factoring Trinomials of the Type ax2 + 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.

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

Check (using FOIL): (2n + 1)(3n + 7) = 6n2 + 14n + 3n + 7 = 6n2 + 17n + 7

≠ 6n2 + 23n + 7

Since this combination doesn’t work, switch the first of last terms. Below, the last terms are switched.

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

Check (using FOIL): (2n + 7)(3n + 1) = 6n2 + 21n + 2n + 7 = 6n2 + 23n + 7

S

ALGEBRA 1


Method 2: Use the FOIL Method in Reverse: Look for binomials that have the following characteristics for ax2 + bx + c.


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ALGEBRA 1


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.


S

n2

n2

n2

n2

n2

n2

n n n n n n n

n n n n n n n

n n n n n n n

1111111n n2n + 7

3n +

1

2n + 7

3n +

1

Algebra Tiles Key

n

n n

11

1

n 1

n

n2

1

n

ALGEBRA 1


42n2

21n

23

2n

6n2

72n

21n3n

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

1. Find two numbers whose product is ac and sum is b. These numbers will be the coefficients of the x terms.


1

2n 7

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.

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

ALGEBRA 1

Factor 20x2 + 17x + 3.

Factoring Trinomials of the Type ax2 + bx + cLESSON 8-6

Additional Examples

Method 1: Table

Factors of 1st Term Sum of the Factors Factors of 3rd Term

20 and 1 = 20 20•1 + 1•3 17 1 and 3 = 3

20 and 1 = 20 20•3 + 1•1 17 1 and 3 = 3

10 and 2 = 20 10•1 + 2•3 17 1 and 3 = 3

10 and 2 = 20 10•3 + 2•1 17 1 and 3 = 3

5 and 4 = 20 5•1 + 4•3 = 17 1 and 3 = 3

5 and 4 = 20 5•3 + 4•1 17 1 and 3 = 3

ALGEBRA 1


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.

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

Check (using FOIL): (5x + 1)(4x + 3) = 20x2 + 15x + 4x + 3 = 20x2 + 19x + 3 ≠ 20x2 + 17x + 3

Since this combination doesn’t work, switch the first of last terms. Below, the last terms are switched.

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

Check (FOIL): (5x + 1)(4x + 3) = 20x2 + 5x + 12x + 3 = 20x2 + 17x + 3

ALGEBRA 1

Factor 20x2 + 17x + 3.

20x + 17x + 3

2 • 10 2 • 3 + 1 • 10 = 16 1 • 32 • 10 2 • 1 + 3 • 10 = 32 3 • 1

1 • 20 1 • 3 + 1 • 20 = 23 1 • 31 • 20 1 • 1 + 3 • 20 = 61 3 • 1

factors of a

factors of c

4 • 5 4 • 3 + 1 • 5 = 17 1 • 3

20x2 + 17x + 3 = (4x + 1)(5x + 3)


Additional Examples

Method 2: FOIL

F O I L

ALGEBRA 1

Factor 20x2 + 17x + 3

60x2

12x

17

5x20x2

35x

12x4x

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

Method 3: “X-Box Method”


1

5x 3

Factor 20x2 + 17x + 3.

ALGEBRA 1

Factor 3n2 – 7n – 6.

3n2 –7n –6

(1)(3) (1)(–6) + (3)(1) = –3 (1)(–6)

(1)(3) (1)(1) + (3)(-6) = –17 (-6)(1)

(1)(3) (1)(-3) + (3)(2) = 3 (2)(–3)

(1)(3) (1)(2) + (3)(-3) = –7 (–3)(2)

3n2 – 7n – 6 = (1n – 3)(3n + 2)


Additional Examples

F O I L

ALGEBRA 1


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 20x2 + 80x + 35

Step 1: 20x2 + 80x + 35 = 5(4x2 + 16x + 7) 5 is a common factor of all 3 terms, so factor it out.

Step 2: Factor 4x2 + 16x + 7.

Step 3: Find the correct combination of the factor pairs 2, 2 and 1,7 to equal 4x2

+ 16x + 7.

(2x + 1)(2x + 7)

Check (FOIL): (2x + 1)(2x + 7) = 4x2 + 14x + 2x + 7 = 4x2 + 16x + 7

Factors of 1st Term Sum of the Factors Factors of 3rd Term

1 and 4 = 4 1•1 + 4•7 16 1 and 7 = 7

1 and 4 = 4 1•7 + 4•1 16 1 and 7 = 7

2 and 2 = 4 2•1 + 2•7 = 16 1 and 7 = 7

ALGEBRA 1


Don’t forget to multiply the binomials by the factors you pulled out (in other words, put the common factor in front of the answer).

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

So, the 20x2 + 80x + 35 completely factored is 5(2x + 1)(2x + 7).

ALGEBRA 1

Factor 18x2 + 33x – 30 completely.

18x2 + 33x – 30 = 3(6x2 + 11x – 10) Factor out the common factor.

18x2 + 33x – 30 = 3(2x + 5)(3x – 2) Include the common factor inyour final answer.

6x2 + 11x –10

6x2 + 11x – 10 = (2x + 5)(3x – 2)

(2)(3) (2)(–10) + (3)(1) = –17 (1)(–10)

(2)(3) (2)(1) + (3)(-10) = –28 (–10)(1)

(2)(3) (2)(–5) + (3)(2) = –4 (2)(–5)

(2)(3) (2)(2) + (3)(-5) = –11 (–5)(2)

(2)(3) (2)(–2) + (3)(5) = 11 (5)(–2)


Additional Examples

Factor 6x2 + 11x – 10. F O I L

ALGEBRA 1

Factor each expression.

1. 3x2 – 14x + 11

2. 6t2 + 13t – 63

3. 9y2 – 48y – 36

(x – 1)(3x – 11)

(2t + 9)(3t – 7)

3(3y + 2)(y – 6)


Lesson Quiz

algebra 1 lesson 8-6 warm-up algebra 1 “factoring trinomials of the type ax 2 + bx + c” (8-6)...

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