Factoring 15.3 and 15.4 Grouping and Trial and Error

15.3 Factoring 15.3 Factoring Trinomials (Part 2)Trinomials (Part 2)

Ax2 + bx + c

METHODS

1.Trial and Error (Takes the most time)

2.Factoring by Grouping

3.Box Method (Grouping with a Box)

3

4

5

6

7

8

9

Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 10Copyright © 2011 Pearson Education, Inc.

Method 1: Factoring Trinomials by Trial and ErrorFactoring by Grouping

Factor

Example

23 14 8.x x+ +



Factor 3x2 + 14x + 8, the result will be of the form

(3x +?)(x +?)

Product of the last terms must be 8, so the last terms must be 1 and 8 or 2 and 4

Rule out negative last terms in the factors, because the middle term of 3x2+14x +8 has the positive coefficient 14

Solution



• Decide between the two pairs of possible last terms

by multiplying:

Solution Continued


Method 1: Factoring Trinomials by Trial and ErrorMethod 1: Factoring Trinomials by Trial and ErrorFactoring by Trial and ErrorFactoring by Trial and Error

FactorFactor

ExampleExample22 5 25.x x− −


Method 1: Factoring Trinomials by Trial and ErrorFactoring by Trial and Error

Factor 2x2 – 5 – 25, the result will be of the form

(2x +?)(x +?)

Product of the last terms must be –25, so the last terms must be 1 and –25, 5 and –5, or –1 and 25

Decide amongst the three pairs of possible last

terms by multiplying:

Solution


Method 1: Factoring Trinomials by Trial and ErrorFactoring by Trial and Error

Solution Continued

Factoring Trinomials Review

X2 + 6x + 5 (x + )(x + )

Find factors of 5 that add to 6:

1*5 = 5 1+5 = 6 (x + 1)(x + 5)

Factoring Trinomials where a ≠ 1BY GROUPING! IT’S FASTER! Follow these steps:

1. Find two numbers that multiply to ac and add to b for ax2 + bx + c

2. Replace bx with the sum of the 2 factors found in step 1.

ie: ax2 + bx + c becomes ax2 + mx + nx + c, where m and n are the factors found in step 1.

3. Use grouping to factor this expression into 2 binomials

2x2 + 5x + 2

Step 1: ac = 2*2 = 4 1*4 = 4 1+4 = 5 2*2 = 4 2+2 = 4 m = 1 and n = 4

Step 2: Rewrite our trinomial by expanding bx 2x2 + 1x + 4x + 2

Step 3: Group and Factor (2x2 + 1x) + (4x + 2) x(2x + 1) + 2( 2x + 1) (2x + 1) (x + 2)

2x2 + 5x + 2

Questions for thought:

1. Does it matter which order the new factors are entered into the polynomial?

2. Do the parenthesis still need to be the same?

3. Will signs continue to matter when finding m and n?

4. Does it matter how we group the terms for factoring?

3z2 + z – 2

Step 1: ac = 3*-2 = -6 -1*6 = -6 -1+6 = 5 1* -6 = -6 1+-6 = -5 -2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1

m = -2 and n = 3Step 2: Rewrite our trinomial by expanding bx

3z2 + 3z – 2z – 2 Step 3: Group and Factor

(3z2 + 3z) + (-2z - 2) 3z(z + 1) - 2( z + 1) (z + 1) (3z - 2)

3z2 + z – 2

Step 1: ac = 3*2 = 6 -1*6 = -6 -1+7 = 6 1* -6 = -6 1+-7 = -6 -2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1

m = -2 and n = 3Step 2: Rewrite our trinomial by expanding bx

3z2 + 3z – 2z – 2

Notice that I changed the order of m and n between step 1 and step 2. Why do you think I did this? Do you have to change the order to get the correct answer?

3z2 + z – 2

What are the 3 steps for factoring this quadratic equation? Step 1: Multiply a*c. Find the factors that multiply

to ac and add to b Step 2: Expand bx to equal mx + nx Step 3: Group and Factor

4x3 – 22x2 + 30x

Step 0: Factor out the GCF: 2x

2x(2x2 – 11x + 15) Step 1: a*c = 30

-1*-30 = 30 -1+-30 = -31

-2*-15 = 30 -2+-15 = -17

-3*-10 = 30 -3+-10 = -13

-5*-6 = 30 -5+-6 = -11

4x3 – 22x2 + 30x

Step 0: Factor out the GCF: 2x

2x(2x2 – 11x + 15) Step 1: a*c = 30

-1*-30 = 30 -1+-30 = -31

-2*-15 = 30 -2+-15 = -17

-3*-10 = 30 -3+-10 = -13

-5*-6 = 30 -5+-6 = -11

4x3 – 22x2 + 30x

Step 2: Expand bx to equal mx + nx-11x = -5x + -6x

2x(2x2 – 5x – 6x + 15) Step 3: Group and Factor

2x((2x2 – 5x )(– 6x + 15))

2x(x(2x – 5) -3(2x – 5))

2x(2x – 5) (x – 3)

4x3 – 22x2 + 30x

Step 2: Expand bx to equal mx + nx-11x = -5x + -6x

2x(2x2 – 5x – 6x + 15) Step 3: Group and Factor

2x((2x2 – 5x )(– 6x + 15))

2x(x(2x – 5) -3(2x – 5))

Note: The Parenthesis are the Same

2x(2x – 5) (x – 3)

Practice

1. 3x2 + 5x + 2

2. 6x2 + 7x – 3

3. 6 + 4y2 – 11y

Practice

1. 3x2 + 5x + 2

(3x + 2)(x + 1)

2. 6x2 + 7x – 3

(3x – 1)(2x + 3)

3. 6 + 4y2 – 11y

(4y – 3)(y – 2)

Review

What is Step 0? When do you need to include this step?

When will your factors both be negative? When will you have one negative and one

positive factor? How do you check your answers?

When the leading coefficient is negative, factor out –1 from each term before using other factoring methods.

When you factor out –1 in an early step, you must carry it through the rest of the steps and into the answer.

Caution!

Additional Example 4: Factoring ax2 + bx + c When a is Negative

Factor –2x2 – 5x – 3.

–1(2x2 + 5x + 3) –1( x + )( x+ )

Factor out –1. a = 2 and c = 3;

Outer + Inner = 5Factors of 2 Factors of 3 Outer + Inner

1 and 2 3 and 1 1(1) + 3(2) = 7 1 and 2 1 and 3 1(3) + 1(2) = 5

–1(x + 1)(2x + 3) (x + 1)(2x + 3)

??? Questions ???

Additional Example 4: Factoring ax2 + bx + c When a is Negative

Factor –2x2 – 5x – 3.

–1(2x2 + 5x + 3)

–1( x + )( x+ )

Factor out –1.

a = 2 and c = 3; Outer + Inner = 5

Factors of 2 Factors of 3 Outer + Inner

1 and 2 3 and 1 1(1) + 3(2) = 7 1 and 2 1 and 3 1(3) + 1(2) = 5

–1(x + 1)(2x + 3)

(x + 1)(2x + 3)

15.4 Special Types of Factoring

1. Difference of Squares

2. Perfect Square Trinomials

Difference of Squares

Think back to Chapter 5. What happened when we multiplied a sum and difference?

(a – b)(a + b) = a2 – b2

So, the reverse is also true.

a2 – b2 = (a – b)(a + b)

x2 – 25

Notice that we do not have a bx term. This means that we only have the F and L in foil; therefore, none of the procedures from 6.1, 6.2, or 6.3 will work.

We need to use a2 – b2 = (a – b)(a + b)

where a = x and b = 5 X2 – 25 = (x – 5)(x + 5)

x2 – 36

We need to use a2 – b2 = (a – b)(a + b)

where a = x and b = 6 X2 – 36 = (x – 6)(x + 6)

Practice 4x2 – 9

100 – 16t2

49y2 – 64z2

Practice 4x2 – 9

a = 2x, b = 3

(2x – 3) (2x + 3) 100 – 16t2

a = 10, b = 4t

(10 – 4t) (10 + 4t) 49y2 – 64z2

a = 7y, b = 8z

(7y – 8z) (7y + 8z)

Perfect Square Trinomials

Think back to Chapter 5. What happened when we squared a binomial?

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

So, the reverse is also true.

a2 + 2ab + b2 = (a + b)2

a2 – 2ab + b2 = (a – b)2

x2 + 10x + 25

This can be worked 2 different ways The first way is the simplest, but depends on

whether you recognize the equation as a perfect square trinomial.

a2 + 2ab + b2 = (a + b)2

Where a = x and b = 5

x2 + 10x + 25 = (x + 5)2

x2 + 10x + 25

This can be worked 2 different ways The second way is to use the method we learned in

6.2

x2 + 10x + 25

5*5 = 25 and 5+5 = 10

(x + 5) (x + 5) or (x + 5)2

4x2 - 4x + 1

This can be worked 2 different ways The first way is the simplest, but depends on

whether you recognize the equation as a perfect square trinomial.

a2 + 2ab + b2 = (a + b)2

Where a = 2x and b = 1

4x2 - 4x + 1 = (2x – 1)2

4x2 - 4x + 1

This time we need to use the 6.3 method

4*1 = 4

-2 * -2 = 4 and -2 + -2 = -4

(4x2 – 2x) ( – 2x + 1)

2x(2x – 1) – 1(2x – 1)

(2x – 1) (2x – 1) or (2x – 1)2

Practice x2 – 4xy + 4y2

9a2 – 60a + 100

25y2 + 20yz + 4z2

Practice x2 – 4xy + 4y2

a = x, b = 2y

(x – 2y)2

9a2 – 60a + 100

a = 3a, b = 10

(3a – 10) 25y2 + 20yz + 4z2

a = 5y, b = 2z

(5y + 2z)

Review

What methods can you use to factor a Difference of Squares?

What methods can you use to factor a Perfect Square Trinomial?

What clues should you look for to identify a Difference of Squares?

What clues should you look for to identify a Perfect Square Trinomial?

??? Questions ???