Module 17

The XYZ Factors

This module is about factoring polynomials. This deals on products of polynomials whose terms have a common monomial factor, trinomials which are products of two binomials, trinomials which are squares of a binomial, and products of the sum and difference of two quantities. From time to time, you will be referred to the lessons in the preceding module on special product because special product and factoring are reverse processes. You will see also how factoring can be applied widely in many mathematics problems.

This module includes the following lessons:

Lesson 1

/

Factoring and Common Factors

Lesson 2

/

Factoring Expressions of the Form x2 + bx + c

Lesson 3

/

Factoring Expressions of the Form ax2 + bx +c

Lesson 4

/

Factoring Perfect Square Trinomial

Lesson 5

/

Factoring the Difference Between Two Squares

Lesson 6

/ Sum and Difference of Two Cubes

After working on this module you should able to:

Factor polynomials in each of the following cases:

polynomials whose terms have a common monomial factor

trinomials which are products of binomials

trinomials which are squares of a binomial

difference of two squares

sum and the difference between two cubes

This is your guide for the proper use of the module:

  1. Read the items in the module carefully.
  2. Follow the directions as you read the materials.
  3. Answer all the questions that you encounter. As you go through the module, you will find help to answer these questions. Sometimes, the answers are found at the end of the module for immediate feedback.
  4. To be successful in undertaking this module, you must be patient and industrious in doing the suggested tasks.

5. Take your time to study and learn. Happy learning!

I. Multiple Choice. Choose the letter of the correct answer.

1. If one of the factors of 18x5y + 12x4y – 6x3y + 3x2y is 3x2y, what is the other factor?

a. 6x4 + 4x3 – 2x2 + xc. 6 + 4x – 2x2 + x3

b. 6x3 + 4x2 – 2x + 1d. 6x + 4x2 – 2x3 + x4

2. Which of the following expressions has (x – 4) and (x + 3) as factors?

a. x2 + 7x + 12c. x2 – 7x – 12

b. x2 + x – 12d. x2 – x – 12

3. Which of the following is TRUE?

  1. 25x2 – 120xy + 144y2 = (5x – 12y)2
  2. 9a2 – 24a – 16 = (3a – 4)2
  3. 16x2 + 12xy + 9y2 = (4x + 3y)2
  4. x2 – y2 = (x – y)2

4. Which of the following is a factor of 9a2 – 25b2?

  1. 3a – 5bii. 3a + 5b

a. ic. i and ii

b. iid. cannot be determined

5. Which of the following are factors of a3 – 8b3?

  1. a – 2b; a2 – 4ab + b2
  2. a – 2b; a2 – 2ab + 4b2
  3. a – 2b; a2 + 4ab + 4b2
  4. a – 2b; a2 + 2ab + 4b2

II. Determine the factors of the following polynomials.

A.

  1. 3x+ 12
  2. 10m + 6m2
  3. xy2z3 – x2y3z4
  4. ay + by – cy + 3y
  5. x2 – 49
  6. x4 – y4
  7. (a – b)2 – 16

B. 1. a2 + 14a + 49

2. m2 + 6n + 8

3. t2 – 15t + 16

4. x6 – 27

5. 8a3 + 125

6.

7. 2x2 – 5x – 12

8. 4n2 – 15n + 9

C. Answer these questions:

  1. If the area of a square is 25x2 + 30x + 9, what is the measure of its side?

b. A rectangle has an area of x2 + 12x + 32. If the length is represented by x +8, what is the width of the rectangle?

Lesson I Factoring and Common Factor

In the previous module, you studied about special ways of finding the product of polynomials. The results are called special products. This time, we will do the other way around. You will be given the product and you will determine its factors. This process is factoring. It is the reverse process of multiplying. When we factor expressions, we determine an equivalent expression that is a product of two or more expressions.

Before we proceed further, let us review what you have learned about factors, prime factorization and greatest common factor.

Following are some examples that will enable you to recall some concepts you have studied before.

  1. Factor: a) 36b) 54c) 72

Solution: For each given number, we may write the factorization process in a form of factor tree as shown here:

36 54 72

9 x 4 9 x 6 9 x 8

3 x 3 x 2 x 2 3 x 3 X 2 x 3 3 x 3 x 2 x 2 X 2

Look at the darkened numbers in the last row. They are the prime factorization of 36, 54 and 72.

Why do you think we call these the prime factorization of the given number?

We say that if a given number is written as a product of prime factors, they are factored completely.

From the prime factorization as shown , we can determine the Greatest

Common Factor (GCF) of the numbers 36, 54, and 72.

36 = 3 x 3 x 2 x 2

54 = 3 x 3 x 2 x 3

72 = 3 x 3 x 2 x 2 x 2

GCF = 3 x 3 x 2

= 18

We say that 18 is the largest number that could divide 36, 54 and 72 exactly. It is also called their greatest common divisor.

The concept of GCF can now be extended to monomials and polynomials.

Study the following examples:

1. Find the GCF of 4x2, 8x4 and 12x6

SOLUTION:

We factor each of the given monomials where the highest common factor can be determined among the three monomials.

4x2 = 4x2  1

8x4 = 4x2  2x2

12x 6 = 4x2  3x4

GCF = 4x2

The greatest common factor of 4x2, 8x4, and 12x6 is 4x2.

2. Find the GCF of 8m3n4 – 20m2n3 -10mn2

8m3n4 = 2mn2 x 4m2n2

-20m2n3 = 2mn2 x -10mn

-10mn2 = 2mn2 x - 5

GCF = 2mn2

The processes of prime factorization and finding the GCF will be very useful in solving problems involving factoring.

Remember:

Factoring is the process of obtaining equivalent expression that

is the product of two or more expressions.

A. Common Monomial Factor

Suppose you are given a rectangle whose area is represented by

10x2 +15xy, with a width of 5x, how do we express the length of this rectangle?

A situation like this can be solved with the knowledge of common monomial factor.

Study the illustrative examples below on factoring polynomials whose terms have a common monomial factor.

Recall that in multiplying monomials and polynomials; you use the distributive property. That is, you multiply the monomial to each of the terms in the polynomial factor.

To illustrate:

x ( b + c +d) = bx +cx + dx.

As such, x is a common factor of bx + cx + dx.

This time, we do the reverse. We look into the polynomials product

bx + cx + dx and determine the factor common among the terms.

Examples:

1. Factor 9y4 – 15y3 + 3y2

Solution:

  • Find the monomial of highest degree in the polynomial which can be factored from each term.

Consider both the literal factors and the numerical factors.

a. The lowest degree of the literal factors in the terms 9y4, -15y3 and 3y2 is y2

and therefore y2 is the highest power which can be factored from the literal factors in each term.

b. Find the GCF of the numerical coefficient 9, 15 , 3

GCF (9, 15, 3 ) = 3

c. Find the product of the greatest numerical factor and literal factor.

Therefore, the greatest common monomial factor is 3y2.

9y4 – 15y3 + 3y2

3y2 3y2 3y2 –5y 3y2 1

  • Divide each term by 3y2 to get the terms of the remaining polynomial factor.

3y2 ( 3y – 5y + 1)

Therefore: 9y4 – 15y3 + 3y2 = 3y2 ( 3y2 – 5y + 1 )

2.Factor : 6a + 8a2b – 14a3b2

= (2a  3) + (2a  4ab) – (2a  7a2b2) (Factoring out the greatest common factor 2a)

= 2a (3 + 4ab – 7a2b2) (Dividing each term by the GCF 2a to get the remaining polynomial factor )

3. Factor: 18x5y + 12x4y – 6x3 + 3x2y

Solution:

18x5y + 12x4y – 6x3 + 3x2y

=(3x2y  6x3) + (3x2y 4x2) – (3x2y  2x) + (3x2y 1) ( Factoring out 3x2y)

= 3x2y(6x3 + 4x2 – 2x + 1) (Dividing each term by 3x2)

I. Factor the following expressions.

1. 15x2 + 21y24. 13a2b – 26ab2

2. 12x2 + 20xy5. 14abc + 28a3b3c3

3. 15xy2 + 25xy

II. Answer the following problem:

1. If the area of a rectangle is represented by 6a2 + 15ab, and its width is 3a,

what is the length of this rectangle?

2. The perimeter of the rectangle below is 10a + 6b and the width as indicated . How do you represent the measure of the length of the rectangle?

Lesson 2 Factoring Expressions of the Form x2 + bx + c

In Lesson 1, you were asked to factor polynomials whose terms have common monomial factor. Here, you will befactoring trinomials of the form

ax2 + bx+ c

where a = 1.

If a = 1, the trinomial will also be in the form x2 + bx + c.

We may begin with a review of finding the product of binomial, say :

( x+6)(x+2) = ?

Do you still remember the FOIL Method? It will be useful when factoring polynomials in the form x2 + bx + c as shown here:

F O I L

(x + 6) (x + 2) = x2 + 2x + 6x + 12

= x2 + 8x + 12

If we do the reverse process, that is, given the product, how do we determine the factors?

In the polynomial product, x2 + 8x + 12 , you cannot find any common factor among the 3 terms. Therefore, factoring as discussed in Lesson 1 does not apply here. Instead, another way of factoring this trinomial is applied.

To factor x2 + 8x + 12, think of the reverse of the FOIL Method. Referring to the above example, note the following:

  • x2 is the result of x  x. Therefore, the first term of each binomial factor is x. We can write (x + _ ) (x + _)
  • The coefficient of the middle term of the trinomial are two numbers whose product is 12 and whose sum is 8.

Here, you first think of two numbers when you multiply will give 12.

We may write these pairs of numbers in a table.

Product of 12 / Sum
4, 3 / 7
-4, -3 / -7
6, 2 / 8
-6, -2 / -8
1, 12 / 13
-1, 12 / -13

The right column contains the sum of the pairs of factors of 12. Ask yourself “What are the factors of +12 that will give the sum of 8?”

The 3rd row satisfies the two conditions : product is 12 and sum is 8.

The numbers we need are 6 and 2.

We now write the factorization:

x2 + 8x + 12 = (x +6) (x + 2)

Example 2:

Factor y2 – 9 y + 20

Since the last term is positive and the middle term is negative, it means we need two negative numbers whose product is 20. and whose sum is -9.

Product of 20 / Sum
-1. -20 / -21
1, 20 / 21
, 5 / 9
-4, -5 / -9
2, 10 / 20
-2, -10 / -20

The numbers are -4 and -5.

So, y2 -9y + 20 = (y-4) (y-5)

We may now check :

(y – 4) (y – 5) = y2 – 5y – 4y +20

= y2 – 9y +20

In the preceding examples, C , the constant term is positive or c> 0. However, there are cases when the constant term of the trinomial is negative or C < 0.

Example 3: Factor x2 + 3x -10

In this case, we look for two numbers whose product is the last term, -10 and whose sum is equal to the numerical coefficient of the middle term, +3 . Since the constant term is -10, it has a positive factor and a negative factor. The mental process maybe written this way:

Product -10 / Sum
-1, 10 / 9
1, -10 / -9
2, -5 / -3
-2, 5 / 3

Thus, x2 + 3x – 10 = (x-2)(x+ 5)

We check:

( x – 2 ) ( x +5 ) = x2 + 5x -2x -10

= x2 + 3x - 10

Example 4 : Factor: y2 -17y – 18

Find two numbers whose sum is -17 and whose product is -18

Product -18 / Sum
9, -2 / 7
-9, 2 / -7
-1, 18 / 17
1, -18 / -17
-3, 6 / 3
3, -6 / -3

Thus, y2 -17y – 18 = ( y + 1)(y -18)

I. Factor each of the following:

1. a2 + 8a + 154. y2 + 7y – 18

2. p2 – 3p – 105. b2 – 5b + 6

3. x2 – x – 20

II. Answer these problems:

1. A balloon flies at a speed of ( x – 2) miles per hour . At this rate, how long will it take to fly ( x2 + 6x – 16) miles?

2. If the area of a rectangle is m2 + 4m +3, what two binomials could represent the length and the width of the rectangle?

Lesson 3 Factoring Expressions of the Form ax2 + bx +c

In Lesson 2, you were introduced to factoring trinomials of the form

ax2 + bx + cwhere a = 1.

In the following examples, you will see how to factor trinomial of the form

x2 + bx + c where a ≠ 1 . The skills you have developed in the previous lesson will be used here.

Consider this product:

F 0 I L

( 2x + 1 ) ( 3x + 2 ) = 6x2 + 4x + 3x + 2

Factors = 6x2 + 7x + 2

As you can see, the numerical coefficient of the first term of the two binomial factors is not equal to 1. As such, the numerical coefficient of the first term of the trinomial product is not equal to 1 also. This leads to more combinations of possible factors.

Like in Lesson 2, think of the reverse of the FOIL Method to determine the factors of a given polynomial. To illustrate how these trinomials are factored, here are some examples:

Example 1. Factor 5x2 – 18x – 8

Solution:

Step 1. Determine if there is a factor common to all terms. There is none.

Step 2. Look at the first terms. Determine two numbers whose product is 5.

1, 5–1, –5

Step 3. Now, look for numbers whose products is -8.

–4, 2 ; 4, –2

1, –8 ; 8, –1

  • Since the last term is negative, the second terms in each factor must be of different signs. Here are some possible combinations in the factorization.
  1. ( x + 2) ( 5x – 4)
  2. ( x + 1) ( 5x –8)
  3. ( 5x – 1) ( x + 8)
  4. ( 5x + 1) ( x – 8)
  5. ( x – 1) ( 5x + 4)
  6. ( x – 1) ( 5x + 8 )
  7. ( x – 4) ( 5x + 2 )
  8. ( x + 4) ( 5x – 2)

1

In each case, when we multiply the first terms, the product is 5x2, and the last erm is –8.

Step 4. Look again at the 8 possible combinations. If you apply the distributive property and combine like terms, which pair will give us the middle term –18?

The combination in d gives us the correct middle term. Thus.

5x2 – 18x – 8 = ( x – 4) (5x + 2)

Check:

( x – 4) (5x + 2) = 5x2 + 2x – 20x – 8

= 5x2 – 18x – 8

Example 2:Factor 2x2 – 7x – 4

Think :

For the first terms: Find two numbers whose product is 2.

For the last terms: Find two numbers whose product is –4

Determine possible factorizations among the combination of factors:

1

  1. ( 2x – 4) ( x + 1)
  2. ( 2x – 1) ( x + 4)
  3. ( 2x + 1) ( x – 4)
  4. ( 2x + 2) ( x – 2)
  5. ( 2x – 2) ( x + 2)

From the above combination, select the pair which will satisfy this condition:

The outside product plus the inside product must equal –7x.

Combination c satisfies the given condition.

Example 3:Factor 20 + 6x – 2x2

20 + 6x – 2x2 = 2 ( 10 + 3x – x2)Factoring the common factor 2

Yes, you have already factored. But the product as written may not be completely factored yet. Examine the polynomial factor:

10 + 3x – x2

.

This polynomial can still be factored!

Think:

First terms: Find two numbers whose product is 10

Last terms: Find two numbers whose product is –x2

Determine the possible factorization:

  1. ( 5 + x) ( 2 – x)
  2. ( 5 – x) ( 2 + x)
  3. (10 + x) ( 1 – x)
  4. (10 – x) ( 1 + x)

The outside product plus the middle product must equal 3x.

Combination b satisfies this condition.

Thus, the complete factorization is:

20 + 6x – 2x2 = 2 ( 5 – x) ( 2 + x)

A. The questions below are given to help you factor 4x2 – x – 18. Answer each question.

1. The product of the numerical coefficient of the first term (4x2) and the last term (-18) is -72. What are the possible pairs of factors of -72?

2. The factors -9 and 8 give a sum of -1.

How do you write 4x2 – x – 18 such that -x is expressed in terms of -9 and 8?

Supply the missing part in the statement below.

3. 4x2 - + – 18 = (4x2 – 9x) + (8x – 18)

4. (4x – 9) + ( 4x – 9)

5. (4x – 9) ( + )

B. Factor the following expressions.

1. 2a2 + a – 154. 3x2 + x – 10

2. 3x2 – 8x – 35. 12x2 – 5x – 2

3. 2x2 – 5x - 12

C. Answer this problem:

  1. The area of the rectangle below is 6x2 + 22x + 20 and the other measures are shown in the figure. What are the measures of a and b?

3x 5

Lesson 4 Factoring Perfect Square Trinomial

Let’s recall the process of squaring a binomial.

( x + 4 )2

x2 + 2(4)(x) + 42

= x2 + 8x + 16

( x – 6 )2

x2 – 2(6)(x) + 62

= x2 – 12x + 36

(5x + 4 )2

(5x)2 + 2 (5x)(4) + 42

=25x2 + 40x + 16

In the above examples, we have squared the binomials ( x + 4), ( x – 6), and (5x + 4). When you square a binomial the resulting product is a perfect square trinomial. It is also called a trinomial square. The two factors of perfect square trinomial are exactly alike.

How do we check if a given trinomial is a perfect square?

A perfect square trinomial can be recognized by inspection and the factors can be identified and written at once.

Following are perfect square trinomials:

1. x2 + 14x + 49

2. 4y2 – 12y + 9

3. 25a2 + 20a + 4

4. 9n4 + 30n2 + 25 = ( 3n + 5)2

5. 64 – 16x + x2 = ( 8 – x )2

Examine again the above trinomials.

  • What can you say about the first and the third term in each trinomial?
  • What is the sign of the first and third term in each?
  • How is the middle term of the trinomial related to the factor of the first and third term of the polynomial?

We can now write:

The trinomials given above are perfect squares for the following reasons:

__ __

1. x2 + 14x + 49 because 14x = 2√x2 √49

= 2 (x)(7)

= 14x

__ __

2. 4y2 -12y +9 because -12y = 2√4y2 √9

= 2 (2y)(3)

= 12y

___ __

3. 25a2 +20a +4 because 20a = 2√25a2 √4

= 2(5a)(2)

= 20a

Apply this test to trinomial d and e.

Examine how factoring is done in each polynomial given below

1. x2 + 14x + 49

2

x 7

 Since middle term is +,

x2 + 14x + 49 = (x + 7)2

2. 4y2 – 12y + 9

2

2y 3

 Since the middle term is –,

4y2 -12y + 9 = (2y – 3)2

3. 25a2 + 20a + 4

2

5a 2

Since the middle term is +,

25a2 + 20a + 4 = (5a + 2)2

4. 9n4 + 30n2 + 25 = (3n + 5)2

5. 64 – 16x + x2 = (8 – x )2

Check the factoring done in numbers 2, 3, 4 & 5 by finding the special product (square of a binomial).

Remember:

A. Tell whether each of the following is a perfect square trinomial or NOT. Give reason for your answer.

1. x2 + 12x + 363. 9a2 – 12ab + 4b2

2. x2 + xy + y24 . 4x2 + 10xy + 25y2

5. x2 – x + 1

B. Factor the following polynomials:

1. a2 + 6a + 94. 25z2 + 10z + 1

2. b2 – 16b + 645. x2 – 3x – 4

3. 4y2 – 4y + 1

C. Answer the following problems:

Refer to the figure below:

  1. If the area of a square ABCD is 16a2 + 40a + 25 units, what is the measure of its side? What is the perimeter of this square?

2. Given the side a+3, what is the area of the smaller square?

3. What is the total area of the two squares?

4. What is the total distance around the two squares?

A B

E a + 3F

D C G

Lesson 5 Factoring the Difference Between Two Squares

Here, you will be introduced to another type of factoring. This is the easiest type of factoring. As mentioned at the beginning of this module, factoring is associated with finding the product, that is, the reverse process of multiplying.

1

Recall finding the special product in this form:

( x + 8) ( x – 8 ) = x2 – 64

( 2x + 4) ( 2x – 4) = 4x2 – 16

( 3a + 5) ( 3a – 5) = 9a2 – 25

The product of the binomial on its left can be described as the difference between two squares. You see that the first and the second term are perfect squares and the sign between them is minus.

Likewise, the above examples illustrate the pattern for factoring polynomial expressed as difference between two squares.

Here are other examples:

  1. x2 – 100

2

x 10

 ( x + 10) ( x – 10)

  1. a2 – 144

2

a 12

 (a + 12 ) (a – 12 )

  1. 9m2 – 49 = ( 3m + 7) ( 3m – 7)
  2. 16x4 – 225= ( 4x2 + 15) ( 4x2 – 15)
  1. x8 – 16 = ( x4 + 4) ( x4 – 4 )

= ( x4 + 4) ( x2 + 2) ( x2 – 2) (Factoring further x4 - 4 )

We now generalize how to factor an expression written as difference of two squares:

Factoring difference of two squares lends itself to other numerical computations.

Study the examples below:

68 x 72

( 70 – 2) x ( 70 + 2)

=702 – 4Apply special product on difference between two squares

= 4900 – 4Multiply

= 4896Subtract

Find the product of 68 and 72 by the long method. Compare the results.

Give your own example. Check the results.

I. Factor each of the following

  1. a2 – 16
  2. 9x2 – 4
  3. 64c2 – 1
  4. 100y2 – 49z2
  5. y2 – 81

II. Simplify by applying special product and factoring.

  1. 49 x 51
  2. 17 x 23
  3. 99 x 101
  4. 44 x 56

III. Solve:

  1. A 5m x 5m landscaping is to be done in one corner of a 20m x 20m field. Find the area of the field that was not affected by the project.

5m

5m

20m

20m

Lesson 6 Sum and Difference of Two Cubes

The expressions a3 + b3 and a3 - b3 represent polynomials that are sum of two

cubes and difference of two cubes, respectively.

How do we express a3 + b3 as a product of two factors?

It can be done this way:

Let us divide a3 + b3 by a + b by the usual way of dividing a polynomial by a polynomial:

2 3

3 3

3 2

2

2 2

2 3

2 3

To check, we multiply the quotient and the divisor:

( a + b ) ( a2 – ab + b2) = a3 – a2b + ab2 + a2b – ab2 + b3

= a3 – a2b + ab2 + a2b – ab2 + b3

= a3 + b3

The product gives the dividend in the problem!

Therefore:

(a + b ) ( a2 – ab + b2) = a3 + b3

or

(a3 + b3) = (a + b ) ( a2 – ab + b2)

Similarly, if we divide a3 – b3 by (a – b ), we get

2 2

3 3

3 2

2

2 2

2 3

2 3

To check:

( a - b ) ( a2 + ab + b2) = a3 + a2b + ab2 - a2b – ab2 + b3

= a3 + a2b + ab2 - a2b – ab2 + b3

= a3 - b3

Therefore:

(a - b ) ( a2 + ab + b2) = a3 + b3

or

(a3 - b3) = (a - b ) ( a2 + ab + b2)

Let us go back and summarize the two processes we have done:

a3 + b3 = ( a + b ) ( a2 – ab + b2)

a3 – b3 = ( a - b ) ( a2 + ab + b2)

Take note of the patterns of the factors:

  • In each case, one factor contains two terms, the other factor contains three terms.
  • The first binomial factor is made up of the cube roots of the two cubes in the binomial expression. The sign between them is the same as the sign between the cubes.
  • The second factor, the trinomial, in each case is made up of the squares of the cube roots, with the product of the cube roots as the middle term. Further note that if it is sum of two cubes, the middle term of the trinomial factor is negative ( – ). If it is the difference of the two cubes, the middle term of the trinomial factor is positive ( + ). In either case, the two factors have only one minus ( – ) sign.

To further illustrate these, examine the following examples:

Example 1: Factor : a3 + 8 [ sum of two cubes]

Solution: a3 + 8

3 +

a + 2

opposite sign

( a + 2 ) (a2 – 2a + 22 )

square of the square of the second term

first term of the of the first factor

first factor

negative of product of the first term and

second term of the first factor

= (a + 2 ) ( a2 – 2a + 4)

Example 2: Factor 8x3 + 27y3

Solution: 8x3 + 27y3 = ( 2x)3 + (3y)3

= [ 2x + 3y ][ (2x)2 – (2x)( 3y) + (3y)2]

Opposite sign

= ( 2x + 3y) (4x2 – 6xy + 9y4)

Example 3: Factor 64x3 – 1

Solution:

64x3 – 1 = ( 4x – 1)[ (4x)2 + 4x( 1) + 12]

opposite sign

= (4x- 1) (16x2 + 4x + 1)

I. Factor the following.

1. 1 + 8x3

2. y3 + 64

  1. a3 – 125
  2. 125y3 – 8z3
  3. 27y3 – 125

Lesson 7 Factoring by Methods of Grouping and Combination of Various Types

Some polynomials where terms have no common factor if taken as a whole. A common factor maybe present in two terms and the remaining two terms having another common factor.

For example, ax + x + 4y+ ay

You could see that if taken as 1 group, the four terms do not contain any common factor. However, x is common to both first and second term, and y, common to the third and fourth term. In such a case, the polynomial can be separated into groups of terms which have a common factor.

That is:

ax + 4x + 4y + ay = ( ax + 4x) + ( 4y + ay)

= a ( a+ 4) + y ( 4 + a)Factoring each binomial

= x( a + 4) + y (a + 4)Commutative Property of Addition for 4 + a

= (a + 4)(x + y)Factoring out the common factor a+4

Here are other examples:

1. Factor : x3 – 2x2 – 3x + 6

Solution: