ValladolidPreparatoriaInstitute page9

page 10 FACTORING

CONCEPT

To understand the theoretical concept of this topic, it is necessary to recall what was mentioned on the page-
refers to the name given to the amounts based on the operation they are performing.

It was said that FACTOR is the term used to refer to any quantity, whether in Arithmetic or Algebra.
that "is playing the sport" called MULTIPLICATION. In more technical terms, a factor is
every quantity that is being multiplied by another.

For example, in the operation 23 × 14, as the number 23 "is playing the sport" called multi-
In application, it is called a factor. Technically, since 23 is being multiplied by another number, this
it is a factor. The same can be said of the number 14.

FACTORIZING a quantity or expression means finding its factors, that is, those numbers
that when multiplied give that amount. For example, factoring the number 6 means finding the numbers
which multiplied together give 6. They are 2 and 3, since 6 = 2 × 3. Factoring 6 means writing it as
shape 2 × 3.

When it comes to an algebraic expression, factoring it is also writing it in a way that its
the main operation is multiplication. Observe the following cases: if we have the expre-
sión2x2+ 9x− 5, its main operation is addition (and subtraction), therefore, what is written there
they are terms, not factors. Factored, it looks like (2x - 1)(x + 5), where the main operation
it is multiplication, so there are factors there. That's why it is factored. Of course, the student-
You should not worry at this moment about how to factor an expression like the one before.
Well, that has not been explained yet. Instead, if we have the expression 6ax.− 2bx− 3ay+ byy
this is written in the form 2x(3a-b) -y(3a-b), it has not been factored since in this last expression
The main operation is subtraction, not multiplication.

In Arithmetic, it is relatively easy to factor a number. Thus, to factor 36, which means
the same as asking, 'what numbers multiplied give 36?', it can even be figured out mentally.
that 36 = 2 × 2 × 3 × 3; however, for algebraic expressions, factoring is no longer so obvious.
tion, so a detailed study is required. Certainly, there are very algebraic expressions
elementary expressions that can be easily factored, such as for example, 6a2It is simple to deduce that
it is equivalent to the multiplication of 2 × 3 × a × a × b; however, the matter gets complicated if one asks
What numbers or amounts multiplied together give 2x?2+ 9x - 5 ?

For this reason, it is necessary to classify algebraic expressions into different cases for factorization.
It should be clear that the number assigned to each factorization case is not its universal name.
Go in the language of Mathematics, simply that as they are going to be numbered, some case has to be
the number 1, another the number 2, and so on. In contrast, the name by which each one appears
Of those cases, it does correspond to a universal name.
ValladolidPreparatoryInstitute page11

CASE 1: COMMON FACTOR

"Common" means that they are or belong to everyone. In such a way that the common factor contains the
meaning of the quantity(s) that appear multiplying in all the terms of the expression.
Remember that you are terming what 'plays to the sum', that is, the amount that is being summed.

2×3+7×3

Analysis: There are two terms, meaning there are two quantities being added: one is 2 × 3; the other
is 7 × 3.
In each term there are two factors. In the first term 2 × 3 the factors are 2 and 3 (because
they are multiplying). In the term 7 × 3 the factors are 7 and 3.
In each of the two previous terms, there is a factor that appears in all of them, that is, it is common.
which is the 3.
Therefore, in 2 × 3 + 7 × 3, the common factor is 3.

To factor an expression where at least one common factor appears in all its terms, one must
it has the following rule:

Factorization by common term:

All common factors are identified and written in their highest form.
A parenthesis is written next, and inside it is what remains of the expression.
original after removing the common factors from each term.
In case the common factor is just one of the terms of the original expression,
instead it is placed.

In the previous rule, it must be clear that the statement 'after having removed from each term
The common factors should not be understood as simply erasing them or making them disappear, but rather it is
equivalent to performing a division of each term of the original expression by the common factor, already
what is being multiplied (factor) is removed through its inverse operation, which is precisely
the division.

Example 2: Factor 2a3b plus 7bxy5

Solution: All common factors are located and written down: in this case it is lab.
A parenthesis is written below and inside it what remains of the original expression then.
by removing the common factors from each term:

b(2a3+ 7xy5).

Finally means that 2a3b plus 7bxy5Invalid input3+ 7xy5) .

page 12 FACTORING

Example 3: Factor 4a2b + 6abx5

Solution: All common factors are located and written: in this case it is 2ab.
A parenthesis is written below and inside it what remains of the original expression then
by removing the common factors from each term:

2ab(2a + 3x5).

Finally means that 4a2b + 6abx5= 2ab(2a + 3x5Note that in this last expression,
the main operation is multiplication.

Example 4: Factor 12a4b3c- 6a2b3x7

Solution: All common factors are located and written down: in this case it is 6a2b3.
A parenthesis is written next, and inside it what remains of the original expression.
by removing the common factors from each term:

6a2b3(2a2a-x7)

Finally means that 12a4b3c- 6a2b3x7= 6a2b3(2a2c-x 7Note that in this last
expression, the main operation is multiplication.

Example 5: Factor 5b2cx- 60a2b 2c5x2

Solution: All common factors are located and written down: in this case it is 5b.2cx.
A parenthesis is written next, and inside it what remains of the original expression then
by having removed the common factors from each term. Since the common factor is the entire first
the term of the original expression is replaced by 1:

5b2cx(1 - 12a 2c4x)

* Finally means that 5b2cx- 60a2b 2c 5x 2= 5b2cx(1 - 12a2c 4x) . It is observed that in this last
expression, the main operation is multiplication.

Example 6: Factor 8b220a2b2+ 16ab3c4

Solution: All common factors are identified and written down: in this case it is 4b.2.
A parenthesis is written below and inside it is what remains of the original expression.
by removing the common factors from each term:

4b2(2 - 5a2 + 4abc4).

Finally means 8b2- 20a2b2+ 16ab3c4= 4b2(2 - 5a2+ 4abc4Note that in this
last expression (right of the equal sign), the main operation is multiplication.
ValladolidPreparatoryInstitute page13

EXERCISE 4

Factor the following algebraic expressions:

1) 3b - 9a 2) 2ax - 9x2 3) a2d5x + 2a3a2

4) 6a5b - 9aba + 3b7 5) 12a5-9x2+ 9b3x3 6) 8ba2x + 2ab3a2x34a2b2a2
7) 14a2b2+ 21ab8a + 35a4b7C2 8) 12a3b4x - 9b4x4+ 36x3 9) bc2+ 22ab2c4x - 14a2bc2
10) 40b6+ 8abc- 35a9b4c 11) 26a5b5x- 13b4x + 39a2b7x3 4b2c2+ 4b2c44bc2
13) 6ab6c + 3abc - 3a 6b8c 14) 6a2b4d- 6b3d3x + 16a2b9x2 15) 40b5c5+ 20a6b9c4100a6b5c4
16) 8a7cf+ 3ab2cf- 3a2c 17) ab4d-b7d8x+a2 18) 40c3+ 2b11c- 100

CASE 2: BY GROUPING

The process consists of forming groups or grouping terms in equal quantities (two by two, or by...
three by three, etc.), to then factor each group by common factor and finally refactor
common factor, where the parentheses that must remain repeated in each group is the common factor.

As a practical rule, the sign of the first term of each group is the sign that must be placed in each
factorization by common factor.

Example 1: Factor 2ac + bc + 10a + 5b

Solution: Two groups are formed, one with the first two terms and the other with the other two terms.

2ac+ bc+ 10a+ 5b

Factoring each group by common factor: The first group has accommodation as a common factor, while
that the second group has the 5. Thus it turns out that:

2ac + bc + 10a + 5b = c(2a + b) + 5(2a + b)

Note that in this last expression (the one to the right of the equal sign), the main operation is
the sum, so it is not yet factored.

Going back to factoring by common factor, since the repeated parentheses are that common factor, end-
it is obtained that

2ac + bc + 10a + 5b = (2a + b)(c + 5).

Note that in this last expression (the one to the right of the equals sign), the main operation is
the multiplication, because it is already factored.
page 14 FACTORING

Example 2: Factorize2b35b4- 6a2+ 30b

Solution: Two groups are formed, one with the first two terms and the other with the other two terms.

a 2b 3− 5b4− 6a2+ 30b

Note that the sign left between each group was negative.

Factoring each group by common factor: The first group has ab3as a common factor, while
that the second group has the 6. So it turns out that:

a2b35b46a2+ 30b = b3(a2- 5b)-6(a2- 5b)

Since the sign that had been left previously between each group was negative, that same sign
it is the one that was placed between each factorization. Note that in this last expression (the one on the right
Since the equal sign is not present, the main operation is subtraction, so it is not yet factored.

Re-factoring by common factor, since the repeated parentheses is that common factor, final-
it is obtained that

a2b35b46a2+ 30b= (a2- 5b)(b3- 6)

Note that in this last expression (the one to the right of the equal sign), the main operation is the
multiplication, since it is already factored.

Example 3: Factor 3ab + x2- 21ab2- 7bx2

Solution: Two groups are formed, one with the first two terms and the other with the other two terms.

3ab+ x 2− 21ab2− 7bx2

Note that the sign that remained between each group was negative.

Factoring each group by common factor: The first group has 1 as a common factor, while
that the second group has the 7b. So it turns out that:

3ab + x2- 21ab2- 7bx2= 1(3ab+x2-7b(3ab+x)2)

Since the sign that had remained previously between each group was negative, that same sign
it is the one that was placed between each factorization. Note that in this last expression (the one on the right
The operation is subtraction, so it is not yet factored.

factoring by common factor, since the repeated parenthesis is that common factor, it is obtained that
ValladolidInstitutePreparatory page15

3ab + x221ab2- 7bx2(3ab + x2(1 - 7b)

Note that in this last expression (the one on the right of the equal sign), the main operation is
the multiplication, because it is already factored.

Example 4: Factor 2ab2+b3- 5b2+ 6a + 3b - 15

Solution: In this case, the fact that there are six terms suggests that two groups of three can be formed.
terms each. Two groups are then formed, one with the first three terms and the other with
the other three terms.

2ab2+ b 3− 5b2+ 6a+ 3b− 15

Note that the sign that remained between each group was positive.

Factoring each group by common factor: The first group has ab.2as a common factor, while
that the second group has the 3. Thus it results that:

2ab2+b35b26a + 3b - 15 = b2(2a+b-5) + 3(2a+b-5)

Since the sign that had previously remained between each group was positive, that same sign
it is the one that was placed between each factorization. Note that in this last expression (the one on the right
the equal sign), the main operation is addition, so it is not yet factored.

Returning to factor by common factor, since the repeated parenthesis is that common factor, final-
it is obtained that

2ab2+b35b2+ 6a + 3b - 15 = (2a + b - 5)(b2+ 3)

EXERCISE 5

Factor the following algebraic expressions:

1) ac + bc + 2ax + 2bx 2) 2a2-4ab - 5a + 10b

3) ab - 1 - abx + x 4) 7ab2+ac- 14b2xy - 2cxy
5) 6ab3x - 4b3+ 21ax - 14 6) x2y3+ 5y3- 3x2- 15
7) 2b3+ 3c3- 2b3x - 3c3x 8) b3c32b2c + bc22
9) 2a - 4b + 2c2-axy + 2bxy - c2xy 10) 5ab - 10 - 5x - abc2+ 2c2+c2x
11) 9x - 8y + 7 - 9a2x + 8a2y - 7a2 12) a2x2-b3x2+x2-a2bc+b4c-bc
13) 2x - y - 2 - 8ax + 4ay + 8a 14) 3a2x - 3b2x - 3x - a2+b2+ 1
15) 10a + 15b + 20 - 6ax - 9bx - 12x 16) 10a - 15b - 20 + 6ax - 9bx - 12x
page 16 FACTORING

CASE 3: DIFFERENCE OF SQUARES

On page 5 it was seen that (a+b)(a-b) = a2-b2It is quite obvious that if the equality is reversed.
rior is still the same:2-b2= . a+ b (
a− Seen )(in this) way, in the reverse of the product
Notably, the factorization of a difference of squares is obtained. Note that in (a+b)(a-b),
the main operation is multiplication.

From the above, the following rule can be written:

A difference of squares is factored into two conjugate binomials, formed with the
square roots of the original terms.

It is important to note that it does not matter whether the sum binomial is written first or the subtraction binomial.
since multiplication is commutative.

Example 1: Factor 4a2-x6

Solution: The square root of 4a2is 2ay dex6esx3In such a way that the conjugate binomials that correspond to it
they weigh son (2a+x3)2a-x3) .

Factoring is: 4a2-x6(2a+x3) (2a - x3).

Example 2: Factor 49a4b6100x2

Solution: The square root of 49a4b 6it's 7a2b 3and of 100x2it's 10x. So the conjugate binomials
which correspond to him are (7a2b3+ 10x)(7a2b3- 10x) .

The factorization is: 49a4b6100x2=(7a2b3+ 10x)(7a2b3-x).

Example 3: Factor 1 - 196a4b16

The square root of 1 is 1 and of 196a4b 16it is 14a2b 8Thus, the conjugate binomials that
they correspond to (1 + 14a2b8(1 - 14a2b 8) .

The factorization is: 1 - 196a4b16=(1 + 14a2b8) (1 - 14a2b8) .

ValladolidPreparatoryInstitute page17

( 5+ b ) 7
2

−
49
Example 4: Factor
9 a6

( 5+ b ) 7
2
5+ b 7 49 7
Solution: The square root of is and of is 3 .
9 3 a 6
a
So the conjugate binomials that correspond to it are

⎛ 5+ b 7 7 ⎞ ⎛ 5+ b 7 7 ⎞
⎜ + 3 ⎟
y ⎜ − 3 ⎟.
⎝ 3 a ⎠ ⎝ 3 a ⎠

and the corresponding factorization is:

( 5+ b ) 7
2

−
49 ⎛ 5+ b 7
= +
7 ⎞ ⎛ 5+ b 7 7 ⎞
− 3 ⎟.
⎜ 3 ⎟⎜
9 a 6
⎝ 3 a ⎠⎝ 3 a ⎠

EXERCISE 6

Factor the following algebraic expressions:

1) 36b29 2) 25a4- 9x2 3) c2-a2c2

4) 64a8b249 5) 16a61 6) 81c2- 25x8
7) 144 - 36a4c2 8) 1 - 9b4x4 9) c2144a2b2
10) 400b6- 81 11) x236b4 12) 4b2c2- 4x2
13) 64b6- 169a6 14) a2b4- 36x2 15) 196b16c25100a16
16) 16f16-a2 17) 64b64-d8x12 18) 400 - 100g100
19) 121 -x8y6 4c4d1616 21) 100a6464b100
22) 1 - 100a6b12 23) 144x144- 64y64 24) 9 -a10b20
25) 196a1216 26) 25x14y6- 64r9 27) 81 -a81b20
28) 36a5016b26 29) x24y69h9 30) 1 - 49d49b2
4 x2 49b16
31) − 121 32) − 64 33) −9
b4 y2 a4

25 1 1 4 9x 8
34 25− 35) − 8 36) −
36a6 4 a 9w12 4

144
−
( a+ b )2 ( 4− x )2
4

−
1 ( 9+ x y )
4 6

−
1
37) 38) 39)
( a− b ) 2
49 9 c8 16 x16
page 18 FACTORING

CASE 4: TRINOMIALS OF THE FORM x2+ bx + c

The form of these trinomials is that there must be a single squared x. The letter represents in ge-
general to any number that goes with lax; and lac represents any number that goes without it
x.

The factorization procedure for these cases consists of finding two numbers, which
They will call each other, which must meet the requirements set out in the following rule:

To factor a trinomial of the form x^22+ bx + c, looking for two integer numbers
what:

Added in b
Multiplied denc.
Each of those found numbers is placed one in each parenthesis,
the following way:

x2+bx+c= (x+m)(x+n)

Example 1: Factor x2+ 5x + 6

Solution: In this case, b = +5 and c = +6.

Two numbers are sought that add up to +5 and multiply to +6. They are +3 and +2.
The sought factors are (x + 3) and (x + 2).
Finally means quex2+ 5x + 6 = (x + 3)(x + 2). Note that in this last expression, the
the main operation is multiplication.

Example 2: Factor x2+ 5x - 6

Solution: In this case, b = + 5 and c = - 6.

Two numbers are sought that add up to +5 and multiply to -6. They are +6 and -1.
The sought factors are (x + 6) and (x - 1).

Finally means quex2+ 5x - 6 = (x + 6)(x - 1). Note that in this last expression, the
the main operation is multiplication.

Example 3: Factor x2-x- 20

In this case, b = -1 and c = -20.

VALLADOLIDINSTITUTEHIGHSCHOOL page19

Two numbers are sought that add up to -1 and multiply to -20. They are -5 and +4.
The sought factors are (x - 5) and (x + 4).

It means quex2-x- 20 = (x- 5)(x+ 4). Note that in this last expression, the operation
the main thing is multiplication.

Example 4: Factorizarx2- 2x - 24

Solution: In this case, b = -2 and c = -24.

Two numbers are being sought that, when added, equal -2 and when multiplied, equal -24. They are +4 and -6.
The sought factors are (x + 4) and (x - 6).

Finally means quex2- 2x - 24 = (x + 4)(x - 6). Note that in this last expression, the
the main operation is multiplication.

Example 5: Factor x2- 17x + 66

Solution: In this case, b = -17 and c = +66.

Two numbers are sought that when added equal -17 and when multiplied equal +66. They are -6 and -11.
The sought factors are (x - 6) and (x - 11).

Finally it means quex2- 17x + 66 = (x - 6)(x - 11). Note that in this last expression, the
the main operation is multiplication.

Example 6: Factorize2-16a + 48

Solution: In this case, b= -16 and c= +48.

Two numbers are being sought that, when added, give -16 and when multiplied, give +48. They are -4 and -12.
The sought factors are (a - 4) and (a - 12).

Finally it means that2- 16a + 48 = (a - 4)(a - 12). Note that in this last expression,
the main operation is multiplication.
page 20 Factorization

EXERCISE 7

Factor the following algebraic expressions:

1) x2+ 3x + 2 2) x2+ 7x + 12 3) x2+ 7x + 10

x2+x - 12 5) x2- 5x - 6 6) x2- 5x - 14
7) y2-5y - 24 8) h2- 30 +h 9) x2- 24 + 2x
10) x2+ 5x - 24 11) d2- 8d- 20 12) x2-10x - 24
13) x2+ 23x - 24 14) x2- 25x + 24 15) k2- 35k- 36
16) 5x - 36 + x2 17) 13x + 36 + x2 18) w2- 12w + 36
19) 25 +y210y 20) x2- 2x - 48 21) 8x + x2+ 16
22) 36 - 37a + a2 23) 36 +b220x 24) r2+ 16r - 36

CASE 5: TRINOMIALS OF THE FORM ax2+ bx + c

The difference with the previous one is that there had to be only one square X in that one.
whereas in this there must be more than one. The letter generally represents any number that
go to lax2(indicate how many squared Xs there are); the letter represents any number that
go together with lax (indicate how many x's there are); and lac represents any number that goes without lax.

For example, the trinomial 49x2- 25x+ 121 is of the mentioned form, where a = 49; b = -25;
c = + 121.

There are several procedures to factor trinomials of this form, of which only
Two will be studied in this course.

FIRST PROCEDURE:

The first procedure involves searching for two numbers, which will be called one and
In another, those who must meet the requirements given in the following rule:

To factor a trinomial of the form ax2+ bx + c, we are looking for two integers
what:

Summed gives = b, that is quem + n = b;

multiplied in the product of ac, that is, quemn = ac;
the second term, namely the linear term bx, is split into the sum of mx + nx,
that is what2+ bx + c = ax2+ mx + nx + c.
It is factored by grouping.
ValladolidInstituteHighSchool page21

Example 1: Factor 2x2+ 5x - 3

In this case, a = 2; b = +5 and c = -3.

Two numbers are sought that add up to +5 and that when multiplied give what results from ac, that is
(2)(- 3) = - 6. Add + 6 and - 1.
The linear term (the 2nd term), which is 5x, is divided into the sum of those obtained numbers, that is
in 6x-x, so it turns out that

2x2+ 5x - 3 = 2x2+ 6x - x - 3

It is factored by grouping:

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

Finally 2x2+ 5x - 3 = (2x - 1)(x + 3). Note that in the expression on the right side of the sign
Likewise, the main operation is multiplication, which means that (2x - 1) and (x + 3) are factors;
That is why a factorization was obtained.

Example 2: Factor 6x2+ 7x + 2

In this case, a = 6; b = 7 and c = 2.

Two numbers are sought that add up to +7 and, when multiplied, yield the result of ac, that is to say
(6)(2) = 12. Son + 4y + 3.
The 2nd term, which is 7x, is split into the sum of those obtained numbers, that is, into 4x + 3x.
that results in

6x27x + 2 = 6x24x + 3x + 2

It is factored by grouping:

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

6x2+ 7x + 2 = (3x + 2)(2x + 1).

Example 3: Factor 4x2+ 21x - 18

In this case, a = 4; b = 21 and c = -18.

Two numbers are sought that add up to +21 and multiplied give the result of ac.
say (4)(-18) = -72. They are +24 and -3.
The 2nd term is divided in the sum of 24x - 3x, so it turns out that

4x2+ 21x - 18 = 4x2+ 24x - 3x - 18

It is factored by grouping:

4x2+ 24x - 3x - 18 = 4x(x + 6) - 3(x + 6) = (x + 6)(4x - 3)

4x2+ 21x - 18 = (x + 6)(4x - 3).
page 22 FACTORING

Example 4: Factor 6x2- 43x + 72

In this case, a = 6; b = -43 and c = 72.

Two numbers are sought that add up to -43 and multiply to give what results from ac, that is
of (6)(72) = 432. They - 27 and - 16.
The 2nd term is divided into -27x - 16x, so it turns out that

6x243x + 72 = 6x2-27x - 16x + 72

It is factored by grouping:
6x2- 27x - 16x + 72 = 3x(2x - 9) - 8(2x - 9) = (2x - 9)(3x - 8)
6x2- 43x + 72 = (2x - 9)(3x - 8)

Example 5: Factor 12y2+ 35y - 3

In this case, a = 12; b = 35 and c = -3.

Two numbers are sought that add up to -35 and that when multiplied give the result of ac, that is,
de (12)(-3) = -36. His +36 and -1.
The 2nd term is split in the sum of +36y - 1y, so it turns out that

12y2+ 35y - 3 = 12y2+ 36y - 1y - 3

It is factored by grouping:

12y2+ 36y - y - 3 = 12y(y + 3) - 1(y + 3) = (12y - 1)(y + 3)

Finally 12y2+ 35y - 3 = (12y - 1)(y + 3). As in the expression on the right side of the equals sign
the main operation is multiplication, meaning that (12y - 1) and (y + 3) are factors; that's why it
obtained a factorization.

Example 6: Factor 3h - 15 + 6h2

Solution: First, the trinomial must be ordered, that is to say, write it as 6h.2+ 13h- 15.

In this case, a = 6; b = 13 and c = -15.

Two numbers are being sought that add up to 13 and that when multiplied give the result of ac, that is to say
de (6)(-15) = - 90. It adds + 18 and - 5.
The 2nd term is split into the sum of + 18h - 5h, which results in

6 hours2+ 13h - 15 = 6h2+ 18h- 5h- 15

It is factored by grouping:
6 hours2+ 18h - 5h - 15 = 6h(h + 3) - 5(h + 3) = (6h - 5)(h + 3)
6h2+ 13h - 15 = (6h - 5)(h + 3).
VALLADOLIDINSTITUTEPREPARATORY page23

SECOND PROCEDURE:

The other procedure to factor trinomials of the form x2+ bx + cse in the following rule:

To factor a trinomial of the form x2+ bx + c:

It is multiplied and divided at the same time (so that it is not altered, according to the only property-
the original polynomial by the coefficient adex2writing
I hurt in the following way:

* The product is given a name, for example, that is, y = axde mane-
that the previous expression transforms into:

* The trinomial2by + acse factorizes the same as in case 4, page 18, to obtain
what

* The variable y is replaced by its equivalent, namely by x.

* The factor that has aa as a common factor is identified to simplify it with the.
denominator.

Example 1: Factor 2x2+ 5x - 3

In this case, a = 2; b = +5 and c = -3.

It is multiplied and divided at the same time (so that it does not change, according to the only property of the fractions-
the original polynomial by the coefficient dex2, that is, by 2, writing it in the following way:

2 (2x2+ 5x− 3 )
2
2x + 5x− 3=
2
page 24 FACTORIZATION

( 2x+) 2 5 2x−( 6 )
=
2

The product 2x is given a name, for example, say y = 2x, so that the expr-
the previous session is transformed into:

( 2x+) 2 5 2x−( 6 ) 1
2
=
2
( y 2+ 5x− 6 )

The new trinomial2+ 5y - 6 factors the same as case 4, page 18, to obtain that

1 1
2
( y 2+ 5y− 6=)
2
( y+ 6y−
)(1 )

The variable y is replaced again by its equivalent, that is, by 2x:

1 1
( y+ 6y−
)( 1= ) ( 2x+ 6 ) 2x−
( 1 )
2 2

The factor that has 2 as a common factor is located in order to simplify it with the denominator. This
factor is (2x + 6), so that finally it is obtained

1
( 2x+
) ( 3 2x−) (1= x+ 3t)mies2x− (1 )( )
2

The desired factorization is 2x 2+ 5x− 3= x+( 3 2x−

)(1 )

Example 2: Factor 6x2+ 7x + 2

In this case, a = 6; b = +7 and c = +2.

It is both multiplied and divided at the same time (so that it does not change, according to the only property of fractions-
the original polynomial by the coefficient dex2, that is by 6, writing it in the following way:

6 (6x2+ 7x+ 2 )
2
6x + 7x+ 2=
6

( 6x+
( ) 2 7) 6x+ 2
=
6
VALLADOLIDINSTITUTEPREPARATORIA page25

The product 6x is given a name, for example, y = 6x, so that the express-
the previous session is transformed into:

( 6x+) 2 7 6x+( 12) 1

6
=
6
( y 2+ 7 y + 12 )

The trinomial2+ 7y + 12 factors the same way as case 4, page 18, to obtain that

1 1
6
( y 2+ 7 y + 12=)
6
( y+ 4y)+( 3 )

The variable y is replaced again by its equivalent, that is, by 6x:

1 1
( y+ 4y)(+ 3= ) ( 6x + 4 )6x
(+3 )
6 6

The factor (the parenthesis) that has 6 as a common factor is located to simplify it with the denominator.
nador. In this case, that common factor 6 is distributed in the two factors as follows

1 1
⎡ (2)(3x
) + (2⎤) (⎡ 3⎦ ⎣2x+ 1⎤ = )⎦ ( 6 ) (3x+ 2 2x
) (+ 1 )
6 ⎣ 6

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

The sought factorization is ( 2 2x+) (1

6x 2+ 7x+ 2= 3x+ )

Example 3: Factor 4x2+ 21x - 18

In this case, a = 4; b = 21 and c = -18.

It is both multiplied and divided at the same time (so that it does not change, according to the only property of the fractions-
the original polynomial by the coefficient dex2, that is by 4, writing it in the following way:

4 (4x2+ 21x− 18 )
4x 2+ 21x− 18=
4

( 4x( )+2 21)4 times− 72

=
4
page 26 FACTORIZATION

The product 4x is given a name, for example y, that is to say y = 4x, so that the expression-
the previous session is transformed into:

( 4x )+2 21 4x(− 72) 1

4
=
4
( y 2+ 21 years−)72

The trinomial2+ 21y - 72 factors the same way as case 4, page 18, to obtain that

1 1
4
( y 2+ 21 years−)72= ( y+ 24
4
)years
( − 3)

The variable y is replaced by its equivalent, that is, by 4x:

1 1
( y+ 24 )( )
years− 3= ( 4x + 24 ) (4x− 3 )
4 4

The factor that has 4 as a common factor is identified to simplify it with the denominator. This
Factor is (4x + 24) such that it is eventually obtained.
1
( 4x)+( 6 4x−)(3= x+ 6) 4x(− 3 )( )
4

The sought factorization is 4x 2+ 21x− 18= x+( 6 4x−)(3 )

EXERCISE 8

Factor the following algebraic expressions:

2x2+ 7x + 5 2) 3x2- 4x + 1 3) 6x2+x- 1

4) 10x2- 21x - 10 5) 4x2- 17x - 15 6) 7x2+ 8x + 1
7) 3y2- 25y - 50 8) 4h224h+ 35 9) 14x2-9x - 8
10) 11x2+ 21x + 10 11) 9d2- 29d- 28 12) 10 times251x + 5
13) 812- 36x + 4 14) 4x2+ 44x + 121 15) 49k270k+ 25
16) 8x2-10x - 25 15j2- 47j + 6 18) 12t2- 73t + 6

Trinomials of the form ax2+ bxy + cy2

These trinomials are similar to those of the form ax.2+bx+c, only adding them the
variable, therefore, after factoring under the same procedure as case 5, page 20,
the variable and the second term of each factor is also added.
VALLADOLIDPREPARATORYINSTITUTE page27

Example 1: Factor 2x2+ 5xy - 3y2

In this case, a = 2; b = +5 and c = -3.

Two numbers are sought that when added give +5 and when multiplied give the result of ac, that is
(2)(-3) = -6. Add 6 and -1.

The 2nd term is split in the sum of 6xy - xy, so it results in

2x2+ 5xy - 3y2= 2x2+ 6xy - xy - 3y2

It is factored by grouping:

2x2+ 6xy - xy - 3y2= 2x(x+ 3y) -y(x+ 3y) = (2x-y)(x+ 3y)

Finally means that 2x2+ 5xy - 3y2(2x - y)(x + 3y)

Example 2: Factor 6d2- 13th + 6th2

In this case, a = 6; b = -13; and c = 6.

Two numbers are sought that add up to -13 and that when multiplied give what results from 'deac', that is
(6)(6) = 36. Son - 4 y - 9.
The 2nd term is split in the sum of -4de - 9de, so it turns out that

6d213th + 6d2= 6d2- 4de- 9de+ 6e2

It is factored by grouping:

6d24th - 9th + 6th2= 2d(3d- 2e) - 3e(3d- 2e) = (3d- 2e)(2d- 3e)

Finally means that 6d2- 13th + 6th2(3d - 2e)(2d - 3e)

EXERCISE 9

Factor the following algebraic expressions:

1) 2x2+ 7xy + 5y2 3x24xy + y2 6x2+xy-y2

4) 10j2- 21jk- 10k2 4m2- 17mn- 15n2 6) 7a2+ 8ac+c2
7) 3a2- 25ab- 50b2 8) 4h2- 24hi+ 35i2 9) 14x2- 9xz - 8z2
10) 9a2+ 12ab + 4b2 11) 9d2- 30df + 25f2 100x + 20xy+y2
2

13) 81x2+ 36xy + 4y2 14) 4c2+ 20cg+ 25g2 15) 64k2- 48kr+ 9r2
page 28 FACTORING

CASE 6: SUM OF CUBES

If (a is multiplied2-ab + b2(a+b) is obtained

a 2− ab+ b 2
a+ b
a 3− a 2b+ ab2
+ a 2b− ab2+ b3
a3 + b3

that is to say, that (a2-ab + b2(a+b)

=a3+b3Obviously, if you reverse the previous equality, what
the result is undoubtedly true, so it can be stated that

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

which amounts to stating that the factorization of a3+b3is (a2-ab + b2(a+b), or, since the
multiplication is commutative, a3+b3(a+b)(a2-ab + b2) .

From the above, the following rule is derived:

A sum of cubes is factored into two factors, in the following way:

The first factor is a binomial formed with the sum of the cube roots of the
original terms;
the second factor is a trinomial that is formed from the previous factor of the si-
following way:

YSquare of the first term (of the previous factor);

Yminus the product of the first term (of the previous factor) by the second;
Yplus the square of the second term (of the previous factor).

Example 1: Factor x3+ 1

Solution: The cubic root of dex3the cube root of 1 is 1.

The first factor is the sum of those cubic roots, that is, it is (x + 1).
The second factor is formed from the previous one, that is, from (x + 1):
Ysquare of the first term: (x)2=x2;
Yminus the product of the first by the second: - (x)(1) = -x;
ValladolidPreparatoryInstitute page29

Yplus the square of the second term: (1)2= 1 .

So the second factor is (x2-x + 1)
Finally, the dex factorization3+ 1 is

x3+ 1 = (x + 1)(x2-x + 1

Example 2: Factor 8x3+ 27

Solution: The cube root of 8x3it's 2x; the cube root of 27 is 3.

The first factor is the sum of those cube roots, that is, it is (2x + 3).
The second factor is formed from the previous one, that is, from (2x + 3):
Ysquare of the first term: (2x)2= 4x2;
Yless the product of the first by the second: - (2x)(3) = - 6x;
Yplus the square of the second term: (3)2= 9 .
Thus, the second factor is (4x2- 6x + 9)
Finally, the factorization of 8x3+ 27 is

8x3+ 27 = (2x + 3)(4x2-6x + 9

Example 3: Factor 64b3+ 27x6

Solution: The cube root of 64b3is 4b; the cube root of 27x6is 3x2.
The first factor is the sum of those cube roots, that is, it is (4b + 3x2) .
The second factor is formed from the previous one, that is from (4b + 3x2) :
Ysquare of the first term: (4b)2= 16b2;
Yminus the product of the 1st by the 2nd: - (4b)(3x)2) = - 12bx2;
Yplus the square of the second term: (3x2)2= 9x4.
Thus, the second factor is (16b2- 12bx2+ 9x4) .
Finally, the factorization of 64b3+ 27x6is

64b3+ 27x6(4b + 3x2(16b2- 12bx2+ 9x4)

Example 4: Factor 125a6b9+ 27

The cube root of 125a6b 9it's 5a2b3the cube root of 27 is 3.

The first factor is the sum of those cubic roots, that is, it is (5a2b3+ 3) .
The second factor is formed from the previous one, that is from (5a2b3+ 3) :
Ysquare of the first term: (5a2b3)2= 25a4b6;
Yminus the product of the 1st by the 2nd : - (5a2b3(3) = - 15a2b3;
Yplus the square of the second term: (3)2= 9 .
Thus, the second factor is (25a4b615a2b3+ 9) .
Finally, the factorization of 125a6b9+ 27 is

125a6b9+ 27 = (5a2b3+ 3)(25a4b6- 15a2b3+ 9)

page 30 FACTORIZATION

1 x3
Example 5: Factorize +
x6 8

1 1 x3 x
Solution: The cube root ofis 2 the cube root of it .
x 6
x 8 2

⎛ 1 + x⎞
The first factor is the sum of those cube roots, that is, ⎜ x2 .
⎝ 2 ⎟⎠

⎛ 1 + x⎞
The second factor is formed from the previous one, that is from
⎜ :
⎝ x
2
2 ⎟⎠
2
⎛ 1 ⎞ = 1
Y square of the first term: ⎜ x2 ⎟ ;
⎝ ⎠ x4

Y minus the product of the 1st by the 2nd:

− ⎛⎜ 2 ⎞⎟ ⎛⎜ ⎞⎟ = −
1 x x 1
= − ;
⎝ x ⎠⎝ 2 ⎠ 2x 2 2x
2
⎛ x ⎞ = x2
Y plus the square of the second term: ⎜ 2⎟
⎝ ⎠ 4

⎛ 1 1 x2 ⎞
So the second factor is ⎜ 4 − + ⎟ .
⎝ x 2x 4 ⎠

⎛ 1 x3 ⎞
Finally, the factorization of ⎜ 6 + ⎟ it
⎝ x 8 ⎠

⎛ 1 x3 ⎞ ⎛ 1 x ⎞⎛ 1 1 x2 ⎞
⎜ 6 + ⎟ = + ⎜ − + ⎟
⎝ x 8 ⎠ ⎜⎝ x 2 2 ⎟⎠ ⎝ x 4 2x 4 ⎠
VALLADOLIDPREPARATORYINSTITUTE page31

EXERCISE 10

Factor the following algebraic expressions:

1) x3+ 8 2) 27x3+ 1 3) 64 +x6

4) a3x3+ 27 5) 64x6+ 27y3 6) 1 + 125a6b9
7) 27y12+ 125x3 8) h9+ 125a6b24 9) 729x9+ 8
10) 8a30x3+ 27 11) 64b12x3+ 27 12) 125 + 27a24d6
13) 27y9+ 1 14) 8d9+ 27a18c12 15) 729y21+ 27a6x3
1 8 27x 3
16) x 3+ 17) a 6 b 3+ 18) + y9
8 x3 8

1 a3 x12 1 x3 b6
19) + 20) + 6 21) +
8 27 y6 b 1000 x3

a3 b3 x9 8 b 3c 6 a3
22) + 23) + 3 24) +
b6 a6 8 y a3 b 3c 6

CASE 7: DIFFERENCE OF CUBES

If you multiply (a2+ab+b2(a-b) is obtained

a 2+ ab+ b 2
a− b
a 3+ a 2 b+ ab 2
− a 2b− ab 2− b 3
a3 − b3

that is to say, that (to2+ab+b2(a-b)

= a3-b3Obviously, if you reverse the previous equality, what
it is certainly true without a doubt, so it can be stated that

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

which is equivalent to stating that the factorization of ao3-b3is (a2+ab+b 2(a-b), or rather, since the
Multiplication is commutative.3-b3(a-b)(a2+ab+b2) .
page 32 FACTORING

The following rule is derived from the above:

A difference of cubes is factored into two factors, in the following way:

The first factor is a binomial formed with the difference of the cube roots of the
original terms;
the second factor is a trinomial that is formed from the previous factor of the si-
next way:

YSquare of the first term (of the first factor obtained before);
Yplus the product of the first term (of the previous factor) by the second;
Yplus the square of the second term (of the previous factor).

Example 1: Factorize31

Solution: The cube root of a3The cube root of 1 is 1.

The first factor is the subtraction of those cubic roots, that is, it is (a - 1).
The second factor is formed from the previous one, that is, from (a-1):
Ysquare of the first term: (a)2=a2;
Yplus the product of the first by the second: (a)(1) = a;
Yplus the square of the second term: (1)2= 1 .
So the second factor is (a2+a+ 1) .
Finally, the factorization of31 is

a3- 1 = (a - 1)(a2+a+ 1)

Example 2: Factor 8x327

Solution: The cube root of 8x3it is 2x; the cube root of 27 is 3.

The first factor is the difference of those cubic roots, that is, it is (2x - 3).
The second factor is formed from the previous one, that is (2x - 3):
Ysquare of the first term: (2x)2= 4x2;
Yplus the product of the first by the second: (2x)(3) = 6x;
Yplus the square of the second term: (3)2= 9 .
So the second factor is (4x2+ 6x + 9)
Finally, the factorization of 8x327 is

8x3- 27 = (2x - 3)(4x2+ 6x + 9

VALLADOLIDPREPARATORYINSTITUTE page33

Example 3: Factor 64b3- 27x6

Solution: The cube root of 64b3it is 4b; the cube root of 27x6is 3x2.
The first factor is the subtraction of those cubic roots, that is, it is (4b - 3x2) .
The second factor is formed from the previous one, that is from (4b - 3x)2) :
Ysquare of the first term: (4b)2= 16b2;
Yplus the product of the first by the second: (4b)(3x)2) = 12bx2;
Yplus the square of the second term: (3x2)2= 9x4.
So the second factor is (16b2+ 12bx2+ 9x4) .
Finally, the factorization of 64b3- 27x6is

64b3- 27x6(4b - 3x22+ 12bx2+ 9x4)

1 a3
Example 4: Factorization −
x6 27

1 1 a3 a
Solution: The cubic root 6 of
is the cube root of is .
x x2 27 3

⎛ 1 − a ⎞
The first factor is the sum of those cube roots, that is, ⎜ x2 .
⎝ 3 ⎟⎠
The second factor is formed from the previous one:

2
⎛ 1 ⎞ = 1
Ysquare of the first term: ⎜ x2 ⎟ ;
⎝ ⎠ x4

⎛ 1 ⎞⎛ a ⎞ = a
Yplus the product of the 1st by the 2nd: ⎜ x2 ⎟ ⎜ 3 ⎟ ;
⎝ ⎠⎝ ⎠ 3x 2
2
⎛ a ⎞ = a2
Yplus the square of the second term: ⎜ 3⎟ .
⎝ ⎠ 9

⎛ 1 a a2 ⎞
So the second factor is ⎜ 4 + + ⎟ .
⎝ x 3x 2
9 ⎠
Finally, the sought factorization is

1
−
a3
= ⎛ 1 − a ⎞ ⎛ 1 + a + a2 ⎞
⎜ ⎟
x6 27 ⎜⎝ x 2 3 ⎟⎠ ⎝ x 4 3x 2 9 ⎠
page 34 FACTORING

EXERCISE 11

Factor the following algebraic expressions:

x627 2) 27x31 3) 64 -x3

4) a6x327 5) 64x327 years3 6) 1 - 125a3b9
7) 27 years12- 125x3 8) h 9- 125a3b24 9) 729x927
10) 27a18y327 11) 64b12x327 12) 125 - 27a24d3
13) 27x61 14) 27d927a18c12 15) 729y2127a3x3
16) 125h99 17) 125f6-b18y21 18) 216x218a3k33
x3 8 a6 b9 27
19) −1 20) − 21) −
y3 a6 27 9 6
dx 8

1 125 w 21 h6
22) 1000− 23) − 21 24) 1−
x12 27 weeks 8 m9

8−
( a+ b ) 3 (1− x ) 6
− 125
( 8− b )
2
9

−8
25) 26) 27)
8 b6 y 12

28) (1− x −) a +(7

3
6
9
) 12

TOTAL OR COMPLETE FACTORIZATION

If it factors4-y4, which belongs to case 3, difference of squares, page 16, it is obtained that

x4-y4= (x2+y2)x2-y2)

It is observed that the second factor (x2-y2) it is again a difference of squares and, therefore,
can be factored again in x2-y2= (x+y)(x-y). Then, the total or complete factorization of
the original expression4-y4

x4-y4= (x2+y2(x+y)(x-y)

Thus, when a complete factorization is required, it is necessary to analyze each factor that results.
trying the previous factorization to see if it belongs or not to one of the previous factorization cases
seen mind, since if this is the case, the factorization process must continue. In other words, it
It says that a factorization is totally complete if none of the obtained factors can be.
re-factor. When a factor, or expression, can no longer be factored it is said to be irreducible.
Thus, it can also be said that a total factorization is one in which all of its
factors are irreducible.
ValladolidInstitutePreparatory page35

Example 1: Factor completely 4a34

Solution: The first possible factorization is by common factor, case 1, page 11:

4a3- 4 = 4(a3- 1)

The second factor is a difference of cubes, case 7, page 31:

a3- 1 = (a - 1)(a2+a+ 1)

So the total or complete factorization is:

4a3- 4 = 4(a - 1)(a2+a+ 1)

Example 2: Completely factor2c-bc2- 2a2+ 2b2

Solution: The first possible factorization is by grouping, case 2, page 13:

a2c-bc2- 2a2+ 2b2 =a(2-b2) - 2(a2-b2)

= (a2-b2)(c- 2)

The first factor is a difference of squares, case 3, page 16.

a2-b2(a-b)(a+b)

So the total or complete factorization is:

a2c-bc2- 2a2+ 2b2(a-b)(a+b)(c-2)

Example 3: Fully factor2b + 3ab - 10b

Solution: The first possible factorization is by common factor, case 1, page 11:

a2b + 3ab - 10b = b(a2+ 3a - 10)

The second factor is a trinomial of the formax.2+bx + c, case 4, page 18:

a2+ 3a - 10 = (a - 2)(a + 5)

So the complete factorization is:

a2b + 3ab - 10b = b(a - 2)(a + 5)

page 36 FACTORIZATION

Example 4: Fully factor 10a2x + 5ax - 105x - 2a2-a+ 21

Solution: The first possible factorization is by grouping, case 2, page 13:

10a2x + 5ax - 105x - 2a2-a + 21 = 5x(2a2+a- 21) - 1(2a2+a- 21)

= (2a2+a- 21)(5x- 1)

The first factor is a trinomial of the form ax2+bx + c, case 5, page 20:

2a2+a - 21 = 2a2+ 7a - 6a - 21
a(2a + 7) - 3(2a + 7)
(2a + 7)(a - 3)

So the complete or total factorization is:

10a2x + 5ax - 105x - 2a2-a + 21 = (2a + 7)(a - 3)(5x - 1)

EXERCISE 12

Totally factor the following algebraic expressions:

1) 2ax2- 2a 2a2b - 2b - a2+ 1 48 - 6x6

4) 3a2b2- 3ab290b2 5) 10b2xy - 55bxy - 30xy 64b3- 8x6
7) 2a + 2ab3- 5x - 5b3x 8) 3a6b - 3bx2+a6-x2 9)x8-y8

10) 4a2x - 4ax - 3x - 8a2+ 8a + 6 11) b12-c6

12) ax2- 5ax + 6a - 4x2plus 20x minus 24 13) x4-y4
14) 2a3- 3a2b + 10a2x - 15abx 15) 6a2+ 9a - 60
a ac a2 a3x b3x
16) + + 17) −
b x y y y

x4 y6 1 1 1 1
18) − 19) + + +
b2 b2 ac ad bc bd