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2.5 Special Products

In the previous sections, the distributive property was used in multiplying

polynomials. For example, the product of two binomials bax+ and dcx+ usingthe distributive property is as follows:

( )( ) ( ) ( )( )( ) ( )( ) ( )( ) ( )( )dbcxbdaxcxax

dcxbdcxaxdcxbax

+++=+++=++

Take note that a shortcut for finding the product of two binomials is to add

the products of four pairs of terms of the binomials namely: the product of the

first terms, the productof the outer terms, the product of the inner terms, and the

productof the last terms. Thisspecial orderormethodof multiplying binomials is

known as the FOIL method. The word FOIL is formed from the first lettersof the

productsof the terms of the two binomials to be added.

The following table illustrates the FOIL method of finding the product oftwo binomials, 2x + 1 and 3x 5.

FOIL Method

F stands for the

product of the First terms

(2x+ 1) (3x 5)

( 2x) (3x) = 6x2 F

O stands for the

product of the Outer terms

(2x + 1) (3x 5 )

( 2x) (5 ) = 10x O

I stands for the

product of the Inner terms

(2x+ 1) (3x 5 )

(1) (3x) = 3x I

L stands for the

product of the Last terms

(2x + 1) (3x 5 )(1) (5) =5 L

F O I L

The product of ( ) ( ) 5310653122 +=+ xxxxx

5762 = xx Combine similar

terms

When using the FOIL method in finding certain types of products, a specificpattern is observed. Suchpatterns lead to thespecial product formulas and can be used

to find the products of binomials which are called special products. The following are the

different types of special products.

Types of Special Products

B. Product of the Sum and Difference of Two Numbers

The product of the sum and difference of two numbers is the difference of twosquares. The difference of two squares is the square of the 1st number minus the

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square of the 2nd number. Thus,

( ) ( ) 22 bababa =+

Example. Find the product of the following binomials.

( )( ) ( ) ( )

22222222

333)1 yxyxyx =+ 449 yx =

( )( ) ( ) ( ) 22232323 252525)2 srsrsr =+

=46 425 sr

( )( ) ( ) ( ) 22222222222 747474)3 cbacbacba =+

=

444

4916 cba ( )( ) ( ) ( ) 23223232 656565)4 nnnnnn yxyxyx =

=nn yx 64 3625

( ) ( ) ( )[ ] ( )

( )

( ) 162251625

454545)5

22

2

2222

++=

+=

+=+++

yxyx

yx

yxyxyx

1625502522 ++= yxyx

C. Square of a Binomial

The square of a binomial is a perfect trinomial square. A perfect trinomial

square is the square of the first number plus or minus twice the product of the two

numbers plus the square of the second number.

( ) ( ) ( )( ) ( )

( ) ( ) ( )( ) ( )222

222

2 bbaaba

bba2aba

+=

++=+

Example. Find the product of the following binomial.

( ) ( ) ( ) ( ) ( ) 222 7762676)1 yyxxyx ++=+ 22

498436 yxyx +=

( ) ( ) ( )( ) ( )222323223 4432343)2 yyxxyx ++=+ 4236 16249 yyxx +=

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( )[ ] ( )[ ] ( ) ( ) ( ) ( ) ( )

( ) ( ) 2222

15228

53121085432)5

zyxzyxyx

zzyxzyxzyxzyxzyx

++=

+=+

222

15228168 zyzxzyxyx ++=

E. Cube of a Binomial

The cube of a binomial is a quadrinomial with the following terms:1st term is the cube of the 1st number

2nd term is plus or minus thrice the square of the first number times the

second number

3rd term is plus thrice the first number times the square of the secondnumber

4th term is plus or minus the cube of the second number

( )

( ) 32233

32233

33

33

babbaaba

babbaaba

+=

+++=+

Example. Find the product of each of the following binomials.

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) 64169912274433433343)1

23

32233

+++=

+++=+

ttt

tttt

6414410827 23 +++= ttt

( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 643629

3222322333223

8415256125

2253253525)2

yyxxyx

yyxyxxyx

+++=

+++=+

643269 860150125 yyxyxx +++=

( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 642244266

32222222223223222

6416214912343

4473473747)3

ccbabacba

ccbacbabacba

+=

+=

642224466

64336588343 ccbacbaba +=

( )[ ] ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 25632 5441 5223235523523252)4

223223

32233

+++++++=

++++=+

bmbm bmbbmbmm

bmbmbmbm

125150756060158126223223

+++++= bmbbmmbmbbmm

F. Product of Binomialand Trinomial

The product of a binomialand a trinomialin the form )((22

babab)a + is thesum or difference of two cubes.

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3322

3322))((

ba)babb)(a(a

bababa

=++

+=++ ba

Example. Find the product of each of the following.

1)363322242 64125)4()5()162025)(45( bababbaaba +=+=++

2222 )4()4)(5()5( bbaa +

2) 27343)3()7()92149)(37(9233423623 ==++ xyxyyxxyx

222323 )3()3)(7()7( yyxx +

( )( ) ( ) ( )

( ) ( )( ) ( )22

333322

3322

2783296432)3

bbaa

bababbaaba

yyxx

yxyxyyxxyx

+

+=+=++

( )( ) ( ) ( )

( ) ( )( ) ( )2222

363322242

4433

647294343443343)4

yyxx

yxyxyxyyxxyx

++

===++

G. Product of Trinomials and Quadrinomials

The special product formulas can be used to find the products of trinomials

and quadrinomials by grouping the terms into binomials.

Example 1. Find the product of the following trinomials.

a) ( ) ( )dcadca 323323 ++

( ) ( )[ ] ( ) ( )[ ]dcadca 323323 += After grouping, the 1st term is 3a and the2ndsecond terms are (2c 3d) for both binomials. Use the formula for the

product of the sum and difference of

two numbers

( ) ( )22

323 dca = Expand ( )2

3d2c . Use the formula for

( ) ( )( ) ( )222 33262269 dda += the square of a binomial.

( )222

91249 dcdca ++= Simplify222 91249 dcdca +=

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Another solution ofExamplea:

a) ( ) ( )dcadca 323323 ++

( ) ( )[ ] ( ) ( )[ ]dcadca 323323 ++= After grouping, the 1st two

terms ( )ca 23 and( )ca 23 + are the 1st terms ofthe binomials. The 3rd terms,

3d are the 2nd terms of binomials. Use the FOIL

method

( ) ( ) ( ) ( ) ( ) ( ) 232332332332323 dcadcadcadcaca ++++= Multiply the 1st term)23)(23( ca + ca .

Use the formula for theproduct of the sum

and difference of twonumbers.

22291249 dcdca +=

Note: In Example a, both the formula for the product of the sum and

difference of two numbers and the FOIL method can be used.

b) ( ) ( )pnmpnm 325525 + ( ) ( )[ ] ( ) ( )[ ]pnmpnm 325525 += After grouping, (5m + 2n)

and (5m 2n) are the 1st terms while

5p and 3p are

the 2nd terms of thebinomials. Use the

FOIL method.

( ) ( ) ( ) ( ) ( ) ( )ppnmpnmpnmnm 352532552525 +++= Multiply the 1st term

(5m+2n)(5m-2n).

( ) ( )222

15615102525 ppnpmpnpmnm ++= Use the formula for theproduct of

the sum and difference of

two numbers.

222

156151025425 ppnpmpnpmnm ++= Combine similar terms

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222 15440425 ppnpmnm ++=

Note: Unlike Example a, Example b can be expressed only as the product of

binomials after grouping. Thus, the FOIL method was used.

c) ( )2

546 zyx

( ) ( )[ ]2

546 zyx += After grouping, the 1st term is 6xand

( ) ( ) ( ) ( ) 22 5454626 zyzyxx +++= the 2nd term is (4y + 5z). Use theformula for the

square of a binomial ( square of the

difference of two numbers ).

( ) ( ) 22 54541236 zyzyxx +++= Expand (4y+5z)2. Use the formula forthesquare of

the sum of two numbers.

( ) ( ) ( ) ( ) ( ) 222 55424541236 zzyyzyxx ++++= Simplify

222 254016604836 zyzyxzxyx +++=

Note: The square of a trinomials can always be expressed as the square of a

binomial after grouping the terms of the given trinomial. Thus, the formula

for the square of a binomial is used.

Example 2. Find the product of the following quadrinomials.

a) ( )( )432432 ++++ zyxzyx

( ) ( )[ ] ( ) ( )[ ]432432 +++= zyxzyx After grouping, the 1st term is (2x+ 3y)and

the 2nd term is (z 4) for bothbinomials. Use

the formula for the product of the

sum and

difference of two numbers.( ) ( ) 22 432 += zyx Expand (2x + 3y)2 and (z - 4)2. Use

the formula

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2222 44233222 +++= zzyyxxfor the square of a binomial.

( ) ( )1689124 222 +++= zzyxyx Simplify

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1689124222 +++= zzyxyx

Note: If after grouping, the product of the given quadrinomials cannot be

expressed as the product of the sum and difference of two expressions , the

FOIL method is used.

b) ( )2

5234 + zya ( ) ( )[ ]25234 = zya After grouping, the 1st

term is (4a3y

and the 2nd term is (2z

5). Use theformula for thesquare

of a binomial.

( ) ( ) ( ) ( ) 22 525234234 += zzyaya Expand (4a 3y)2 and

(2z 5)2

. Use the formula for thesquare of a

binomial.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2222 552225234233424 +++= zzzyayyaa Multiply (4a 3y) and(2z 5). Use

FOIL method.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2222 552221 52 068233424 ++++= zzyay za zyyaa Simplify252043040121692416 222 +++++= zzyayzazyaya

Note: The square of a quadrinomial can always be expressed as the square of a

binomial after grouping the terms of the given quadrinomial. Thus the

formula for the square of a binomial is used.

G. Square of a Polynomial

The square of a polynomial is equal to the sum of the squares of each term of thepolynomial and twice the product of any combination of two terms. This method of

finding the square of a polynomial is useful if the polynomial contains more than

four terms.

Example: Find the product of the following polynomials.

1) ( )2

6357 ++ pnm

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( )( )( )( ) 3322

3322

babababa

babababa

=++

+=++

Product of Binomial and Trinomial

Square of a Polynomial

The square of a polynomial is equal to the sum of the squares of each term

of the polynomial and twice the product of any combination of two terms.

2.6 Factoring

Factoring is the reverse of multiplication. It is the process of expressing a

given polynomial as a product of its factors.

A polynomial with integral coefficients is said to be prime if it has no

monomial or polynomial factors with integral coefficients other than itself and one.

Thus, a polynomial with integral coefficients is said to be completely factoredwheneach of its polynomial factors isprime.

Types of Factoring

H. Removal of the Highest Common Factor (HCF)

A common factoris a factor contained in every term of a polynomial.

The highest common factor is the product of the greatest common factors

of the numerical coefficients and the literal coefficients having the leastexponents in every term of a polynomial.

Example: The HCF of a) 5x and 15x2is 5xb) 3ab2 and 6a is 3ac) 8x2y2z3, 16x3y3z2 and 24x4yz4 is 8x2yz2

If every term of a polynomial contains a common factor, then thepolynomial can be factored by removing the HCF. Thus, the factors are the HCF

and the quotient obtained by dividing the given polynomial by the HCF.

( )

( ) ( )dcbax

x

dxcxbxaxxdxcxbxax

+=

+=+

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Example. Factor the following completely:

+=+

xy

xyyxxyxyyx

8

5616)8(5616)1

7772

( ) ( )66 728 yxxy +=

( )

+=+

rqp

rqprqprqprqprqprqpryp

32

342533423234253342

5

2520155252015)2

( )( )22232 5435 prpqqrrqp +=

( ) ( ) ( ) ( )

( )( ) ( )

( )

=

=+

ha

hayhaxha

hayhaxahyhax

3

363

3636)3

( ) ( )yxha = 23

I. Difference of Two Squares

The factors of the difference of two squares are the sum and difference of

the square roots of the two squares.

( ) ( )bababa22 +=

Note: The sum of two squares (a2 + b2) is a prime polynomial, hence, it is notfactorable.

Example. Factor the following completely.

( )( ) ( ) ( )[ ]222

4242

3

99)1

yxb

yxbbybx

==

( )( )( )22 33 yxyxb +=

( ) ( ) ( ) ( )2222

252525)2 tratara = Factor out (5a 2)

( ) ( ) ( )trtra += 25

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( ) ( ) ( )

( )[ ] ( ) ( ) ( )[ ]22333

444

464)3

++=

=

bababa

baba

( ) ( )164424 22 +++= babababa

( )

( ) ( )[ ]33

3333

23

83243)4

mn

mnmn

yx

yxyx

=

=

( )( )mmnnmn yyxxyx 22 4223 ++=

( )

( ) ( )[ ]3232632842

523

1258337524)5

yxxy

yxxyxyxy

=

=

( )( )42222 25104523 yxyxyxxy ++=

( )

( ) ( )[ ]( )( )422422232

6667

4164242

6422128)6

bbaabaabaa

baaaba

++==

=

( )( )( )4224 416222 bbaababaa +++=

( ) ( )

( ) ( ) ( )[ ]

( )( )( )84482222

24442444

34341212)7

yyxxyxyx

yyxxyx

yxyx

+++=

++=

=

( )( )( )( )844822 yyxxyxyxyx ++++=

Note: Example 7 can also be factored as the difference of two squares

( ) ( )( )( )

( ) ( ) ( ) ( )( )( )( )( )422422422422

32323232

6666

26261212

yyxxyxyyxxyx

yxyx

yxyx

yxyx

++++=

+=

+=

=

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))()()()((

4224422422

yyxxyxyxyyxxyx +++++= D. Perfect Trinomial Square (PTS)

A trinomial is a perfect trinomial square if the first and last terms are

perfect squares and the middle term is plus or minus twice the product of the

square roots of the 1st and the last terms.The factors of a PTS are the square of the sum or the difference of

the square root of the 1st and the last terms of the given trinomial.

( )

( ) 222

222

bab2aba

bab2aba

=+

+=++

Example. Factor the following completely.

( ) ( ) ( ) ( )2222 1162611236)1 ++=+ xyxyxyyx

( )2

16 += xy

( ) ( )( ) ( ) 2222 5532325309)2 bbaababa ++=++

( )2

53 ba +=

( ) ( )( ) ( ) ( ) ( ) ( )[ ]22

22

2222

442882)3

++=

++=++

aab

aabbabba

( ) ( )2

22 += ab( ) ( )123363)4 223 +=+ aaaaaa

( ) ( )2

13 = aa

( ) ( ) ( ) ( ) 22222 2272742849)5 zzxyxyzxyzyx +=+

( ) 227 zxy =

( ) ( )( ) ( )

( )

( ) ( )[ ] 2

22

222224

22

4

442168)6

+=

=

+=+

aa

a

aaaa

( ) ( ) 22 22 += aa

( ) ( ) 2222 119 ++= xxxx

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( )

( ) ( )( ) ( )[ ]( )

( ) ( )[ ]

222

232

23232

362258

119

19

1129

1299189)7

++=

+=

++=

++=+

xxxx

xx

xxx

xxxxxx

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 2222 5532325309)8 yxyxyxyxyxyxyxyx +++=+++ ( ) ( )[ ]

( )

( )

( )[ ] 2

2

2

2

42

82

5533

53

yx

yx

yxyx

yxyx

=

+=

++=

+=

( )2

44 yx =

E. General Trinomial

A general trinomial of the form22 )( bdyxybcadacx +++ can be factored

into the product of two binomials ))(( dycxbyax ++ . a, b, c, d can beobtained by using the trial and error method.

Trial: a) The first terms of both binomials are factors of the first term

of the given trinomial.

b) The second terms of both binomials are factors of the lastterm of the given trinomial.

To check if trialis correct, the sum of the products of the outer and inner

terms must be equal to the middle term of the given trinomial.

Example. Factor the following completely.

( ) ( )yxyxyxyx 25294845)122 +=

( )352412208)2 223 +=+ yyyyyy

( ) ( )1324 = yyy

( ) ( )72321132)3 2 ++=++ xxxx

( ) ( )174342521)4 22 += cdcdcddc

( )( )1232344)5 2 += nnnn xxxx

( ) ( ) ( )[ ] ( )[ ]5225325526)6 2 +++=++ babababa Simplify

( )( )524536 +++= baba

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( ) ( ) ( ) ( )[ ]( )[ ] ( )[ ]yxzyxz

yxyxzzyxyxzz

+=+=+

743

734321912)72222

( ) ( )yxzyxz ++= 7743

F. Factoring by Grouping

Factoring by grouping is used to factor polynomials consisting of

more than three terms.

In factoring by grouping, the terms of the given polynomial are grouped toform a binomial or a trinomial that are both factorable.

Example. Factor the following completely.

( ( )( ) ( )

( ) ( )53353

51535153)1

2

2

3232

+=++=

++=+

xx

xxx

xxxxxx

( ) ( )53 2 += xx

( ) ( )

( ) ( )

( )[ ] ( ) ( )[ ]dcbacba

dcba

dcdcbabacdabdcba

+=

=

++=++

2424

24

4481648416)2

22

22222222

( ) ( )532 += xx

( ) ( )

( ) ( )

( )[ ] ( )[ ]zyxzyx

zyx

zyxyxzyxyx

3232

32

944944)3

22

222222

+++=

+=

++=++

( ) ( )zyxzyx 3232 +++=

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( ) ( )

( ) ( )

( ) ( )[ ] ( ) ( )[ ]zyxzyx

zyx

zyzyxzyzyx

+=

=

+=+

55

5

225225)4

22

222222

( )( )zyxzyx ++= 55

( ) ( )

( ) ( )yxyx

yxyxyxyyxyx

22

244424)5

2

2222

+=

++=++

( ) ( )122 += yxyx

( ) ( )

( ) ( ) ( )bababa

babaabba

6526565

1210362512103625)6 2222

++=

+=+

( )( )26565 ++= baba

( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )yxxyxyxyx

xyxyxyxyx

xxyyxxxyyx

++=

+++=

+=+

23242

23242

638638)7

22

22

233233

( ) ( )xyxyxyx 324222 ++=

( ( ) (

( ) ( ) ( )( ) ( )[ ] 2

22

2222

32

3262

912644912644)8

yx

yxyx

yyxyxxyyxyxx

+=+++=

++++=+++

( )2

23 += yx

G. Factoring by Completing the Square

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Factoring by completing the square is applicable to binomials whose

terms are both perfect squares and to trinomials with at least two terms that

are perfect squares.The method consists of adding and subtracting a term that is a perfect

square that will make the given binomial or the trinomial a PTS. The

resulting expression is then factored as the difference of two squares.

Example. Factor the following completely.

22yxyx 4))(22(22 ++=+ ])2()[(4)1 222244 yxyx

2222

)2()2( xyyx +=

)22)(22(2222 xyyxxyyx +++=

)22)(22(2222 yxyxyxyx +++=

22

yy 44 +++=++ ])5(6)[(256)2 222224 yyyy

22 )2(])5(y10)[( 222 yy ++=

222 )2()5( yy +=

)25)(25(22 yyyy +++=

)52)(52(22 +++= yyyy

2222

baba 99 +++=++ ])2(15)6[(41536)3 2222222224 bbaabbaa 2222222 )3(])2(24)6[( abbbaa ++=

2222 )3()26( abba +=

)326)(326( 2222 abbaabba +++= )236)(236(

2222 babababa +++=

22

xx ++=+ ])3(7)[(97)4 222224 xxxx

22222 )(])3(6)[( xxx +=

222

)()3( xx =

)3)(3(22 xxxx +=

)3)(3(22 += xxxx

22

ss 44 ++=+ ])5(14)[(2514)5 222224 ssss

2222 )2(])5(s10)[( ss += 2

222 )2()5( ss =

)25)(25(22 ssss +=

)52)(52(22 += ssss

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)644

9x9x ++=+ ])6(21)[(3621 242448 xxxx

222424

)3(])6(12)[( xxx +=

2224

)3()6( xx =

)36)(36( 2424 xxxx += )63)(63(

2424 += xxxx

H. Factoring by Synthetic Division

Factoring by synthetic division is applicable to a polynomial in one variable

whose degree is higher than two.

Factor TheoremThefactor theorem is used to determine if a binomialx c is a factor of the

given polynomial.

Example 1. Determine if:

a) 1+x is a factor of 65223 + xxx

( ) 65223 += ccccP

( ) ( ) ( ) ( ) 6151211 23 +=P Replace c with1 6521 ++= Perform the indicated operations

77 +=

( ) 01 =P Since P(-1) = 0, then x + 1 is a factor of x3 + 2x2 5x 6.

b) 3y is a factor of 155323 + yyy

1553)(23 += ccccP

( ) ( ) ( ) ( ) 15353333 23 +=P Replacec with 3= 27 + 27 15 15 Perform the indicated operation

and simplify.

( ) 30543 =P

( ) 243 =P Since P(3) = 24, y 3 is not a factor of y3 + 3y2 5y 15.

Example 2. Factor the following using synthetic division.

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1

1) 343423 + xxx

1

034

34

0314

314

3434

++

+++

( ) ( ) ( )34xand1x,1x +34x3x4x

23 +

The factors of are

2) 304919234 + xxxx

25

3

011

22

0231

10155

0101321

303963

30491911

++

+++

++++

( ) ( ) ( ) ( )1xand2x,5x,3x +3049x19xxx

234 +

The factors of are

3) 2019223 + xxx

45

011

44

0431

20155

201921

++

+++

( ) ( ) ( )1xand4x,5x +

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2019x2xx23 +

The factors of are

4) 7236261322345 +++ xxxxx

3

322

021

420441

882

08421

241263

02420251

72606153

7236261321

+ ++

++++

++++

++++++

( ) ( ) ( ) ( )2xand2x,3x,3x 2 ++

7236x26x13x2xx 2345 +++ The factors of are

I. Factoring the Sum or Difference of Two Prime Odd Powers

The sum or difference of two prime odd powers can be performed using the

following formulas:

( )( )( )( )1n23n2n1nnn

1n23n2n1nnn

yyxyxxyxyx

yyxyxxyxyx

++++=

+++=+

Example. Factor the following completely.

( ) ( )

( )( )

( ) ( )16842222222

232)1

234

432234

555

+++=

+++=

+=+

xxxxx

xxxxx

xx

( ) ( )

( )( )65423324567777)2

babbabababaaba

baba

++++=

+=+

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( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( )( ) ( )6542332456

6524334256

7777

6432168422

2222222

2128)3

yxyyxyxyxyxxyx

yxyxyxyxyyxxyx

yxyx

++++++=

++++++=

=

GENERAL PROCEDURE FOR FACTORING1. Factorout any common factor.

2. If the polynomial is a binomial, factor it as the difference of two squares (a2 b2) or

thesum or difference of two cubes (a3 b3). Thesum of two squares (a2 + b2) isprime.

3. If the polynomial is a trinomial, factor it as aPTS (a2 2ab + b2) or by trial anderror method.

4. If the polynomial has more than three terms, tryfactoring by grouping.

5. If the polynomial is a binomial whose terms are both perfect squares or a trinomial

with at least two terms that are perfect squares but is not a PTS, use factoring by

completing the square.

6. If the polynomial is in one variable and has a degree higher than two, use synthetic

division

7. If the binomial is the sum or difference of two prime odd powers, apply the formulas

8. Check if all the factors are prime.

Summary of Special types of Factoring

Special Types of Factoring

Removal of the Highest Common Factor (HCF)

( ) ( )dcbaxdxcxbxax +=+Difference of Two Squares

( ) ( )bababa +=22

Sum and Difference of Two Cubes Squares

( )(( ) ( )2233

2233

babababa

babababa

+++=

++=+

Perfect Trinomial Square (PTS)

( )( )222

222

22

babababababa

=++=++

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