2.5 Special Products In the previous sections, the distributive property was used in multiplying polynomials. For example, the product of two binomials ax + b and cx + d using the distributive property is as follows:

( ax + b )( cx + d ) = ax( cx + d ) + b( cx + d ) = ( ax)( cx ) + ( ax)( d ) + ( b )( cx) + ( b )( d )

Take note that a shortcut for finding the product of two binomials is to add the products of four pairs of of the binomials namely: the product of the first , the product of the outer , the product of the inner , and the product of the last . This special order or method of multiplying binomials is known as the FOIL method. The word FOIL is formed from the first letters of the products of the of the two binomials to be added.

The following table illustrates the FOIL method of finding the product of two binomials, 2x + 1 and 3x – 5. FOIL Method F stands for the product of the First O stands for the product of the Outer I stands for the product of the Inner L stands for the product of the Last

(2x + 1) (3x – 5) ( 2x) (3x) = 6x2 F (2x + 1) (3x – 5 ) ( 2x) (–5 ) = –10x O (2x + 1) (3x – 5 ) (1) (3x ) = 3x I (2x + 1) (3x – 5 ) (1) (–5) = –5 L

F

O

I

L

The product of ( 2 x +1)( 3 x − 5) = 6 x −10 x + 3 x − 5 Combine similar = 6x 2 − 7 x − 5 2

When using the FOIL method in finding certain types of products, a specific pattern is observed. Such patterns lead to the special 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 Product of the Sum and Difference of Two Numbers The product of the sum and difference of two numbers is the difference of two squares. The difference of two squares is the square of the 1st number minus the square of the 2nd number. Thus,

( a + b )( a − b ) = a 2 − b 2 Example. Find the product of the following binomials. 1)( 3x 2 + y 2 )( 3x 2 − y 2 ) = ( 3x 2 ) − ( y 2 ) 2

2

= 9x 4 − y 4

1

2)( 5r 3 + 2 s 2 )(5r 3 − 2 s 2 ) = ( 5r 3 ) − ( 2s 2 ) 2

2

25 r 6 − 4 s 4

=

3)( 4a 2 b 2 + 7c 2 )( 4a 2 b 2 − 7c 2 ) = ( 4a 2 b 2 ) − ( 7c 2 ) 2

2

= 16 a 4 b 4 − 49 c 4

4)(5 x 2 n − 6 y 3n )(5 x 2 n − 6 y 3n ) = (5 x 2 n ) − ( 6 y 3 n ) 2

[

][

= 25 x 4 n −36 y 6 n

]

5) 5( x + y ) + 4 5( x + y ) − 4 = [ 5( x + y ) ] − ( 4) 2

2

2

2

= 25( x + y ) − 16

2

2

(

)

= 25 x 2 + 2 xy + y 2 − 16 =

25 x + 50 xy + 25 y 2 −16 2

A. 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.

( a + b ) 2 = ( a ) 2 + 2( a )( b ) + ( b ) 2 ( a − b ) 2 = ( a ) 2 − 2( a )( b ) + ( b ) 2

Example. Find the product of the following binomial. 1)( 6 x + 7 y ) = ( 6 x ) + 2( 6 x )( 7 y ) + ( 7 y ) 2

2

=

2

36 x 2 −84 xy + 49 y 2

2)(3 x 3 + 4 y 2 ) = (3 x 3 ) + 2(3 x 3 )( 4 y 2 ) + ( 4 y 2 ) 2

2

=

2

9 x 6 − 24 x 3 y 2 +16 y 4

3)(5m 4 − 2n5 ) = (5m 4 ) − 2(5m 4 )( 2n5 ) + ( 2n5 ) 2

2

=

2

25 m 8 − 20 m 4 n 5 + 4n 10

4)( 5a 2 b 3 − 3c 4 ) = (5a 2 b 3 ) − 2( 5a 2 b 3 )( 3c 4 ) + (3c 4 ) 2

2

=

25 a 4 b 6 − 30 a 2 b 3 c 4 + 9c 8

5)(3.4 x − 5) = (3.4 x ) − 2(3.4 x )( 5) + ( 5) 2

2

2

2

= 9( 4 2 x ) − 30 ( 4 x ) + 25

6)[ 5a ( 2b − c ) + 3] = [ 5a ( 2b − c ) ] − 2( 5a )( 2b − c )( 3) + ( 3) 2

2

2

= 25 a 2 ( 2b − c ) − 30 a ( 2b − c ) + 9 2

(

)

= 25 a 2 4b 2 − 4bc + c 2 − 60 ab + 30 ac + 9 = 100 a 2 b 2 −100 a 2 bc + 25 a 2 c 2 −60 ab +30 ac +9

B. Product of Any Two Binomials

2

The product of any two binomials is a general trinomial and is obtained by using the FOIL method.

( ax + b )( cx + d ) = acx 2 + adx + adx + bcx = acx 2 + ( ad + bc ) x + bd

+ bd

Example. Find the product of the following binomials.

1)( 2 x 2 − 4 )(3 x 2 + 5) = ( 2 x 2 )(3 x 2 ) + ( 2 x 2 )( 5) − ( 4 ) (3 x 2 ) − ( 4 )( 5) = 6 x 4 +10 x 2 −12 x 2 − 20 =

6 x 4 − 2 x 2 − 20

2)( 2 p 2 q 3 − r 2 )( 5 pq − r ) = ( 2 p 2 q 3 )( 5 pq ) − ( 2 p 2 q 3 )( r ) − ( 5 py ) ( r 2 ) + ( r 2 )( r ) = 10 p 3 q 4 − 2 p 2 q 3 r − 5 pqr 3)(7 a + 3b c 2

3

3

)(6a

+r 3

2

(

)

= 10 p 3 q 4 − pqr 2 pq 2 + 5r + r 3 2

+ 5b c 3

3

) = (7a )(6a ) + (7a )(5b c ) + (3b c )(6a ) + (3b c )(5b c ) 2

2

2

3

3

3

3

2

3

3

3

3

= 42 a 4 + 35 a 2b 3c 3 +18 a 2b 3c 3 +15 b 6 c 6 = 42 a 4 + 53 a 2 b 3 c 3 +15 b 6 c 6

4)(35 x + 53 y )(33 x − 52 y ) = (35 x )(33 x ) − (35 x )(52 y ) + (53 y )(33 x ) − (53 y )(52 y ) = 38 x − (35 x )(52 y ) + (53 y )(33 x ) − 55 y

= 38 x − (33 x )(52 y )(32 x − 5 y ) − 55 y

5)[ 2( x − y ) + 3z ][ 4( x − y ) − 5 z ] = 8( x − y ) −10 z ( x − y ) +12 z ( x − y ) − ( 3z )( 5 z ) 2

(

)

= 8 x 2 − 2 xy + y 2 + 2 z ( x − y ) −15 z 2

= 8 x −16 xy +8 y + 2 xz − 2 yz −15 z 2 2

2

C. Cube of a Binomial The cube of a binomial is a quadrinomial with the following : 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 second number 4th term is plus or minus the cube of the second number

( a + b ) 3 = a 3 + 3a 2b + 3ab 2 + b3 ( a − b ) 3 = a 3 − 3a 2b + 3ab 2 − b3 Example. Find the product of each of the following binomials. 1)( 3t + 4 ) = ( 3t ) + 3( 3t ) ( 4 ) + 3( 3t )( 4 ) + ( 4 ) 3

3

2

2

3

( )

= 27 t 3 +12 9t 2 + 9t (16 ) + 64

(

=

2) 5 x 3 + 2 y 2

27 t +108 t 2 +144 t + 64 3

)

2

(

= 5x 3

)

3

(

) ( 2 y ) + 3(5x )( 2 y ) + ( 2 y ) ( 25 x ) +15 x ( 4 y ) + 8 y

+ 3 5x 3

= 125 x 9 + 6 y 2

2

2

2 2

3

6

3

4

2 3

6

= 125 x 9 +150 x 6 y 2 + 60 x 3 y 4 + 8 y 6

3)( 7 a 2 b 2 − 4c 2 ) = ( 7 a 2 b 2 ) − 3( 7a 2 b 2 ) ( 4c 2 ) + 3( 7 a 2 b 2 )( 4c 2 ) − ( 4c 2 ) 3

3

2

2

= 343 a 6 b 6 − 12 c 2 ( 49 a 4 b 4 ) + 21a 2 b 2 (16c 4 ) − 64 c 6 =

343 a 6 b 6 − 588 a 4 b 4 c 2 + 336 a 2 b 2 c 4 − 64 c 6 3

3

4)[ ( m + 2b ) − 5] = ( m + 2b ) − 3( m + 2b ) ( 5) + 3( m + 2b )( 5) − ( 5) 3

3

2

2

( )

3

(

)

= m3 + 3 m 2 ( 2b ) + 3( m )( 2b ) + ( 2b ) − 15 m 2 + 4mb + 4b 2 + 25( 3m + 6b ) − 125 2

3

= m + 6bm +12 b m +8b −15 m − 60 bm − 60 b + 75 m +150 b −125 3

D.

2

2

3

2

2

Product of Binomial and Trinomial The product of a binomial and a trinomial in the form (a ± b) (a 2 ab + b 2 ) is the sum or difference of two cubes. (a +b)( a 2 − ab + b 2 ) = a 3 + b 3 (a − b)(a

2

+ ab + b 2 ) = a 3 − b 3

Example. Find the product of each of the following. 1) (5a 2 + 4b)( 25 a 4 − 20 a 2b +16 b 2 ) = (5a 2 )3 + (4b)3 = 125 a 6 + 64 b3 (5a 2 ) 2 − (5a 2 )( 4b) + ( 4b) 2

2) (7 x3 − 3 y 2 )( 49 x 6 + 21 x 3 y 2 + 9 y 4 ) = (7 x3 )3 − (3 y 2 ) = 343 x9 − 27 (7 x 3 ) 2 − (7 x 3 )( 3 y 2 ) + (3 y 2 ) 2

(

)( ) ( ) + (3 y ) = (2 x ) − (2 x )(3 y ) + (3 y ) )(3 + 3 4 + 4 ) = (3 ) − (4 ) = 3 − 4 (3 ) + (3 )(4 ) + (4 )

3) 2 x a + 3 y b 4 x 2 a − 6 x a y b + 9 y 2b = 2 x a a 2

(

4) 32 x − 4 y

4x

a

2x

2x 2

2y

2x

y

b 3

8 x 3a + 27 y 3b

b 2

b

y

3

2x 3

y 3

6x

3y

= 729 x − 64 y

y 2

E. Product of Trinomials and Quadrinomials The special product formulas can be used to find the products of trinomials and quadrinomials by grouping the into binomials. Example 1. Find the product of the following trinomials. a) ( 3a − 2c + 3d )( 3a + 2c − 3d ) = [ ( 3a ) − ( 2c − 3d ) ][ ( 3a ) + ( 2c − 3d ) ]

= ( 3a ) − ( 2c − 3d ) 2

[

2

= 9a 2 − ( 26) − 2( 26)( 3d ) + ( 3d )

(

2

= 9a 2 − 4c 2 + 12 cd + 9d 2

)

2

]

After grouping, the 1st term is 3a and the 2ndsecond are ±(2c –3d) for both binomials. Use the formula for the product of the sum and difference of two numbers Expand ( 2c − 3d ) 2 . Use the formula for the square of a binomial.

Simplify

= 9a 2 − 4c 2 +12 cd − 9d 2

4

Another solution of Example a: a) ( 3a − 2c + 3d )( 3a + 2c − 3d ) = [ ( 3a − 2c ) + ( 3d ) ][ ( 3a + 2c ) − ( 3d ) ]

After grouping, the 1st two ( 3a − 2c ) and ( 3a + 2c ) are the 1st of the binomials. The 3rd , ±3d are the 2nd of binomials. Use the FOIL method 2 = ( 3a − 2c )( 3a + 2c ) − 3d ( 3a − 2c ) + 3d ( 3a − 2c ) + 3d ( 3a + 2c ) − ( 3d ) Multiply the 1st term (3a − 2c )( 3a + 2c ) . 2 2 = ( 3a ) − ( 2c ) − 9ad + 6cd + 9ad + 6cd − 9d 2 Use the formula for the product of the sum and difference of two numbers. 2 2 2 Combine similar = 9a − 4c − 9ad + 6cd + 9ad + 6cd − 9d =

9a 2 − 4c 2 + 12 cd − 9d 2

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) ( 5m + 2n − 5 p )( 5m − 2n − 3 p ) = [ ( 5m + 2n ) − ( 5 p ) ][ ( 5m − 2n ) − ( 3 p ) ]

After grouping, (5m + 2n) and (5m – 2n) are the 1st while –5p and –3p are the 2nd of the binomials. Use the FOIL method. = ( 5m + 2n )( 5m − 2n ) − 5 p( 5m − 2n ) − 3 p ( 5m + 2n ) + ( 5 p )( 3 p ) Multiply the 1st term (5m+2n)(5m2n). 2 2 2 = ( 5m ) − ( 2n ) − 25 pm +10 pn −15 pm − 6 pn +15 p Use the formula for the product of the sum and difference of two numbers. = 25 m 2 4n 2 − 25 pm +10 pn −15 pm −6 pn +15 p 2 Combine similar =

25 m 2 − 4n 2 − 40 pm + 4 pn + 15 p 2

Note: Unlike Example a, Example b can be expressed only as the product of binomials after grouping. Thus, the FOIL method was used. c) ( 6 x − 4 y − 5 z ) 2 2 = [( 6x) − ( 4 y + 5z ) ]

= ( 6 x ) − 2( 6 x )( 4 y + 5 z ) + ( 4 y + 5 z ) 2

= 36 x 2 −12 x( 4 y + 5 z ) + ( 4 y + 5 z )

2

After grouping, the 1st term is 6x and the 2nd term is –(4y + 5z). Use the formula for the square of a binomial ( square of the difference of two numbers ).

2

Expand (4y+5z)2. Use the formula for the square of the sum of two numbers. 2 2 2 Simplify = 36 x − 12 x( 4 y + 5 z ) + ( 4 y ) + 2( 4 y )( 5 z ) + ( 5 z )

[

=

]

36 x 2 − 48 xy − 60 xz + 16 y 2 + 40 yz + 25 z 2

5

Note: The square of a trinomials can always be expressed as the square of a binomial after grouping the 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) ( 2 x + 3 y − z + 4 )( 2 x + 3 y + z − 4) = [ ( 2 x + 3 y ) − ( z − 4 ) ][ ( 2 x + 3 y ) + ( z − 4 ) ]

= ( 2 x + 3 y ) − ( z − 4) 2

[

2

= ( 2 x ) + 2( 2 x )( 3 y ) + ( 3 y ) 2

2

After grouping, the 1st term is (2x + 3y) and the 2nd term is ±(z – 4) for both binomials. Use the formula for the product of the sum and difference of two numbers. Expand (2x + 3y)2 and (z - 4)2. Use the formula 2 ) 2 the square of a binomial. − ( z ) − 2( z )( 4) + ( 4for

] [

]

= ( 4 x 2 + 12 xy + 9 y 2 ) − ( z 2 − 8 z + 16 ) =

Simplify

4 x +12 xy + 9 y − z + 8 z −16 2

2

2

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) ( 4a − 3 y − 2 z + 5) 2 2 = [ ( 4a − 3 y ) − ( 2 z − 5) ] = ( 4a − 3 y ) − 2( 4a − 3 y )( 2 z − 5) + ( 2 z − 5) 2

[( )

2

]

[

= 4a 2 − 2( 4a )( 3 y ) + ( 3 y ) − 2( 4a − 3 y )( 2 z − 5) + ( 2 z ) (2z – 5). Use

[

= ( 4a ) − 2( 4a )( 3 y ) + ( 3 y ) 2

2

2

]

2

After grouping, the 1st term is (4a–3y and the 2nd term is –(2z – 5). Use the formula for the square of a binomial. Expand (4a – 3y)2 and (2z – 5)2. Use the formula for the square of a binomial. 2 − 2( 2 z )( 5) + ( 5) Multiply (4a – 3y) and

]

FOIL method. 2 2 − 2( 8az − 6 yz − 20a + 15 y ) + ( 2 z ) − 2( 2 z )( 5) + ( 5) Simplify

[

]

= 16 a 2 − 24 ay + 9 y 2 −16 az +12 yz + 40 a −30 y + 4 z 2 − 20 z + 25

Note: The square of a quadrinomial can always be expressed as the square of a binomial after grouping the 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 the polynomial and twice the product of any combination of two . This method of finding the square of a polynomial is useful if the polynomial contains more than four . 6

Example: Find the product of the following polynomials. 1) ( 7 m + 5n − 3 p + 6) 2 = ( 7m ) + ( 5n ) + ( − 3 pn ) + ( 6) + 2( 7m )( 5n ) + 2( 7m )( − 3 p ) + 2( 7m )( 6) + 2( 5n )( − 3 p ) + 2( 5n )( 6) + 2( − 3 p )( 6) 2

=

2

2

2

49 m 2 + 25 n 2 + 9 p 2 + 36 + 70 mn − 42 mp + 84 m −30 np + 60 n −36 p

2) ( 2a + 3b − 4c + d − 2 ) 2

= ( 2a ) + ( 3b ) + ( − 4c ) + ( d ) + ( − 2 ) + 2( 2a )( 3b ) + 2( 2a )( − 4c ) + 2( 2a )( d ) + 2( 2a )( − 2 ) + 2

2

2

2

2

2( 3b )( − 4c ) + 2( 3b )( d ) + 2( 3b )( − 2 ) + 2( − 4c )( d ) + 2( − 4c )( − 2 ) + 2( d )( − 2 ) =

4a 2 + 9b 2 +16 b 2 + d 2 + 4 +12 ab −16 ac + 4ad + 8a − 24 bc + 6bd −12 b − 8cd +16 c − 4d

Summary of the Special Product Formulas. Special Product Formulas Product of the Sum and Difference of Two Numbers = Difference of Two Squares

( a + b )( a − b ) = a 2 − b 2 Square of a Binomial = Perfect Trinomial Square

( a + b) 2 ( a − b) 2

= a 2 + 2ab + b 2 = a 2 − 2ab + b 2

Product of any Two Binomials = General Trinomial

( ax + b )( cx + d ) = acx 2 + ( bc + ad ) x + bd

Cube of a Binomial = Quadrinomial

( a + b) 3 ( a −b) 3

= a 3 + 3a 2 b + 3ab 2 + b 3 = a 3 − 3a 2 b + 3ab 2 − b 3

Product of Binomial and Trinomial

( a + b ) ( a 2 − ab + b 2 ) = a 3 + b 3 ( a − b ) ( a 2 + ab + b 2 ) = a 3 − b 3

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 .

2.6 Factoring Factoring is the reverse of multiplication. It is the process of expressing a given polynomial as a product of its factors.

7

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 factored when each of its polynomial factors is prime.

Types of Factoring A.

Removal of the Highest Common Factor (HCF) A common factor is 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 least exponents in every term of a polynomial. Example: The HCF of a) 5x and 15x2 is 5x b) 3ab2 and 6a is 3a c) 8x2y2z3, 16x3y3z2 and 24x4yz4 is 8x2yz2 If every term of a polynomial contains a common factor, then the polynomial 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.  ax − bx + cx − dx  ax − bx + cx − dx = ( x )   x   = ( x )( a − b + c − d )

Example. Factor the following completely: 16 x 7 y +56 xy 7 1) 16 x 2 y +56 xy 7 = (8 xy )   8 xy 

=

( 8 xy ) ( 2 x 6 + 7 y 6 )

 Factor out HCF, 8xy   

Factor 15 p 2 q 4 r 3 + 20 p 3q 5r 2 − 25 p 4 q 3r  2)15 p 2 y 4 r 3 + 20 p 3q 5r 2 − 25 p 4 q 3r = (5 p 2 q 3r ) out HCF,  5 p 2 q 3r 2 3  5p q r  2 3 2 2 2 = (5 p q r )(3qr + 4 pq r − 5 p ) 3) 6 x( a − h ) + 3 y ( h − a ) = 6 x( a − h ) − 3 y ( a − h )

 6 x( a − h ) − 3 y ( a − h )  = 3( a − h )   3( a − h )   =

Multiply 3y(h – a) by -1

Factor out HCF, 3(a – h)

3( a − h )( 2 x − y )

B. 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. a 2 − b 2 = ( a + b )( a − b ) 8

Note: The sum of two squares (a2 + b2) is a prime polynomial, hence, it is not factorable. Example. Factor the following completely. 1)bx 2 − 9by 4 = b( x 2 − 9 y 4 )

[

Factor out HCF, b

]

= ( b) ( x ) − (3 y ) = ( b ) ( x + 3 y 2 )( x − 3 y 2 ) 2 2

2

(

2)( 5a − 2) r 2 − ( 5a − 2) t 2 = ( 5a − 2) r 2 − t 2

)

( 5a − 2 )( r + t )( r − t )

=

3) 25 a 2 −16 b 2 = ( 5a ) − ( 4b ) 2

2

4)( 2a − 5b ) −16 = ( 2a − 5b ) − ( 4) = [ ( 2a − 5b ) + 4][ ( 2a − 5b ) − 4] = ( 2a − 5b + 4 )( 2a − 5b − 4) 2

Factor out (5a – 2)

2

2

Simplify

5)( 3 x − y ) − ( 2 z + 1) = [ ( 3 x − y ) + ( 2 z + 1) ][ ( 3 x − y ) − ( 2 z + 1) ] 2

2

( 3x − y + 2 z +1)( 3x − y − 2 z −1)

=

6) x 4 − 81 = ( x 2 ) − ( 9) 2

Simplify

2

= ( x 2 + 9 )( x 2 − 9 )

(x

=

)

+ 9 ( x + 3)( x − 3)

2

C. Sum and Difference of Two Cubes The factors of the sum or difference of two cubes consist of a binomial factor and a trinomial factor. The binomial factor is the sum or difference of the cube roots of the two of the given binomial. The trinomial factor is obtained from the binomial factor and has the following : First term is the square of the 1st term of the binomial factor. Middle term is the product of the two of the binomial factor multiplied by (–1). Last term is the square of the 2nd term of the binomial factor.

( = ( a − b ) (a

a 3 + b 3 = ( a + b ) a 2 − ab + b 2 3

a −b

3

2

+ ab + b

2

) )

Example. Factor the following completely. 1)9a 2 b 2 − 9a 2 b 5 = (9a 2 b 2 )(1 − b 3 ) Factor out HCF, 9a2b2 2 2 2 = 9a b (1 − b ) 1 + b + b 3 3 Factor out HCF, 8 2)216 x + 8 y = 8 27 x 3 + y 3

[

(

(

= 8 ( 3x ) + ( y ) 3

) )

3

(

(

]

= 8( 3x + y ) 9 x 2 − 3 xy + y 2

)

) 9

3)( a − b ) − 64 = ( a − b ) − ( 4) 3

3

3

[

= [ ( a − b ) − 4] ( a − b ) + 4( a − b ) + ( 4)

=

4)3 x

3n

− 24 y

(

2

( a − b − 4) ( a

= 3 x − 8y

3m

3n

[(

3m

) (

)

)

n 3

]

− 2ab + b + 4a − 4b + 16 )

2

3

2

Simplify

2

]

Factor out HCF, 3

= 3 x − 2ym = 3( x n − 2 y m )( x 2 n + 2 x n y m + 4 y 2 m ) Factor out HCF, 3y2x 5)24 y 2 x 4 − 375 y 8 x = 3 y 2 x 8 x 3 − 125 y 6

(

[

= 3 y 2 x ( 2 x ) − (5 y 2 3

) )] 3

= 3 xy 2 ( 2 x − 5 y 2 )( 4 x 2 +10 xy 2 + 25 y 4 )

(

6)128a 7 − 2b 6 a = 2a 64a 6 − b 6

[(

)

) − (b )] = 2a ( 4a − b )(16a

= 2a 4a

2 3

Factor out HCF, 2a

2

2

2

4

+ 4a 2 b 2 + b 4

)

Factor (4a2 – b2) as the difference of two squares

2a ( 2a + b )( 2a − b ) (16 a 4 + 4a 2 b 2 + b 4 )

=

7) x12 − y 12 = ( x 4 ) − ( y 4 ) 3

[

3

= ( x4 − y4 ) ( x4 ) + x4 y4 + ( y4 ) 2

2

Factor as the difference of two cubes

]

Factor (x4 - y4) as the difference of two squares

= ( x 2 + y 2 )( x 2 − y 2 )( x 8 + x 4 y 4 + y 8 )

(x

=

2

+ y 2 )( x + y )( x − y ) ( x 8 + x 4 y 4 + y

Factor (x2 – y2) as the difference of two squares 8

)

Note: Example 7 can also be factored as the difference of two squares Factor as the difference of two squares 12 12 6 2 6 2 x

−y

( ) −(y ) = ( x + y )( x − y ) = ( x ) + ( y )  ( x ) − ( y )      = ( x + y )( x − x y + y )( x − y )( x = x

6

6

2 3

2

6

6

2 3

2

4

2 3

2

2

Factor as the sum and difference of two cubes

2 3

4

2

2

4

+ x2 y2 + y4

)

Factor (x2 – y2) as the difference of two squares

= ( x 2 + y 2 )( x 4 − x 2 y 2 + y 4 )( x + y )( x − y )( x 4 + x 2 y 2 + y 4 )

D. Perfect Trinomial Square (PTS) A trinomial is a perfect trinomial square if the first and last are perfect squares and the middle term is plus or minus twice the product of the square roots of the 1st and the last . The factors of a PTS are the square of the sum or the difference of the square root of the 1st and the last of the given trinomial. a 2 + 2ab + b 2 = ( a + b ) a 2 − 2ab + b 2 = ( a − b ) 10

2 2

Example. Factor the following completely. 1)36 x 2 y 2 12 xy +1 = ( 6 xy ) + 2( 6 xy )(1) + (1) 2

2

( 6 xy +1) 2 2 2 = ( 3a ) + 2( 3a )( 5b ) + ( 5b ) 2 = ( 3a + 5b )

=

2)9a 2 + 30 ab + 25b 2

(

3)2a 2 b + 8ab + 8b = ( 2b ) a 2 + 4a + 4

[

)

= ( 2b ) ( a ) + 2( a )( 2) + ( 2 ) 2

2

Factor out HCF, 2b

]

( 2b )( a + 2) Factor out HCF, 3a 4)3a − 6a + 3a = ( 3a ) ( a 2 − 2a + 1) 2 = ( 3a )( a − 1) 2 2 2 2 5)49 x y − 28 xyz + 4 z 2 = ( 7 xy ) − 2( 7 xy )( 2 z ) + ( 2 z ) 2 = ( 7 xy − 2 z ) 2 2 6)a 4 − 8a 2 +16 = ( a 2 ) − 2( a 2 )( 4) + ( 4) 2

=

3

2

(

= a2 −4

)

2

Factor as difference of two squares

= [ ( a + 2 )( a − 2 ) ]

=

(

2

Simplify

( a + 2) 2 ( a − 2) 2

) = 9 x [( x ) + 2( x ) ( 1) + (1) = 9 x ( x + 1) = 9 x [ ( x + 1) ( x − x + 1) ] = 9 x ( x + 1) ( x

7)9 x 8 18x 5 + 9 x 2 = 9 x 2 x 6 + 2 x 3 + 1 2

2

3 2

3

3

2

Factor out HCF, 9x2

]

2

2

2

Factor as the sum of two cubes

2

2

2

2

− x + 1)

Simplify 2

[

]

8)9( x + y ) − 30( x + y )( x − y ) + 25( x − y ) = 3( x + y ) − 2( 3)( x + y )( 5)( x − y ) + [ 5( x − y ) ] 2 = [3( x + y ) − 5( x − y ) ] Apply distributive property 2

2

2

2

= ( 3 x + 3 y − 5 x + 5 y ) Combine similar 2 = ( − 2 x +8 y ) Factor out HCF, –2 2

= [ − 2( x − 4 y ) ] =

4( x − 4 y )

2

2

Simplify

E. General Trinomial A general trinomial of the form acx 2 + (ad + bc ) xy + bdy 2 can be factored into the product of two binomials (ax + by )( cx + dy ) . a, b, c, d can be obtained by using the trial and error method. Trial: a) The first of both binomials are factors of the first term of the given trinomial. b) The second of both binomials are factors of the last term of the given trinomial.

11

To check if trial is correct, the sum of the products of the outer and inner must be equal to the middle term of the given trinomial. Example. Factor the following completely. 1)45 x 2 − 8 xy − 4 y 2 =

( 9 x + 2 y )( 5 x − 2 y )

(

)

Factor out HCF, 4y

2)8 y 3 − 20 y 2 +12 y = 4 y 2 y 2 − 5 y + 3 =

4 y ( 2 y −3)( y −1)

( x + 3)( 2 x + 7 ) − 25 cd − 4 = ( 3cd − 4 )( 7cd +1)

3) 2 x 2 +13 x + 21 = 2

4) 21c d

2

(2x

5) 4 x 2 n − 4 x n − 3 =

n

)(

)

− 3 2 x n +1

6)6( 2a + b ) − 5( a + b ) − 25 = [ 3( a + b ) + 5][ 2( 2a + b ) − 5] 2

( 6a + 3b + 5)( 4a + 2b − 5)

=

Simplify

[

7)12 z 2 + 9 z ( x − y ) − 21( x − y ) = 3 4 z 2 + 3z ( x − y ) − 7( x − y ) = 3[ 4 z + 7( x − y ) ][ z − ( x − y ) ] 2

= 3( 4 z + 7 x − 7 y )( z − x + y )

2

]

Factor out HCF, 3 Simplify

F. Factoring by Grouping Factoring by grouping is used to factor polynomials consisting of more than three . In factoring by grouping, the of the given polynomial are grouped to form a binomial or a trinomial that are both factorable. Example. Factor the following completely.

(

)

1)3 x 2 + x 3 −15 − 5 x = 3 x 2 + x 3 − (15 + 5 x ) = x ( 3 + x ) − 5( 3 + x ) 2

(

= (3 + x) x 2 − 5

=

)

Factor out HCF of each grouped Factor out HCF, (3 + x) Simplify

( x + 3) ( x 2 − 5)

(

) (

) grouped as a PTS

2 Factor 2) 16 a 2 + b 2 − 4c 2 − d 2 − 8ab + 4cd = 16 a 2 − 8ab + b 2 − 4c 2 − 4 cd + deach

= ( 4a − b ) 2 − ( 2c − d ) 2

Factor as the difference of two square

= [( 4a − b + 2c ) ][ ( 4a − b ) − ( 2c − d ) ]

=

( x + 3) ( x 2 − 5)

12

Simplify

(

) (

3) 4 x 2 + 4 xy + y 2 − 9 z 2 = 4 x 2 + 4 xy + y 2 − 9 z 2 = ( 2 x + y ) − ( 3z ) 2

)

Factor the 1st first grouped as a PTS

2

= [ ( 2 x + y ) + 3z ][ ( 2 x + y ) − 3z ] = ( 2 x + y + 3 z )( 2 x + y − 3 z )

(

) (

4) 25 x 2 − y 2 + 2 yz − z 2 = 25 x 2 − y 2 − 2 yz + z 2 =( 5 x ) 2 −( y − z )

)

Factor the 2nd grouped as a PTS

2

= [ ( 5 x ) + ( y − z ) ][ ( 5 x ) − ( y − z ) ] = ( 5 x + y − z )( 5 x − y + z )

(

Factor as the difference of two squares as a PTS Simplify

)

5) x 2 − 4 xy + 2 y + 4 y 2 = x 2 − 4 xy + 4 y 2 + ( x − 2 y )

Factor as the difference of two squares as a PTS Simplify Factor the 1st grouped as a PTS

= ( x −2 y) +( x −2 y) Factor out HCF, (x -2y) = ( x − 2 y )( x − 2 y +1) 6) 25 a 2 − 36 b 2 +10 b −12 a = ( 25 a 2 − 36 b 2 ) + (10 a −12 b ) Factor the 1st grouped as 2

= ( 5a + 6b )( 5a − 6b ) + 2( 5a − 6b )

= ( 5a − 6b )( 5a + 6b + 2 ) 7)8 x − y + 3xy − 6 x = (8 x − y ) + ( 3xy − 6 x 2 ) 3

3

2

3

difference of two squares and the 2nd grouped by removal of HCF, 2 Factor out HCF, (5a – 6b)

3

( = ( 2x − y)(4x

Factor the 1st grouped as difference of two cubes and the 2nd grouped by removal of HCF, 3x Simplify

) ) − ( 3x )( 2 x − y )

= ( 2 x − y ) 4 x 2 + 2 xy + y 2 + 3x( y − 2 x )

=

2

+ 2 xy + y

2

( 2 x − y ) ( 4 x 2 + 2 xy + y 2 − 3 x )

(

)

Factor out HCF, (2x –y)

(

8) x 2 + 4 x + 4 − 6 xy −12 y + 9 y 2 = x 2 + 4 x + 4 − ( 6 xy +12 y ) + 9 y 2

= ( x + 2) 2 − 6 y ( x + 2) + ( 3 y ) 2 = [( x + 2) − ( 3 y ) ] 2

=

( x − 3 y + 2)

2

) Factor the 1st

Grouped as a PTS and the 2nd grouped by removal of HCF, 6y

Factor as a PTS Simplify

G. Factoring by Completing the Square Factoring by completing the square is applicable to binomials whose are both perfect squares and to trinomials with at least two that are perfect squares. 13

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. 1) x 4 + 4 y 4 = [( x 2 ) 2 + 2( x 2 )(2 y 2 ) + ( 2 y 2 ) 2 ] − 4 x 2 y 2 = ( x 2 + 2 y 2 ) 2 − ( 2 xy ) 2 = ( x 2 + 2 y 2 + 2 xy )( x 2 + 2 y 2 − 2 xy ) = ( x 2 + 2 xy + 2 y 2 )( x 2 − 2 xy + 2 y 2 ) 2) y 4 + 6 y 2 + 25 = [( y 2 ) 2 + 6 y 2 + 4 y 2 + (5) 2 ] − 4 y 2 = [( y 2 ) 2 +10 y 2 + (5) 2 ] − ( 2 y ) 2 = ( y 2 + 5) 2 − ( 2 y ) 2 = ( y 2 + 5 + 2 y )( y 2 + 5 − 2 y ) = ( y 2 + 2 y + 5)( y 2 − 2 y + 5) 3) 36 a 4 +15 a 2 b 2 + 4b 2 = [( 6a 2 ) 2 +15 a 2 b 2 + 9a 2 b 2 + ( 2b 2 ) 2 ] − 9a 2 b 2 = [( 6a 2 ) 2 + 24 a 2 b 2 + ( 2b 2 ) 2 ] − (3ab ) 2 = (6a 2 + 2b 2 ) 2 − (3ab ) 2 = (6a 2 + 2b 2 + 3ab )( 6a 2 + 2b 2 − 3ab ) = (6a 2 + 3ab + 2b 2 )( 6a 2 − 3ab + 2b 2 ) 4) x 4 − 7 x 2 + 9 = [( x 2 ) 2 − 7 x 2 + x 2 + (3) 2 ] − x 2 = [( x 2 ) 2 − 6 x 2 + (3) 2 ] − ( x) 2 = ( x 2 − 3) 2 − ( x ) 2 = ( x 2 − 3 + x)( x 2 − 3 − x ) = ( x 2 + x − 3)( x 2 − x − 3) 5) s 4 −14 s 2 + 25 = [( s 2 ) 2 −14 s 2 + 4 s 2 + (5) 2 ] − 4 s 2 = [( s 2 ) 2 −10 s 2 + (5) 2 ] − ( 2 s ) 2 = ( s 2 − 5) 2 − (2 s ) 2 = ( s 2 − 5 + 2 s )( s 2 − 5 − 2 s ) = ( s 2 + 2 s − 5)( s 2 − 2 s − 5)

6) x 8 − 21 x 4 + 36 = [( x 4 ) 2 − 21 x 4 + 9x 4 + (6) 2 ] − 9x 4 = [( x 4 ) 2 −12 x 4 + (6) 2 ] − (3 x 2 ) 2 = ( x 4 − 6) 2 − (3 x 2 ) 2 = ( x 4 − 6 + 3 x 2 )( x 4 − 6 − 3 x 2 ) = ( x 4 + 3 x 2 − 6)( x 4 − 3 x 2 − 6)

H. Factoring by Synthetic Division Factoring by synthetic division is applicable to a polynomial in one variable whose degree is higher than two. 14

Factor Theorem The factor theorem is used to determine if a binomial x – c is a factor of the given polynomial. Factor Theorem A polynomial P(x) has a factor x – c, where c is any real number if and only if P(c) = 0 (remainder is 0). Example 1. Determine if: a) x +1 is a factor of x 3 + 2 x 2 − 5 x − 6 P ( c ) = c 3 + 2c 2 − 5c − 6

P ( −1) = ( −1) + 2( −1) − 5( −1) − 6 3

2

Replace c with –1 Perform the indicated operations

= −1 + 2 + 5 − 6 = −7 + 7

P ( −1) = 0

Since P(-1) = 0, then x + 1 is a factor of x3 + 2x2 – 5x –6. b) y −3 is a factor of y 3 + 3 y 2 − 5 y −15 P (c ) = c 3 + 3c 2 − 5c −15

P( 3) = ( 3) + 3( 3) − 5( 3) − 15 3

2

Replace c with 3 Perform the indicated operation and simplify.

= 27 + 27 – 15 –15 P ( 3) = 54 −30 P ( 3) = 24

Since P(3) = 24, y – 3 is not a factor of y3 + 3y2 – 5y – 15. Example 2. Factor the following using synthetic division. 1) 4 x 3 − 3 x 2 − 4 x + 3 4 −3 +4 4 +1 −4 4 −3

−4 +3 +1 − 3 −3 0 +3 0

1

−1

The factors of 4x 3 − 3x 2 − 4x + 3 are ( x − 1) , ( x + 1) and ( 4x − 3 ) 2) x 4 − x 3 −19 x 2 + 49 x − 30

1 − 1 − 19 + 49 − 30 + 3 + 6 − 39 + 30 1 + 2 − 13 + 10 − 5 + 15 − 10 1 −3 + 2 +2 − 2 1 −1

0

0

3 −5

0

2

4 3 2 The factors of x − x − 19x + 49x − 30 are ( x − 3 ) , ( x + 5 ) , ( x − 2 ) and ( x − 1)

15

3) x 3 − 2 x 2 −19 x + 20

1 − 2 −19 + 20

5

+ 5 +15 − 20 1 +3 − 4

0

−4 + 4 1 −1

−4

0

The factors of x 3 − 2x 2 − 19x + 20 are ( x − 5 ) , ( x + 4 ) and ( x − 1) 4) x 5 + 2 x 4 −13 x 3 − 26 x 2 + 36 x + 72

1 + 2 − 13 − 26 + 36 + 72

3

+ 3 + 15 + 6 − 60 − 72 1 + 5 + 2 − 20 − 24 0 − 3 − 6 + 12 + 24 1 + 2 − 4 − 8 + 2 + 8 + 8 1 + 4 + 4 − 2 − 4 1+ 2 0

−3

0

2

0

−2

2 The factors of x 5 + 2x 4 − 13x 3 − 26x 2 + 36x + 72 are ( x − 3 ) , ( x + 3 ) , ( x + 2 ) and ( x − 2 )

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:

( = ( x − y )( x

x n + y n = ( x + y ) x n −1 − x n −2 y + x n −3 y 2 −  + y n −1 n

x −y

n

n −1

+x

n −2

y+x

n −3

2

y + + y

n −1

) )

Example. Factor the following completely. 1) x 5 + 32 = ( x ) + ( 2 ) 5

5

= ( x + 2) ( x 4 − 2 x 3 + 2 2 x 2 − 2 3 x + 2 4 ) =

( x + 2) ( x 4 − 2 x 3 + 4 x 2 − 8 x + 16)

2)a 7 + b 7 = ( a ) + ( b ) 7

=

7

( a + b ) ( a 6 − a 5 b + a 4 b 2 − a 3 b 3 + a 2 b 4 − ab 5 + b 6 )

3) x 7 − 128 y 7 = ( x ) − ( 2 y ) 7

7

(

= ( x − 2 y ) x 6 + 2 yx 5 + ( 2 y ) x 4 + ( 2 y ) x 3 + ( 2 y ) x 2 + ( 2 y ) x + ( 2 y ) =

( x − 2 y)( x

2

6

3

4

5

+ 2 x y + 4 x y + 8 x y + 16 x y + 32 xy + 64 y 5

4

2

3

3

GENERAL PROCEDURE FOR FACTORING 16

2

4

5

6

)

6

)

1. Factor out any common factor. 2. If the polynomial is a binomial, factor it as the difference of two squares (a2 – b2) or the sum or difference of two cubes (a3 ± b3). The sum of two squares (a2 + b2) is prime. 3. If the polynomial is a trinomial, factor it as a PTS (a2 ± 2ab + b2) or by trial and error method. 4. If the polynomial has more than three , try factoring by grouping. 5. If the polynomial is a binomial whose are both perfect squares or a trinomial with at least two 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) ax − bx + cx − dx = ( x )( a − b + c − d )

Difference of Two Squares

a 2 − b 2 = ( a + b )( a − b )

Sum and Difference of Two Cubes Squares

( = ( a + b ) (a

a 3 + b3 = ( a + b ) a 2 − ab + b 2 3

a −b

3

2

+ ab + b

Perfect Trinomial Square (PTS)

2

) )

a 2 + 2ab + b 2 = ( a + b ) 2 a 2 − 2ab + b 2 = ( a − b ) 2

17