MATH 102 – College Algebra
Chapter 2
SPECIAL PRODUCTS AND FACTORING
TYPES OF SPECIAL PRODUCTS
Some products of polynomials can be solved without applying the distributive property. These cases of products of
polynomials have been classified because of the special forms of the factors and can be obtained by applying the Special
Products Formulas.
Type 1: Product of Two Binomials
2
2
(ax + by) (cx + dy) = acx + (ad + bc)xy + bdy
The product of a binomial by another binomial is obtained as follows:
•
•
•
The first term of the product is the product of the first terms of the multiplicand and the multiplier;
The second or the middle term of the product is obtained by taking the cross product or adding the product
of the extreme terms to the product of the adjacent terms.
The last term of the product is the product of the last terms of the multiplicand and the multiplier.
Examples:
1. (3x – 2y)(4x + 3y)
=
=
3(4)x2 + [(3)(3)+(-2)(4)]xy + (-2)(3)y2
12x2 + xy – 6y2
2. (m3 – 5) (m3 – 2)
=
=
(m3m3) + (- 5 – 2)m3 + 10
m6 - 7m3 + 10
3. (3a – 5x)(6b + 7y)
=
=
(3a)(6b) + (3a)(7y) + (-5x)(6b) + (-5x)(7y)
18ab + 21ay – 30bx – 35xy
Type 2: Square of Binomials
2
2
2
(x + y) = x + 2xy + y
2
2
2
(x – y) = x – 2xy + y
•
The result of a square of a binomial is a perfect trinomial square which comprises the following terms:
o square of the first term;
o twice the product of the first term and the second term;
o square of the second term.
Examples:
1. (2x – 3y)2
=
=
4x2 + 2(2x)(-3y) + 9y2
4x2 -12xy + 9y2
2. (5x + 2)2
= 25x2 + 2(5x)(2) + 4
= 25x2 + 20x + 4
3. (3ab – 4m)2 = (3ab)2+2(3ab)(-4m)+(-4m)2
= 9a2b2 – 24abm + 16m2
4. (4x – 3y + 2)2 = [(4x – 3y) + 2]2
= (4x– y)2+2(4x–3y)(2)+ 22
= [16x2+2(4x)(-3y)+(-3y)2]
+ 4(4x – 3y) + 4
= 16x2–24xy+9y2+1x–12y+4
1
Prepared by Mrs. Koni Gutierrez Cruz
Assistant Professor I
| Bataan Peninsula State University
MATH 102 – College Algebra
Type 3: Product of the Sum and Difference of the Same Two Terms
2
2
(x + y) (x – y) = x – y
•
The product of the sum and difference of the same two terms is equal to the difference of their squares.
Examples:
1. (2x – 7y) (2x + 7y)
=
=
(2x)2 – (7y)2
4x2 – 49y2
2. (3m3 – 4n4) (3m3 + 4n4)
=
=
(3m3)2 – (4n4)2
9m6 – 16n8
3. (x + 2y – 5) (x – 2y + 5)
=
=
=
=
[x + (2y – 5)] [x – (2y – 5)]
(x)2 – (2y – 5)2
x2 – (4y2 – 20y + 25)
x2 – 4y2 + 20y - 25
4. (3a – 5b + 4) (3a – 5b – 4)
=
=
=
[(3a – 5b) + 4] [(3a – 5b) – 4]
(3a – 5b)2 – 42
9a2 – 30ab + 25b2 – 16
EXERCISES:
EXERCISES: Find the product of the following using
using special product formulas if necessary:
A.
B.
1.
2.
3.
4.
5.
6.
7.
8.
9.
(5x – 2y)(2x – 3y)
(3m – 2n)(6m – 5n)
(2r2 – 5s)(3r2 – s)
(ab + 3)(ab – 7)
(3a2b – 5xy)(6a2 – 2xy)
(6m – 3n)(4r + 2t)
(5mn – 2de)(3gh + rs)
(abc + 1)(def – 4)
(3kn + 2)(4kn – 4)
1
 3

10.  xy − z  + 2 z 
2
 2

C.
1.
2.
3.
4.
5.
6.
7.
(x – 6)2
(3x – 5)2
(2m – n)2
(3b + 4c)2
(2ab – 5) (2ab – 5)
( - 3 + 4y)2
(6xy – 5z)2
2
1.
2.
3.
4.
5.
6.
7.
2

 x − 5
3

9. [(x – 2) + 4] [(x – 2) + 4]
10. [(2x – 5) – (x + 1)]2
8.
(x + 6) (x – 6)
(3d – 5)(3d +5)
(4x – 9y)(4x + 9y)
(3m2n + 2pq2) (3m2n – 2pq2)
(2m2 – 3n)(2m2 +3n)
(3b2d + 10e)(3b2d – 10e)
(2k3n2 + 7)( 2k3n2 – 7)
3
 3

8.  x 2 − y  x 2 + y 
4
4



9. [(2x – 3) + 4] [(2x – 3) – 4]
10. [(2x – 5) + (x + 1)]
[(2x – 5) – (x + 1)]
2
Prepared by Mrs. Koni Gutierrez Cruz
Assistant Professor I
| Bataan Peninsula State University
MATH 102 – College Algebra
Type 4: Cube of a Binomial
3
3
2
2
3
(x + y) = x + 3x y + 3xy + y
3
3
2
2
3
(x – y) = x – 3x y + 3xy – y
•
The cube of a binomial is a four-term polynomial which comprises the following terms:
o cube of the first term;
o thrice the square of the first term times the second term;
o thrice the first term times the square of the second term;
o cube of the second term.
Examples:
1. (x2 – 3)3
= (x2)3 + 3(x2)2(-3) + 3(x2)(-3)2 + (-3)3
= x6 – 9x4 + 27x2 – 27
2. (3m4 + 2n2)3
= (3m4)3 + 3(3m4)2(2n2) + 3(3m4)(2n2)2 + (2n2)3
= 27m12 + 54m8n2 + 36m4n4 + 8n6
3. (d2e3 – f2g3h4)3
= (d2e3)3 + 3(d2e3)2(-f2g3h4) + 3(d2e3) (-f2g3h4)2 + (-f2g3h4)3
= d6e9 – 3d4e6f2g3h4 + 3d2e3f4g6h8 – f6g9h12
Type 5: Special Case of Product of Binomial and Trinomial
2
2
3
3
(x – y)(x + xy + y ) = x – y
2
2
3
3
(x + y)( x - xy + y ) = x + y
The definite forms of the binomial and trinomial lead to a very simple product called the sum or difference of two
cubes.
negative product of the 1st & 2nd
terms of the binomial
Examples:
1. (3a – 4b)( 9a2 + 12ab + 16b2 )
=
(3a)3 – (4b)3
=
square of the
first term of the
binomial
27a3 – 64b3
square of the
last term of the
binomial
the product is obtained
by taking the
cube of the first tem
plus or minus the
cube of the last term
of the binomial.
2. (3a + 4b) (9a2 - 12ab + 16b2 )
=
(3a)3 – (4b)3
=
27a3 – 64b3
3. (6x2 + 5y) (36x4 – 30x2y + 25y2)
=
(6x2)3 + (5y)3
=
216x6 + 125y3
4. (2ab3 – 7x2y) (4a2b6 + 14ab3 x2y + 49x4y2) =
5. (x2 + y3) (x4 + x2y3 + y6)
=
(2ab3)3 – (7x2y)3 =
8a3b9 – 343x6y3
**this does not fit the given formula, thus cannot apply the special
product formula.
**it can be solved by applying the distributive property.
x6 + x4y3 + x2y6 + x4y3 + x2y6 + y9 =
x6 + 2x4y3 + 2x2y6 + y9
3
Prepared by Mrs. Koni Gutierrez Cruz
Assistant Professor I
| Bataan Peninsula State University
MATH 102 – College Algebra
Type 6: Square of Trinomial
2
2
2
2
(x + y + z) = x + y + z + 2xy + 2xz + 2yz
Examples:
1. (2x – y – z)2
=
[2x + (-y) + (-z)]2
=
=
(2x)2+(-y)2+(-z)2+2(2x)(-y)+2(2x)(-z)+2(-y)(-z)
4x2 + y2 + z2 – 4xy – 4xz + 2yz
2. (2a + 4b + c)2
=
=
(2a)2 + (4b)2 + (c)2 + 2(2a)(4b) + 2(2a)(c) + 2(4b)(c)
4a2 + 16b2 + c2 +16ab + 4ac + 8bc
3. (5x2 – 2x + 1)2
=
=
=
(5x2)2 + (-2x)2 + (1)2 + 2(5x2)(-2x) + 2(5x2)(1) + 2(-2x)(1)
5x4 + 4x2 + 1 – 20x3 + 10x2 – 4x
(arranging & combining terms)
5x4 – 20x3 + 14x2 – 4x+ 1
** this does not fit the given formula because the given is not a
trinomial.
** can be solved by grouping and applying Type 2 formula.
4. (a – 2b + c – d)2
=
=
=
[(a – 2b) + (c – d)]2
(a – 2b)2 + 2(a – 2b)(c – d) + (c – d)2
a2 – 4ab + 4b2 + 2ac – 2ad – 4bc + 4bd + c2 – 2cd + d2
EXERCISES:
EXERCISES: Find the product of the following using Types
Types 4 to 6 formulas if necessary:
B.
A.
1.
2.
3.
4.
5.
6.
7.
8.
9.
2y)3
(x –
(3m – n)3
(2r2 – 5s)3
(ab + 3)3
(3a2b – 5xy)3
(6m – 3n)3
(5mn – 2de)3
(abc + 1)3
(3kn + 2)3
1
10.  xy − z 

2
3
C.
1.
2.
3.
4.
5.
6.
7.
8.
9.
6)(x2 +
(x –
6x + 36)
(2 + 5y)( 4 – 10y + 25)
(2b – 1) (4b2 + 2b + 1)
(3x + y)(9x2 – 3xy + y2)
(m – n)(m2 + mn +n2)
(3b + 4c)(9b2–12bc+16c2)
(2ab–5) (4a2b2+10ab+25)
(- 3 + 4y)(9 + 12y + 16y2)
(6xy – 5z)(36x2y2 – 25z2)
10
2
 4

10.  x − 5  x 2 + x + 25 
3
3
 9

1.
2.
3.
4.
5.
6.
7.
(x + y – 6)2
(3d + 2e – 5)2
(4x – 9y + 2)2
(3m2n + 2pq2 – 5)2
(2m2 – 3m – 2)2
(3b2d + 10e + f)2
(2k3 + 7n2 – 5)2

3
8.  x 2 − y + 1 
4
2

9. [(2x – 3) + 4]2
10. (2x – 5 – x + 1)2
4
Prepared by Mrs. Koni Gutierrez Cruz
Assistant Professor I
| Bataan Peninsula State University