This document provides information about polynomial operations including: 1) Defining polynomials as algebraic expressions involving integer powers of a variable and real number coefficients. 2) Examples of adding, subtracting, and multiplying polynomials by using vertical or FOIL methods. 3) Important formulas for polynomial operations such as (a + b)(a - b) = a2 - b2 and (a + b)2 = a2 + 2ab + b2. 4) Worked examples of applying these formulas and methods to polynomials involving single and multiple variables.

1. Polynomial Operations
Chapter 6
p.333
1

2. 2

3. 3

4. What is polynomial?
A polynomial of a letter is an algebraic expression that is the sum
of the products between real numbers and the non-negative integer
powers of the letter.
Examples, Suppose that the letter is x, then
3 x + 2, 2 x 2 − 3 x + 5, x 3 + 2 x 2 − 4 x, x100 are all polynomials of x.
Note:
1. The Polynomial letter could be any letter.
3 y + 2, 2t 2 − 3t + 5, r 3 + 2r 2 − 4r , p100 are all polynomials .
2. Any number is also considered as a polynomial. This is because
5 = 5x 0 , which is 5 times the zero power of x.
3. Each product is called the term of the polynomial
2 y 2 − 3 y + 5 has three terms: 2 y 2 , − 3 y, 5.
4

5. 4. Each number of the product is called the coeficient of the polynomial
3 x + 2 has 2 coeficients 3 and 2
2 y 2 − 3 y + 5 has 3 coeficients 2, − 3 and 5.
x 3 + 2 x 2 − 4 x − 3, has 4 coeficients 1, 2, − 4 and − 3.
x100 has one coeficient 1.
5. The heighest power exponent of x is called the degree of the polynomial.
3 x + 2 has degree =1,
2 y 2 − 3 y + 5 has degree =2
x3 + 2 x 2 − 4 x − 3
x100 has degree =100
5 has degree = 0.
6. A singleton term polynomial is called monomial, such as
x100 , 2x, 5
5

6. 7. A two terms polynomial is called binomial, such as
2 x + 5, 9 p 5 + 7, x 2 − 4.
8. A three terms polynomial is called trinomial, such as
2 x 2 − 3 x + 5, 9 p 5 + 7 p 2 + p, x 2 + 4 x + 4.
9.
2 x 2 − 3x + 5 x ,
1
+ 4 x + 4,
2
x
x +1
x−2
are not polynomials. Why?
10. Polynomial could have more than one letters, if all letters have
non-negative powers, such as
2 x 2 y − 3 xy 3 + 5 x 6 y 2 , 8m 2 p 5 + 9m3 p 2
They are called multi-varialbles polynomials. In the above examples,
we also can consider that only one letter as variable and other letters
as numbers.
6

7. Polynomial Operations
1. Add and subtract: We only do on the same power terms.
ax n ± bx n = ( a ± b ) x n
3x 3 + 5 x 3 = ( 3 + 5 ) x 3 = 8 x 3
3m 5 − 7 m 5 = ( 3 − 7 ) m 5 = − 4m 5
when we have multiple terms, we just add or subtract terms with
the same powers
Examples:
(2 x 4 − 3 x 2 + 5 x) + ( x 4 + 2 x 2 + 4 x)
= (2 + 1) x 4 + (−3 + 2) x 2 + (5 + 4) x = 3x 4 − x 2 + 9 x
(3 x 2 + 7 x − 6) − (5 x 2 − 4 x + 8)
= (3 − 5) x 2 + (7 − (−4)) x + (−6 − 8) = −2 x 2 + 11x − 14
Can we do addition 2 x 4 + 3 x 2 ?
7

8. Do operation vertically
(3 x 2 + 7 x − 6) − (5 x 2 − 4 x + 8) = − 2 x 2 + 11x − 14
3x 2 + 7 x − 6
5 x 2 − 4 x + 8 (−
− 2 x 2 + 11x − 14
or
(3 x 3 + 7 x 2 − 6) + (5 x 3 − 4 x + 8) = 8 x 3 + 7 x 2 − 4 x + 2
3 x 3 + 7 x 2 + 0 ×x − 6
5 x3 + 0 ×x 2 − 4 x + 8 (+
8 x3 + 7 x 2 − 4 x + 2
Align the same power terms vetically. Put 0s for the missing power
terms. Then do numbers operations veritcally.
8

9. 2. Multiplication: Use formula
( ax ) ×( bx )
m
n
= abx m + n
( 3x ) ( 5 x ) = ( 3 ×5 ) x = 15x
( 3m ) ( −7m ) = ( 3 ×( −7 ) ) m
2
3+ 2
3
5
2
5
5+ 2
= − 21m7
When two polynomials have multiple terms, then every term of the
1st polynomial must multiply to every term of the 2nd polynoimial.
Examples:
(3 x − 4)(2 x 2 − 3 x + 5)
= (3 x)(2 x 2 − 3 x + 5) + (−4)(2 x 2 − 3 x + 5)
= (3 x)(2 x 2 ) + (3 x)(−3 x) + (3 x)(5) + ( −4)(2 x 2 ) + (−4)(−3 x) + (−4)(5)
= 6 x 3 − 9 x 2 + 15 x − 8 x 2 + 12 x − 20
= 6 x 3 − 17 x 2 + 27 x − 20
9

10. Do multiplication vertically, put 0 for missing power terms
(3 x − 4)(2 x 2 − 3x + 5) = 6 x 3 − 17 x 2 + 27 x − 20
2x2
6 x3
6 x3
−8 x 2
−9 x 2
−17 x 2
−3 x
3x
+12 x
+15 x
+27 x
+5
−4 (×
−20
(+
−20
(2 x + 3)( x 2 − 5) = 2 x 3 + 3x 2 − 10 x − 15
x2
2 x3
2 x3
3x 2
0 x2
+3 x 2
0 ×x
2x
0x
−10 x
−10 x
−5
+3 (×
−15
(+
−15
10

11. Multiplication vertically example2
( x 2 + 2 x − 3)(2 x 2 − 3x + 5) = 2 x 4 + x 3 − 7 x 2 + 19 x − 15
x2
+ 2x
−3
2x2
−3 x
+ 5 (×
5x2
+10 x −15
−3 x 3
−6 x 2
+9 x
2 x4
+4 x 3
−6 x 2
2 x4
+ x3
− 7 x2
+ 19 x
Multiply by
5
Multiply by
−3x
Multiply by 2x2
− 15
11

12. Vertical multiplication with numbers only
(3 x − 4)(2 x 2 − 3 x + 5) = 6 x3 − 17 x 2 + 27 x − 20
(2 x + 3)( x 2 − 5) = 2 x 3 + 3 x 2 − 10 x − 15
12

13. ( x 2 + 2 x − 3)(2 x 2 − 3 x + 5) = 2 x 4 + x 3 − 7 x 2 + 19 x − 15
13

14. Two binomial multiplication. Use FOIL rule
First terms, Outside terms, Inside terms, Last terms
(2 x − 3)(6 x + 5) = ( 2 x ) ( 6 x ) + ( 2 x ) ( 5 ) + ( −3) ( 6 x ) + ( −3) ( 5 )
1 24 124 1 24 124
4 3 4 3
4 3 4 3
F
O
I
L
= 12 x 2 + 10 x − 18 x − 15
= 12 x 2 − 8 x − 15
or simply vertical way
2x
6x
12 x 2
12 x 2
−3
+5
(×
10 x −15
−18 x
− 8x
− 15
14

15. Exercises
1. (3 x 2 − 4 x + 5) + ( −2 x 2 + 3 x − 2)
2. (4m3 − 3m 2 + 5) + ( −3m3 − m 2 + 5)
3. 2(12 x − 8 x + 6) − 4(3 x − 4 x + 2)
2
2
4. − (8 x + x − 3) + (2 x + x) − (4 x − 1)
2
2
2
5. 2 x 3 (3 x 2 − 5 x + 2)
6. ( x 2 + 5)(3 x − 2)
7. (4 x + 5)(3 x − 2)
8. (3 x 2 − 4 x + 5)(3 x + 1)
9. (3 x 2 − 4 x + 5)(2 x 2 + x − 2)
15

16. Some important formulas
1. ( x + y )( x − y ) = x 2 − y 2
(sum and difference product)
Use FOIL ( x + y )( x − y ) = x 2 − xy + xy − y 2
= x2 − y 2
or
(a + b)(a − b) = a 2 − b 2
2. ( x + y ) 2 = x 2 + 2 xy + y 2 (square of sum )
Use FOIL ( x + y ) 2 = ( x + y )( x + y ) = x 2 + xy + xy + y 2
= x 2 + 2 xy + y 2
or
(a + b) 2 = a 2 + 2ab + b 2
3. ( x − y ) 2 = x 2 − 2 xy + y 2
(square of difference )
Use FOIL ( x − y ) 2 = ( x − y )( x − y ) = x 2 − xy − xy + y 2
= x 2 − 2 xy + y
or
(a − b) 2 = a 2 − 2ab + b 2
16

17. 4. ( x + y )( x 2 − xy + y 2 ) =x3 + y 3
(sum of cubic powers)
This is because
( x + y )( x 2 − xy + y 2 ) = x( x 2 − xy + y 2 ) + y ( x 2 − xy + y 2 )
(
) (
)
= x 3 − x 2 y + xy 2 + yx 2 − xy 2 + y 3 = x3 + y 3
Eexamples
( a + 1)( a 2 − a + 1) = a 3 + 1
(a + 2)(a 2 − 2a + 4) = ( a + 2)( a 2 − 2 ×a + 2 2 ) = a 3 + 23 = a 3 + 8
(a + 3)(a 2 − 3a + 9) = (a + 2)(a 2 − 3 ×a + 32 ) = a 3 + 33 = a 3 + 27
5. ( x − y )( x 2 + xy + y 2 ) =x 3 − y 3 (difference of cubic powers)
This is because, we can use − y to replace y in the above formula
( x + ( − y ) ) ( x 2 − x ( − y ) + ( − y ) 2 ) =x 3 + ( − y ) 3
which is
Examples
( x − y )( x 2 + xy + y 2 ) =x 3 − y 3
(a − 1)(a 2 + a + 1) = a 3 − 1
(a − 2)(a 2 + 2a + 4) = (a − 2)(a 2 + 2 ×a + 2 2 ) = a 3 − 23 = a 3 − 8
(a − 3)(a 2 + 3a + 9) = (a − 2)(a 2 + 3 ×a + 32 ) = a 3 − 33 = a 3 − 27
17

18. Example
1. (3 p + 11)(3 p − 11) = ( 3 p ) − 112 = 9 p 2 − 121
2
2. (5m − 3)(5m + 3) = ( 5m
3
3
)
3 2
− 32 = 25m6 − 9
3. (9k − 11r )(9k + 11r ) = ( 9k ) − ( 11r
3
2
3
)
3 2
= 81k 2 − 121r 6
4. (2m + 5) 2 = ( 2m ) + 2( 2m)(5) + ( 5 ) = 4 m 2 + 20 m + 25
2
( x + y )2
5. ( 3x − 7 y
2
x2
) = ( 3x )
4 2
( a −b ) 2
a2
y2
2 xy
2
− 2(3 x)(7 y )+ ( 7 y
4
2 ab
)
4 2
b2
= 9 x 2 − 42 xy 4 + 49 y 8
6. ( 3 x − 2 y ) ( 9 x + 6 xy + 4 y ) = ( 3 x ) − ( 2 y ) = 27 x 3 − 8 y 6
1 24 144 2444 {
4 3
4
3
1 3
2
2
(a − b)
2
2
a 2 + ab + b 2
3
4
a3
2 3
b3
18

19. Exercises
1. (3 x + 5)(3 x − 5)
2
2
2. (2m3 + n)(2m3 − n)
3. ( 5r + 4t
)
2 2
4. ( 2 x − 3 y )
4
2
5. (3 p + 5) 2
6. (4 − x)(16 + 4 x + x 2 )
7. (2a + 3b)(4a 2 − 6ab + 9b 2 )
19

20. Higher Power of binomial
We have (a + b) 2 = a 2 + 2ab + b 2 what is (a + b)3 ?
After calculating
( a + b) 3 = (a + b)( a + b) 2 = (a + b)( a 2 + 2 ab + b 2 )
= a (a 2 + 2ab + b 2 ) + b( a 2 + 2ab + b 2 )
= (a 3 + 2a 2b + ab 2 ) + (a 2b + 2a 2b + b3 )
= a 3 + 3a 2b + 3ab 2 + b3
We can see the powers of a is decreasing and the powers of b is
increasing. The coefficients are 1, 3, 3, 1. Similarly,
(a + b) 4 = (a + b)(a + b)3 = (a + b)(a 3 + 3a 2b + 3a 2b + b 3)
= a (a 3 + 3a 2b + 3a 2b + b3) + b(a 3 + 3a 2b + 3a 2b + b 3)
= a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4
The coefficients are 1, 4, 6, 4, 1.
20

21. Pascal Triangle
We see that
( a + b ) has coefficients
1, 1
(a + b) 2 = a 2 + 2ab + b 2 has coefficients 1, 2, 1
( a + b)3 = a 3 + 3a 2b + 3a 2b + b3 has coefficients 1, 3, 3,1
(a + b) 4 = a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4 has coefficients 1, 4, 6, 4, 1
So we can arrage them into a triangle like
1
1
1
1
1
2
3
4
1
3
6
1
4
1
Every number is the sum of two numbers on its shoulder.
21

22. 1
1
1
1
1
1
2
3
4
5
1
1
3
6
10
1
4
10
1
5
1
It we continue to calculate the numbers on the next line, we
will get numbers 1, 5, 10, 10, 5, 1, which are coefficients
of power (a + b)5 . Therefore we get
(a + b)5 = a 5 + 5a 4b + 10a 3b 2 + 10a 2b3 + 6ab 4 +b 5
This triangle is called the Pascal Triangle.
22

23. Powers of (a − b)
n
Similarly, we have
(a − b) 2 = a 2 − 2ab + b 2
(a − b)3 = a 3 − 3a 2b + 3a 2b − b3
(a − b) 4 = a 4 − 4a 3b + 6a 2b 2 − 4ab3 + b 4
(a − b)5 = a 5 − 5a 4b + 10a 3b 2 − 10a 2b3 + 6ab 4 − b5
The coeficient is negative if the power exponent of b is odd number
Practice Exercises
1. ( x + 3)3
2. ( x − 2) 4
3. ( x − 1)5
4. ( x + 1)6
23

24. 3. Division
Case 1: If the devisor is monomial
4 x 3 − 8 x 2 + 6 x 4 x3 8x 2 6x
=
−
+
= 2x2 − 4x + 3
2x
2x
2x
2x
Sometimes, we may have remainder
4 x 3 − 8 x 2 + 6 x + 3 4 x3 8x 2 6x
3
3
=
−
+
+
= 2x2 − 4x + 3 +
2x
2x
2x
2x
2x 2 x
Case 2: If the devisor is binomial
Use factoring
ab + ac = a (b + c) or
ab − ac = a (b − c )
4x2 + 2x
2 x ×2 x + 2 x 2 x (2 x + 1)
=
=
= 2x
2x +1
2x +1
2x +1
2
4x2 − 6x − 3 4x2 + 2x − 8x − 4 + 1 ( 4 x + 2x ) − ( 8x + 4) + 1
=
=
2x +1
2x +1
2x +1
4x2 + 2x 8x + 4
1
1
=
−
+
= 2x − 4 +
2x +1
2x +1 2x +1
2x +1
24

25. Other methods to get result
4x2 − 6x − 3
1
= 2x − 4 +
2x +1
2x +1
vertical division
25

26. 13
2
4m − 8m + 4m + 6
1
2
= 2m − 3m + +
2m − 1
2 2m − 1
vertical way with number only
3
2
26

27. 3x 3 − 2 x 2 − 150 3 x 3 − 2 x 2 + 0 ×x − 150
12 x − 158
=
= 3x − 2 +
2
2
x −4
x + 0 ×x − 4
x2 − 4
Put 0s for the missing terms
Remainder
27

28. Exercises
Do divisions
−4 x 7 − 14 x 6 + 10 x 4 − 14 x 2
1.
−2 x 2
10 x8 − 16 x 6 − 4 x 4
2.
−2 x 6
Use vertical devision with number only
12 x 3 − 2 x + 5
3.
x −3
6 x 4 + 9 x3 + 2 x 2 − 8 x + 7
4.
3x 2 − 2
5x4 + 2x2 − 3
5.
x2 − x + 1
28

29. Factoring
Factoring is the reverse of polynomial multiplication and based on
ab + ac = a (b + c ) here a could number or formula
Factor the Greatest Common Factor GCF, including the largest
posssible common number factor and lowest power of x or anything
Examples:
9 x 2 + 6 x −12 = (3 × x 2 + 3 × x − 3 × ) = 3(3 x 2 + 2 x − 4)
3
2
4
9 x 5 + 6 x 3 − 12 x 2 = (3 x 2 × x 3 + 3 x 2 × x − 3 x 2 × ) = 3 x 2 (3 x 3 + 2 x − 4)
3
2
4
6 x 2t + 8 xt + 12t = (2t × x 2 + 2t × x + 2t × ) = 2t (3 x 2 + 4 x + 6)
3
4
6
14(m + 1)3 − 28(m + 1) 2 − 7( m + 1)
= ( 7(m + 1) × m + 1) 2 − 7(m + 1) × m + 1) − 7( m + 1) × )
2(
4(
1
= 7(m + 1) ( 2(m + 1) 2 − 4(m + 1) − 1)
29

30. Group Factoring
If there are four terms, we can group the 1st two and the last two terms.
Then do the preliminary factors on two grous and factor again.
x 3 + 2 x 2 + 2 x + 4 = ( x 3 + 2 x 2 ) + ( 2 x + 4 ) = x 2 ( x + 2 ) + 2( x + 2)
= ( x 2 ( x + 2 ) + 2( x + 2) ) = ( x + 2 ) ( x 2 + 2 )
4 x 3 + 2 x 2 − 6 x − 3 = ( 4 x 3 + 2 x 2 ) − ( 6 x + 3 ) = 2 x 2 ( 2 x + 1) − 3(2 x + 1)
= ( 2 x 2 ( 2 x + 1) − 3(2 x + 1) ) = ( 2 x + 1) ( 2 x 2 − 3)
mp 2 + 7m 2 + 3 p 2 + 21m = ( mp 2 + 7m 2 ) + ( 3 p 2 + 21m )
= m ( p 2 + 7m ) + 3 ( p 2 + 7m )
(
)
= m ( p 2 + 7m ) + 3 ( p 2 + 7 m ) ¬ − − We can skip this
= ( p 2 + 7 m ) ( m + 3)
30

31. Factor the following
Exercises
1. 12m + 60
2. 4 p 3 q 4 − 6 p 2 q 5
3. 4k 2 m3 + 8k 4 m3 − 12k 2 m 4
4. 4( y − 2) 2 + 3( y − 2)
5. 6 st + 9t − 10 s − 15
6. 20 z 2 − 8 x + 5 pz 2 − 2 px
31

32. Quadratic polynomials
Factoring is the reverse of polynomial multiplication.
(3x − 4)(2 x + 5) = 6 x 2 + 7 x − 20
is multiplication
6 x 2 + 7 x − 20 = (3x − 4)(2 x + 5) is factoring
How to obtain numbers 3, − 4, 2 and 5 from 6, 7 and − 20?
Because 6 x 2 = 3 x ×2 x so we have 6 = 3 ×2 and − 20= ( −4 ) ×
5
Also 7 x = 3 x × + ( −4 ) ×2 x so we have 7 = 3 × + ( −4 ) ×2
5
5
Therefore we have chcart (answer are 4 corner numbers)
2x+5
3x−4
32

33. Example 2. Factor 6 x 2 − 13 x + 6
This is not match
This is match
2x−3
3x−2
Answer: 6 x 2 − 11x + 6 = (2 x − 3)(3 x − 2)
33

34. Example 3. Factor 4 x 2 − 11xy + 6 y 2
4x−3y
x−2y
Answer: 4 x 2 − 11xy + 6 y 2 = (4 x − 3 y )( x − 2 y )
Example 4. Factor 6 p 2 − 7 p − 5
Answer: 6 p 2 − 7 p − 5 = (2 p + 1)(3 p − 5)
34

35. Example 5. Factor x − 11x + 30
2
Answer: x 2 − 11x + 30 = ( x − 5)( x − 6)
Example 6. Factor a 2 − 5ab − 14b 2
Answer: a 2 − 5ab − 14b 2 = (a + 2b)(a − 7b)
35

36. Note: If the first coefficient is one like x 2 + px + q
then we only need to decompose q to the product of two number
such that their sum is p.
Examples:
1.
x 2 − 11x + 30
because
30 = ( −5)(−6) and
so
2.
3.
x 2 − 11x + 30 = ( x − 5)( x − 6)
a 2 − 5a − 14
because
− 14 = 2 ×(−7) and
so
(−5) + (−6) = −11
2 + (−7) = −5
a 2 − 5a − 14 = (a + 2)(a − 7)
x 2 + 10 x − 39
because
− 39 = 13 ×( −3) and 13 + ( −3) = 10
so
x 2 + 10 x − 39 = ( x + 13)( x − 3)
36

37. Factor the following
Exercises
1. 8h 2 − 2h − 21
2. 3m 2 + 14m + 8
3. 9 y 2 − 18 y + 8
4. 6k 2 − 5kp − 6 p 2
5. 5a 2 − 7 ab − 6b 2
6. 24a 4 + 10a 3b − 2a 2b 2
7. 18 x 5 + 15 x 4 z − 75 x 3 z 2
8.
x 2 + 12 x + 27
9.
x 2 + x − 12
10.
x 2 + 11x − 12
11.
x 2 + 10 x − 24
12.
x 2 − 5 x − 24
13.
x2 − 2 x + 5
37

38. Prime Polynomial
If a integer coefficients polynomial cannot be factored to a product
of polynomials with integer, then it called prime polynomials.
1. Suppose that m and n are positive integers, then mx 2 + n is prime
such as
x 2 + 9, 2 y 2 + 5,
x2 + y2
are all prime.
2. For quadratice polynoimal ax 2 + bx + c
if number
b 2 − 4ac < 0
Example: x 2 + x + 1,
then ax 2 + bx + c is prime
x 2 + 2 xy + 3 y 2 are all prime.
if number b 2 − 4ac is not a square number then ax 2 + bx + c is prime
Example: x 2 + 3 x + 1, b 2 − 4ac = 32 − 4 × × = 5 not a square number
11
so x 2 + 3 x + 1 is prime
38

39. Use Formulas
We have the following formulas
x 2 − y 2 = ( x + y )( x − y )
difference of squares
x 2 + 2 xy + y 2 = ( x + y ) 2
perfect saqure of sum
x 2 − 2 xy + y 2 = ( x − y ) 2
perfect saqure of difference
x 3 + y 3 = ( x + y )( x 2 − xy + y 2 ) sum of cubic powers
x 3 − y 3 = ( x − y )( x 2 + xy + y 2 ) difference of cubic powers
Examples
1. 4m 2 − 9 = (2m) 2 − 32 = (2m + 3)(2m − 3)
1 3 1 3
2
2
2
2
x
y
x+ y
x− y
2. 4 x 2 − 9 y 2 = (2 x) 2 − (3 y ) 2 = (2 x + 3 y )(2 x − 3 y )
39

40. 3. 256k − 81m = ( 16k
4
4
= ( 16k 2 + 9m 2
) − ( 9m ) = ( 16k
) ( (4k ) − (3m) )
2 2
2 2
2
2
+ 9m 2 ) ( 16k 2 − 9m 2 )
2
= ( 16k 2 + 9m 2 ) ( 4k + 3m ) ( 4k − 3m )
4. (a + 2b) 2 − 4c 2 = ( a + 2b) 2 − (2c) 2 = [ ( a + 2b) + 2c ] [ ( a + 2b) − 2c ]
= (a + 2b + 2c)(a + 2b − 2c)
5. x 2 + 2 x + 1 = ( x + 1) 2
6. x 2 + 2 xy + y 2 = ( x + y ) 2
7. x 2 − 6 x + 9 = ( x − 3) 2
8. 25 y 2 − 10 y + 1 = ( 5 y ) − 2 × y × + 12 = (5 y − 1) 2
5 1
2
9. 4m + 28m + 49 = ( 2m ) + 2 ×2m ×7 + (7) 2 = (2m + 7) 2
2
2
40

41. 10. x − 6 x + 9 − y = ( x − 3) − ( y
) = ( x −3+ y ) ( x −3− y )
11. ( m + 14m + 49 ) − ( y − 10 y + 25 ) = (m + 7) − ( y − 5)
2
4
2
2
2 2
2
2
2
2
2
= [ (m + 7) + ( y − 5) ] [ (m + 7) − ( y − 5) ]
= (m + y − 2)(m − y + 12)
12. x 3 + 27 = x 3 + 33 = ( x + 3)( x 2 − x × + 32 ) = ( x + 3)( x 2 − 3 x + 9)
3
13. m − 64n = m − ( 4n ) = ( m − 4n)  m 2 + m ×4n + ( 4n) 2 


3
3
3
3
= (m − 4n) ( m 2 + 4mn + 16n 2 )
14. 8q + 125 p = ( 2q
6
9
= (2q
2
= (2q 2
) +(5p )
+ 5 p ) ( 2q ) − 2q × p + (5 p )
5


+ 5 p ) ( 4q − 10q p + 25 p )
2 3
3 3
2 2
3
3
4
2
2
3
3
3 2



6
41

42. Exercises
Factor the following by formulas
1. 9m 2 − 12m + 4
2. 16 p 2 + 40 p + 25
3. 36 x 2 − 60 xy + 25 y 2
4. 9 x 2 − 6 x3 + x 4
5. 4 x 2 y 2 + 28 xy + 49
6. ( a − 2b) 2 − 6( a − 2b) + 9
7. 9m 2 n 2 − 4 p 2
8. ( a + 2b) 2 − 25( a − 3b) 2
9.
8 x3 + 27
10.
x 4 − 81
11.
27 − ( m + 2n)3
12. x 4 − 16
13.
x4 − 5x2 + 4
42