1078 Basic Algebra questions: factoring, expanding,
and simplifying
The last five people to edit this page are
Edward Morey October 6, 2006
(note that the questions are numbered in groups)
(note that this page is a work in progress
Please email your instructor with suggestions and corrections - thanks
_________________________________________________________
1. Simplify and find the answer of the following expressions in fractional form.
(A)
3 1 1
  
10 3 5
4
1 1
(B)  2    0 
5
3 6
(C)
2
1
5
3
2
1
1
3
4
2
Factor the following expressions:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
4 x 2 y  6 xy 2  0
3x3 y  6 x 2 y 2  3xy3
2 x2  5x  3
2 x 2  3xy  2 y 2
6 x 2  25 xy  4 y 2
m3  n3 and m3  n3
z3 1
2m2  mn  n 2
x 2  16 y 2
3m4  24mn3
11. 18 x 3  8 x
Simplify the following expressions:
x 2  3x
1.

2 x2  5x  3
x3  1
2.
=
x2 1
2 x2  7 x  3
3.
 ( x  3) =
4 x2  1
 d5 d2  a
4.   2   3 =
 3a 6a  4d
5.
6.
7.
8.
9.
10.
m3  n 3
m3n  m2 n2  mn3

=
2m2  mn  n 2
2m 3 n 2  m 2 n 3
x 1
1 =
x 1
3
2

=
a 1 1 a
2
5
=
 2
2
( x  1)
x x2
y2
1 2
x =
y
1
x
1
1
1
1
1
x
Try to find the LCD (least common denominator) for these fractions:
2 3 11 5
  
5 7 25 28
7
6
2


2.
ab b(a  b) (a  b)(a  b)
1.
Open up the parenthesis and gather the like terms:
1. 4(x-2y) – 3(2x-7y)
2. (4x-1)(3x+1)
3. (2 x  3)(2 x 2  3x  2)
4. 4(2 x4  4 x3  1)  ( x  1)(4 x4  9 x)  4[ x6  (2 x3  4)]
Basic operation of the real number: Simplify the following expressions:
3 1 1
  
7 3 5
4
1 1
2  0 
2.
5
3 6
2
1
5
2 
3. 3
1
1
3
4
7
4. a + 2b – (c + 2a – 3b) =
1.
1. Combine terms:
1
8
a.

x y yx
b.
c.
1
2

2x 3y
3x x

2 6
2. Simplify
ax  ay
a.
a
b. [(a  b) x  (a  b) z ]  (a  b)
Expand the expressions
1. 1.  x  2 y   x 2  4 xy  4 y 2
Try to use the perfect square formula:
first term square – double the product of two terms + second term square
Or you can FOIL  x  2 y   x  2 y 
2


2. m  m   m   m  1 
First open up   then    then    . Pay attention that if there is a negative
sign in front, we need to switch the sign inside.


m  m   m   m  1   m  m   m  m  1  m  m  1  m  m  1  1


3.
2 x  3 x  2  x   x  5   1  2 x  3x  2  x  x  5  1  2 x  3x  2  5  1
4.
 2 x  3  2 x 2  3x  2    2 x   2 x 2    2 x 3x    2 x  2   3  2 x 2   33x   6
 2 x  3x  9  2 x  3x  27   x  27
 4 x 3  6 x 2  4 x  6 x 2  9 x  6  4 x 3  13x  6
2
 2 x  12  x  3x  1   4 x 2  4 x  1  3x 2  x    x 2  5 x  1
 


 
2
2
5.   x 2  5 x  1 x 2  5x  1
 x 4  10 x 3  27 x 2  10 x  1
_
1.
d5  d2 a

 3
3a  6a 2 4d
 d5  1  d5
6


   24ad  8d
3
a
24
ad
3
a



2.
2  x  3
2  x  3  x 2 x  6  x
2
1
x
x6






x x  3 x  x  3 x  x  3
x  x  3
x  x  3 x  x  3
3x  8 2 x  1 5
 3 
4 x2
x
8x
 3x  8 2 x   2 x  1 8  5 x 2

8 x3
8 x3
8 x3
3.
6 x 2  16 x 16 x  8 5 x 2


 3
8 x3
8 x3
8x
2
6 x  16 x  16 x  8  5 x 2 x 2  8


8 x3
8 x3
4.
 x  3 2 x  1  1  1
2 x2  7 x  3
  x  3 
2
4x 1
 2 x  1 2 x  1  x  3 2 x  1
x
y x
yx

y  x 2x  y y  x
x
yx  x 2
y
y y
y







y 2x y 2x  y
y
x
y 2x  y y  2x  y 
2

x
x x
x
1
5.
x  x
1   xy
xy  x 2
y  y
or


y 
y
2 xy  y 2
2
2

xy

x 
x
x2
x 1 1


2
x  2 x  1 3x  3 6
6.
x2
x 1
1
2
2



: find LCD which is  x  1  3  2  6  x  1
2
 x  1 3  x  1 6
1

6 x2
6  x  1
2

2  x  1 x  1
6  x  1
2
 x  1

2
6  x  1
2

6 x2  2 x2  2  x2  2 x 1
6  x  1
2

7 x2  2x  3
6  x  1
2
7. In the following, show what is wrong and how it can be corrected?
 x  h
3
 x3
  x  1  x3  3x 2  3x  1
3
h
For first step, we cannot simply cancel out the h in both the numerator and denominator.
We can take this step only if h is a common factor for both numerator and denominator,
hx x
 .
like
hy y
For this question, firstly I will apply the formula: difference of cubes to numerator.
 x  h

3
 x3
h

 x  h   x   x  h    x  h  x  x
2

h  x  h    x  h  x  x2
2
h

2

h
 x  h
2

  x  h  x  x 2  3x 2  3xh  h 2
8.
2  x  1
 x  3
2
x3
 2


x  1 x  1  x  1 x  1  x  1 x  1

2  x  1   x  3 2 x  2  x  3
x 1


 x  1 x  1  x  1 x  1  x  1 x  1
1. Evaluate the following numerical expressions
2
a. 2  4  7  4  8  3


_____10_______
2  4  7  4  8  3  2  4  9  8  3  2  36  8  3  30  3  10
2
b. (8)
2
3
( 8)  ( 8)
1
2
3


2
3
3
____ 4_______
8

2
  2   4
2
7 1

c. 5 2
3
10
7 1 14 5
9


5 2  10 10  10  9  3  9  10  3
3
3
3 10 10 10 3
10
10
10
fully expand the polynomial:
1. ( x  3)( x  7)
______3______
( x  3)( x  7)  x 2  10 x  21
2. [a  (2b  c)][ a  (2b  c)]
Apply “difference of squares”:  x  y  x  y   x 2  y 2
[a  (2b  c)][a  (2b  c)]  a 2   2b  c   a 2  4b 2  4bc  c 2
2
factor the polynomial completely:
1. x 2  2 xy  y 2
Perfect square: x 2  2 xy  y 2   x  y 
2
2. 2 x 2 y  8xy2  8 y 3
2 x 2 y  8xy 2  8 y 3  2 y  x 2  4 xy  4 y 2   2 y  x  2 y 
3. u 4  v 4
u 4  v 4   u 2  v 2  u 2  v 2    u 2  v 2   u  v  u  v 
Factoring Polynomials:
1. 8u 3v  6u 2v 2  4uv3 , all three items include 2uv
 2uv(4u 2  3uv  2v 2 )
2. 7m(2m-3)+5(2m-3), common factor is (2m-3)
=(2m-3)(7m-5)
3. a(3c+d)-4d(3c+d)
=(3c+d)(a-4d)
4. 2 x 2  x  4 x  2
 x(2 x  1)  2(2 x  1)
 (2 x  1)( x  2)
5.
3y2  3y  2 y  2
2
 3 y ( y  1)  2( y  1)
 ( y  1)(3 y  2)
6. wy-wz+xy-xz
=w(y-z)+x(y-z)
=(y-z)(w+x)
7. 2 x 4  24 x3  40 x 2
 2 x 2 ( x 2  12 x  20)
 2 x 2 ( x  10)( x  2)
8. 4 xy 2  12 xy  9 x
 x(4 y 2  12 y  9)
 x(2 y  3)2
9. 16 x 2 y  8 xy  y
 y (16 x 2  8 x  1)
 y (4 x  1)2
10. 6m 2  mn  12n 2
 (2m  3)(3m  4)
11. 6( x  y)2  23( x  y)  4 :think of the whole part (x-y) as a single variable, use ac
test to decompose this two-degree polynomial
 [6( x  y )  1][( x  y )  4]
 (6 x  6 y  1)( x  y  4)
12. 27a 2  a 5b3 : be sensitive that 27 is the cube of 3
 a 2 (27  a3b3 )
 a 2 (ab  3)(a 2b2  3ab  9)
Find the simplest form of a fraction:
a, there is no common factor of numerator and denominator;
b, combine all the like terms of numerator and denominator;
c, There is no fraction form appear in the numerator and denominator;
1.
d5  d2 a 


3a  6a 2 4d 3 

d5  1  d5

24ad  8d 6

3a  24ad  3a
4m  3 3 2m  1


18m3 4m 6m 2
2(4m  3)  9m 2 *3  6m(2m  1)

36m3
8m  6  27 m 2  12m 2  6m

36m3
15m 2  14m  6

36m3
1st denominator: 2.3.3.m.m.m
2nd one: 2.2.m.m
 2.3.3.m.m.m .2  36m3 (LCD)
rd
3 one: 2.3.m.m
(from 1st ) (from 2nd)
x 2  6 x  9 x 2  2 x  15
3.
first step to take is to decompose all the two-degree

x2  x  6
x2  2 x
polynomial using the way you good at (either ac test, or
( x  3)2
x( x  2)


the
( x  2)( x  3) ( x  3)( x  5) some formula like the perfect square), and change
division into multiplication by flipping the 2nd fraction.
x

x5
2.
x 1
1
x 1
x 1 x 1


x 1 x 1
x 1 x 1
2


x 1
x 1
4.
For those mixed operation with fraction and integer, we
should think of all the items as fraction. So firstly, find
the LCD.
2x
1
1
First decompose all the high-order polynomial (order is


2
x y
x y x y
2x
1
1
=
greater than one), then find the LCD using the way


( x  y )( x  y ) x  y x  y
we
use in class.
2 x  ( x  y)  ( x  y)

For those intermediate steps, don’t get rid of the
( x  y )( x  y )
parentheses so early.
2( x  y )
2


( x  y )( x  y ) x  y
5.
6.
2
x2
x 1 1


2
x  2 x  1 3x  3 6
x2
x 1
1



2
( x  1) 3( x  1) 6

x 2 *6  ( x  1) * 2( x  1)  ( x  1) 2
6( x  1) 2

6x  2x  2  x  2x 1
6( x  1) 2

7 x2  2 x  3
6( x  1) 2
2
2
2
First factoring the 2nd order polynomial then find LCD
y
1
2
 2
 2
y  y  2 y  5 y  14 y  8 y  7
y
1
2



( y  1)( y  2) ( y  2)( y  7) ( y  1)( y  7)
y *( y  7)  ( y  1)  2*( y  2)

( y  1)( y  2)( y  7)
7.
2
y2  7 y  y 1  2 y  4
y2  4 y  3

( y  1)( y  2)( y  7)
( y  1)( y  2)( y  7)
( y  1)( y  3)
y 3


( y  1)( y  2)( y  7) ( y  2)( y  7)

I expand the numerator as 3rd equality place as I want to combine the like terms, then try
to factor again and find some new factors(some cases cannot like the above question). I
do not expand out the denominator as I want to cancel some factor if there is similar one
appear upstairs.
 x3  y 3
y  x 2  xy  y 2
8. 


3
y
x

y
y2


( x  y)( x 2  xy  y 2 )  y
y2

 2
y3 ( x  y)
( x  xy  y 2 )
1
1  x4
 3

9. 

 x  2 x 1  x  2
3( x  1)  ( x  2) x  2


( x  2)( x  1) x  4
(2 x  5)( x  2)
(2 x  5)


( x  2)( x  1)( x  4) ( x  1)( x  4)
s2
s
10. s 2 t
t
t
s t
2
2
s2
s ( s  t ) s  s  st

s  t  (s  t )
 s 2 t
t
t (s  t )
t 2  ts  t 2

s t
s t
(s  t )
st ( s  t )


1
( s  t ) ts
Factor completely:
1. x 2  2 x  15
A ( x  3)( x  5)
2. 3 x 2  5 x  2
A (3x  1)( x  2)
3. 9 x 2  36 y 2
A (3x  6 y )(3x  6 y )
4. Factor the polynomial completely:
a) x 2  2 xy  y 2
b) 3x 2 y  12 xy 2  12 y 3
c) u 4  v 4
A a) ( x  y ) 2 b) 3 y( x  2 y) 2
c) u 4  v 4   u 2  v 2  u 2  v 2    u 2  v 2   u  v  u  v 
directly using the formula: difference of squares
1. (3x1/ 2  y1/ 2 )2  9 x  6 x1/ 2 y1/ 2  y
Using formula: perfect square
1.
2.
3.
4.
5.
16m4 n6  3 24 m4 n6  2mn2 2m
x
x  x 2/ 3
x  x 2/ 3
 1/ 3 2/ 3 
 x 2/ 3
3
x
x x x
3
4
3x3 4 3x3 4 3x3 * 22 4 12 x3



4
22
22 * 22
2
1 2

3
x 3 x 2  x1/ 2 x 2 / 3  x 2
1
( x  1)1/ 2  x( )( x  1) 1/ 2
2

x 1
6.
7
 x6  x 6 x
1
2( x  1)  x
)
2 x 1  2 x 1  x  2
x 1
x 1
2( x  1)3/ 2
2
3( x  2)  2 x
( x  2) 2 / 3  x(
)
1/ 3
x6
3( x  2)
3( x  2)1/ 3



4/3
4/3
( x  2)
( x  2)
3( x  2)5/ 3
x  1  x(
2
( x  2) 2 / 3  x( )( x  2) 1/ 3
3
( x  2) 4 / 3
1. Answer True or False by writing the whole word as ‘T’ and ‘F’ can look alike:
5 3 2 5

 
d.
_______T_____
8 6 1 2
5 3 2 5 3 2 5 6 2 20 5

       

8 6 1 8 6 1 8 3 1 8 2
z
1

z  x  y 1 x  y
e.
_______F_____
We can cancel out the common factor for both the numerator and denominator. Z is not a
factor of the denominator in the question. It is a part of it.
z
1

 z  x  y  1  x  y False
Compare the following two cases: 
 z  1
True
 zxy xy
 x 3 y   x 2 ( x)(  y)
f.
______T______
For right hand side:  x 2 ( x)( y )   1 x 2 (1)( x)(1)( y )   x 2 y
Think of –x as –1 times x, and –y as –1 times y.
Polynomial operations
1. 4a-2a[5-3(a+2)], follow the sequence that ( ) prefer to [ ] prefer to { }
= 4a-2a[5-3a-6]
=4a-2a[-3a-1]
=4a+6a2+2a
=6a2 + 6a
2. 2 y  3 y[4  2( y  1)]
 2 y  3 y[4  2 y  2]
 6 y 2  16 y
3.
-10x
-5
(3x  5)(2 x  1)  6 x 2  7 x  5
6x 2
3x
4. (4t  3)(t  2)  4t 2  11t  6
5. (3y+2)(3y-2) (Either POIL it OR Use the difference of squares formula)
= 6 y 2 -4
6. (2m-7)(2m+7)= 4m2  49  (4m)2  (3n)2  16m2  9n2
7. 9 x 2  4 y 2
8. a 3  b3 (difference of cubes)
9. a 3  b3 (sum of cubes)
10. ( x 2  2 xy  y 2 )( x 2  2 xy  y 2 )
 ( x  y)2 ( x  y)2
 [( x  y )( x  y )]2
 ( x 2  y 2 )2
 x4  2 x2 y 2  y 4
11.
(2 x  1)2  (3x  2)(3x  2)
 (2 x  1)2  (9 x 2  4)
 4 x2  4 x  1  9 x2  4
 5 x 2  4 x  5
12.
(m  2)2  (m  2)(m  2)
 m 2  4m  4  (m 2  4)
 4m  8
Problems 3-6 refer to the following polynomials
(A) 3x  4
(B) x  2
(C) 3 x 2  x  8
(D) x 3  8
1
a
(E)  2
a a 1
3. Add (A), (B), (C), and (D) and simplify
4. Subtract the sum of (A) and (C) from the sum of (B) and (D) and simplify
5. Multiply (C) and (D) and simplify
6. Simplify (E)
7. Find answers to the following expression
(A). 5 2 
(B). 33 
(C). 9 
(D). 3 64 
(E). 2 x 2 x 3  3( x 2 ) 3  x(2 x 4 )  9 x 6 
8. Factor the following polynomial
3a 3  6a 2
1. State the real number property that justifies below:
x y  yx
A: Commutative property
2. Answer True or False by writing the whole word as ‘T’ and ‘F’ can look alike:
7 3 2 7

 
g.
____________
8 6 1 2
z
1
h.
____________

z  x  y 1 x  y
a. T
b. F
* note that “b” is true when z = 1 so that it may be an inappropriate question.
3. Evaluate the following numerical expressions
2
a. 12  5  7  4  7  10


____________
2
b. (8) 3
6 2

7
3
c.
2
21
____________
____________
a. 5
b. 4
c. 2
Basics of polynomials
1. What is the degree of the following polynomial?
5x 4  3x 2 y 3  7 xy 2  2 y  5
A Five
Factoring and rational expressions:
1.
3 x 2 y  4 xy 2  4 y 3
y (3 x 2  4 xy  4 y 2 )
y[(3 x 2  6 xy)  (2 xy  4 y 2 )]
y[3 x( x  2 y )  2 y ( x  2 y )]
y (3 x  2 y )( x  2 y )
1. Evaluate the following numerical expressions
2
12  5  7  4  7  10

A5
2.

____________
3p
p3
 2
p  8 p  16 p  4 p
3p
p3

2
p ( p  4)
( p  4)
2
3p2
( p  3)( p  4)

2
p ( p  4)
x ( x  4) 2
3 p 2  ( p  3)( p  4)
p ( p  4) 2
3 p 2  ( p 2  p  12)
p ( p  4) 2
2 p 2  p  12
p ( p  4) 2
Order of operations: polynomials:
1. Indicate true or false for the following two expressions. If false provide a counterexample:
a. a  (b  c)  a  b  a  c
False
a
bc
Counter Example a=6, b=4, c=2
6(4+2) = 66=1
64 + 62 = 3/2 + 3 = 9/2
19/2
b. (a  b)  c  a  c  b  c
True by distribution of addition over multiplication
1
1
1
( a  b)  c  ( a  b)   a   b   a  c  b  c
c
c
c
a  (b  c) 
2. Simplify the following expression:
(3 x  2 y ) 2  (3 x  2 y )(3 x  2 y )
 9 x 2  12 xy  4 y 2  (9 x 2  4 y 2 )
 9 x 2  12 xy  4 y 2  9 x 2  4 y 2
 8 y 2  12 xy
Algebra and Real Numbers
1. State whether the following statements are True or False.(1-12)
______ 1. If xy=1, then it is always the case that either x=1 or y =1.
______ 2. If x(y+1) = 0, then x = 0 or y+1 =0 or both.
______ 3. a  b  b  a .
______ 4. ( a  b)  c  a  (b  c )
______ 5. ( a  b)  c  a  (b  c )
______ 6. (a  b)  c  a  (b  c )
______ 7 –(-a)+b = a+b.
x
1
______ 8

x  y 1 y
Straight lines and parabolas
1. Write the equation for a line with slope of -2/3 and a y-intercept 2 and graph it on
x=[-6,6].
In Slope-Intercept Form: y = -⅔ x + 2
y
6
4
2
x
-6
-4
-2
2
4
6
-2
2. Graph y=x2-2x-3 on x=[-2,4].
y= (x2-2x)-3=(x2-2x+1-1)-3=(x2-2x+1)-4=(x-1) 2-4
y
4
2
x
-2
-1
1
-2
-4
2
3
4
1. Perform the following operation and simplify:
(5 x  7)(3x  4)
A 15 x 2  41x  28
2. Perform the following operation and simplify:
(4 x  3)( 4 x  3)  (3x  1) 2
A 7 x 2  6 x  10
3. Perform the given calculation by fully expanding the polynomial:
a) ( x  2)( x  9)
b) [a  (3b  c)][ a  (3b  c)]
A a) x 2  11x  18
b) a 2  9b 2  6bc  c 2