Elementary Algebra Review
Includes order of operations, algebraic expressions, solving linear equations, story problems,
exponents, factoring, radicals and solving quadratic equations.
A. The Order of Operations
Many times you will encounter problems which combine several arithmetic operations.
Calculate the answer to the following problem:
6+24=?
Did you get an answer of 32 by first adding 6 + 2 to get 8, then multiplying 8  4? Or did you
get an answer of 14 by first multiplying 2  4 to get 8, then adding 8 + 6? As you can see, doing
the problems in a different order produces a different answer. However, both answers cannot be
correct.
Mathematical computations must be done in a standardized order to ensure that everyone gets the
same answer to the same problem. This accepted “order of operations” is as follows:
1) Do all operations which appear inside parentheses.
2) Do any powers and square roots.
3) Do all multiplications and divisions, working from left to right.
4) Do all additions and subtractions, working from left to right.
The sentence Please Excuse My Dear Aunt Sally will help you remember the correct order.
Here are some examples which demonstrate proper application of the order of operations.
Example 1:
3  5 - (5 + 6) = ?
3  5 - 11 =
15 - 11 =
4
Do the parentheses first.
Do the multiplication.
Do the subtraction.
When doing longer problems, it’s often difficult to keep track of what step you’re on. To
prevent errors, be sure to do only one step at a time and write down the results after each step.
Example 2:
36  (3 + 1)  6 = ?
36  4  6 
96=
Do the parentheses first.
Do multiplication and
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54
division left to right.
(In this problem, the division came
first.)
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Example 3:
(3 + 2)2  (17 - 3  5) = ?
(5)2  (17 - 15) =
(5)2  2 =
25  2 =
50
Example 4:
(20 + 25)  5 - 23 + 1 = ?
45  5 - 23 + 1
45  5 - 8 + 1 =
9-8+1=
2
Do the parentheses. (Notice
that inside the parentheses
the order of operations must
still be followed.)
Do the exponents.
Do the multiplication.
Do the parentheses.
Do the powers.
Do the division.
Do the addition and
subtraction from left to right.
(Here the subtraction came
first.)
Here are some practice exercises. Be sure to follow the correct order of operations.
1) 22 - 4  3 =
2) 8  2  3 + 16 =
3) 18 - 32 =
4) 33 - 24 =
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5) 2  2  3 + 1 =
6) 5 + 6  2  4 - 1 =
7) 42 + 2  4 =
8) (8 - 2)2  6 =
9) (5 - 3)2  6  4 =
10) 2  (5 - 3 + 6) + 5 =
11) 68 - 23  5 + 3 =
12) 14  2 - 3  2 + 33 =
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13) 4 + 27  3  2 - 8 =
14) 3  5 + 7  3 - 5  3 =
15) 5  33 - 20  5 =
16) 4  24 + 18  3 =
17) 3  (7 - 5)2 - 6  2 =
18) 5  6 - (5 - 3) + 8  2 =
19) 32  6  9 + 4  2 =
20) 2  (6 - 3)2 - 2 =
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21) -3²
22) (-3)²
23) 5-2(6-7)
3²−6²
24) (3−6)²
8
25) (2)(3)−(1)(6)
26)
2[10−(2)(5)]
3³
27) 5│2-4│
28)
4−│6−8│
│2│
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Answers:
1) 10
2) 28
3) 9
4) 11
5) 4
6) 7
7) 24
8) 6
9) 6
10) 21
11) 31
12) 28
13) 14
14) 21
15) 131
16) 70
17) 0
18) 44
19) 14
20) 16
21) -9
22) 9
23) 7
24) -3
25) Undefined
26) 0
27) 10
28) 1
B)Algebraic Expressions
1. Evaluate the following expressions when x = -3, y = 4, z = -2.
a. xyz
b. x2 + z2
2𝑥+𝑦
c.
𝑧
d. (x+y)(y + z)
e. y – z
f. z – y
g. 5y2 – 2xyz – z2
h. |𝑥 − 𝑧|
Answers
a) 24
b) 13 c) 1
d)2
e)6
f) -6
g)28
h) 1
2. Perform the indicated arithmetic on the following expressions.
a. (2x + 3y) + (5x – 7y)
b. (2x + 3y) – (5x + 7y)
c. 3(x + 5) – 2(7 – x)
d. (3a + 4b)(2a – 3b)
e. (2t + 3s)2
f. (m – n)(m + n)
g. (8x2 + 5x – 7) + (6x2 – 2x -11)
h. (8x2 + 5x -7) – (6x2 – 2x -11)
Answers
a) 7x -4y
b) -3x – 4y
c) 5x + 1
d) 6a2 –ab -12b2
e) 4t2 + 12st + 9s2
f) m2 – n2
g) 14x2 + 3x – 18
h) 2x2 + 7x + 4
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C)Solving linear equations
When solving linear equations, it is important to remember that what you do to one side of the
equation, you must do to the other side. The following examples will illustrate this idea.
Example 1:
Solve 2x + 3 = 11
Step 1: (subtract 3 from both sides)
2x + 3 = 11
-3 -3
2x = 8
Step 2: (divide both sides by 2)
2𝑥
8
= 2
2
x=4
Example 2:
Solve 5x – 4 = 3x + 12
Step 1: (add 4 to both sides)
5x – 4 = 3x + 12
+4
+4
5x = 3x + 16
Step 2: (subtract 3x from both sides)
5x = 3x + 16
-3x -3x
2x = 16
Step 3: (divide both sides by 2)
2x = 16
2
2
x=8
Example 3:
Solve 3(x – 5) – 4(2x + 3) = 13
Step 1: (remove parenthesis by distributing)
3x – 15 – 8x – 12 = 13
Step 2: (combine like terms on the left)
-5x – 27 = 13
Step 3: (add 27 to both sides)
-5x -27 = 13
+27 +27
-5x = 40
Step 4: (divide both sides by -5)
-5x = 40
-5
-5
x = -8
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When equations contain fractions, we will clean out denominators by multiplying by the lowest
common denominator (also called least common multiple).
Example 4:
2
1
Solve 3 𝑥 + 5 = 4 𝑥 −
1
3
Step 1: Each term will be multiplied by 12 or as a fraction
2
3
𝑥+5=
12
1
4
𝑥−
2𝑥
1
12
1
3
12
5
12
1𝑥
12
+1
( 1 ) ( 3 ) + ( 1 ) (1) = ( 1 ) ( 4 ) − ( 1 )( 3 )
Step 2: Cancel where possible and multiply
4
3
4
12
2
12 5
12
1
12 1
1
3
1
1
4
1
( ) ( 𝑥) +
1
( ) = ( )( ) −
1
1
( )
3
1
8x + 60 = 3x – 4
Step 3: Subtract 60 from both sides
8x + 60 = 3x – 4
-60
-60
8x = 3x – 64
Step 4: Subtract 3x from both sides
8x = 3x – 64
-3x -3x
5x = -64
Step 5: Divide both sides by 5
5𝑥
−64
= 5
5
x=
−64
5
4
𝑜𝑟 -125
Practice problems
1. 5x – 32 = -7
2. 3a + 5 = 2 (6 – 2a)
3. 42 – b – 2(3 + b) = 0
4.
3
1
x – 4 = 3x + 5
4
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10
1
3
2
x–5=3
5.
2
6. 8 – 2(x + 4) = 2 – 3(5 – x)
7. 3(y – 1) = 2(y + 2)
8. 4(3x – 1) + 11 = 2(6x + 5) -8
9. 0.4p + 0.2 = 4.2p – 7.8 – 0.6p
10.
3𝑥+2
5
2𝑥
= 10
Answers
1. 5
2. 1
3. 12
4.
5.
6.
108
5
3
or 21 5
38
8
or 2 15
15
13
5
3
or 25
7. 7
8. No solution
9.
5
2
10. -1
D) Linear Equations--Solving for a particular variable.
We will use the same techniques discussed in part C trying to isolate the variable we’re solving
for.
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Example 1:
A=
4𝑏
for x
𝑥
𝑥
Step 1: Multiply both sides by 1 cancelling where possible
1
𝑥
●
1
4𝑏
A=
𝑥
𝑥
●
1
1
Ax=4b
Step 2: Divide by A
𝐴𝑥
𝐴
x=
Example 2:
1
4𝑏
=
𝐴
4𝑏
𝐴
1
1
= 𝑏 - 𝑐 for b
𝑎
Step 1: multiply each term by
1
1
𝑎𝑏𝑐
1
●
1
𝑎
=
𝑎𝑏𝑐
1
1
●
𝑎𝑏𝑐
1
cancelling where possible.
1
1
𝑏
-
𝑎𝑏𝑐
1
1
●
1
𝑐
1
bc = a – c - ab
Step 2: Add ab to both sides (we’re now collecting all the b’s on the same side)
bc = ac – ab
+ab
+ ab
bc + ab = ac
Step 3: Distribute out the b (also called factoring out)
bc + ab = ac
b(c + a) = ac
Step 4: Divide both sides by c + a
𝑏(𝑐+𝑎)
𝑐+𝑎
𝑎𝑐
= 𝑐+𝑎
𝑎𝑐
b = 𝑐+𝑎
Practice problems
1. A = 2πr for r
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2. 2x + 2y = P for x
3. A =
𝑥+𝑦+𝑧+𝑟
4
for y
4. I = prt for p
5. Q =
𝑝−𝑞
2
for q
6. y – hx = 5x for x
7.
𝑎
𝑏
+ = 1 for b
3
4
Answers
𝐴
1. r = 2𝜋
2. x =
𝑃−2𝑦
2
3. y = 4A – x – z – r
𝐼
4. p = 𝑟𝑡
5. q = p – 2Q
𝑦
6. x = 5+ℎ
7. b =
12−4𝑎
3
E) Solving linear inequalities.
Inequalities are solved in the same manner as equalities. The exception is if you multiply or
divide both sides by a negative number, you must change the direction of the inequality.
Example 1:
Solve 3x – 5(x – 2) ≥ 12
Step 1: remove the parenthesis
3x – 5x + 10 ≥ 12
Step 2: combine like terms
-2x + 10 ≥ 12
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Step 3: subtract 10 from both sides
-2x + 10 ≥ 12
-10
-10
-2x ≥ 2
Step 4: Divide both sides by -2 and change the inequality sign to ≤
-2x ≤ 2
-2 -2
x ≤ -1
Since there are infinitely many answers, you may sketch a graph of them on the
number line.
Example 2:
5 – 9x ≥ 33 + 5x
Step 1: subtract 5 from both sides
5 – 9x ≥ 33 + 5x
-5
-5
-9x ≥ 28 + 5x
Step 2: Subtract 5 x from both sides
-9x ≥ 28 + 5x
-5x
-5x
-14x ≥ 28
Step 3: Divide both sides by -14 AND change the inequality sign
-14x ≤ 28
-14
-14
x ≤ -2
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Practice problems
1. 4(x – 3) + 5 > 6(x + 2) – 8
2. -3x – 8 < 10
3. -5(3x + 2) ≥ 20
Answers
1. x < −
11
2
2. x > -6
3. x ≤ -2
F. Solving two equations in two unknowns.
Two equations in two unknowns may be solved by substitution or elimination
(multiplication/addition).
Example 1: Solve 2x + 3y = 6
x=y–2
Step 1: We’ll use substitution to solve this since the second equation is already solved
for x.
2(y – 2) + 3y = 6
Step 2: remove parenthesis
2y – 4 + 3y = 6
Step 3: combine like terms
5y – 4 = 6
Step 4: Add 4 to both sides
5y – 4 = 6
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15
+4 +4
5y = 10
Step 5: Divide both sides by 5
5y = 10
5
5
y=2
Step 6: Now substitute 2 in for y in either of the original equation.
x=y–2
x=2–2
x=0
The solution is x = 0 or y = 2. It may also be represented as an ordered pair (0, 2)
Example 2:
5x + 3y = 17
5x – 2y = -3
Step 1: Use elimination to solve these: first multiply the second equation by -1.
5x + 3y = 17
-5x + 2y = 3
Step 2: Add the two equations to get 5y = 20 and then y = 4. Substitute 4 in for y in
either of the first two equations and solve for x.
5x + 3(4) = 17
5x + 12 = 17
-12 -12
5x = 5
5 5
x= 1
The solution is x = 1 and y = 4. It may be represented as an ordered pair (1, 4)
Practice problems. Solve using either method.
1. y = 2x – 5
3y – x = 5
2. x + 3y = 19
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x – y = -1
3. 3y + 2x = 2
-2y + x = 8
4. 2 x – 3y = -1
3x + 4y = 24
Answers
1. (4, 3)
2. (4, 5)
3. (4, -2)
4. (4, 3)
G. Story Problems.
Solve the following problems by writing an equation or a system of equations.
1. One number is 6 more than twice another number. The sum of the numbers is 36. Find
the numbers.
2. John had test scores of 98, 94, 80 and 86. What does he have to score on his 5th test to
get an average of 90?
3. In Chicago, taxis charge $4 plus 90 cents per mile for an airport pickup. How far from
the airport can Jan travel for $17.50?
4. There were 166 paid admissions to a game. The price was $3.10 each for adults and
$1.75 each for children. The amount take in was $459.25. How many adults and how
many children attended?
5. Solution A is 50% acid and solution B is 80% acid. How many liters of each should be
used in order to make 100 liters of a solution that is 68% acid?
6. Mr. J’s Pizza Parlor charges $3.70 for a slice of pizza and a soda and $9.65 for three
slices of pizza and two sodas. Determine the cost of one soda and the cost of one slice of
pizza.
7. If the radius of a circle is 5cm, find:
a. The diameter
b. The circumference
c. The area
8. If the side of a square box is 4 inches, find:
a. The area of the top
b. The area of all sides including tops and bottom
c. The volume of the box
9. A triangle has an area of 12 square feet and a height of 3 feet. Find the length of the
base.
10. A rectangular box has a length of 8 inches and width of 4 inches and a height of 2 inches.
Find the volume.
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Answers
1. 10 and 26
2. 92
3. 15 mi
4. 125 adults
41 children
5. 40 L of A
60L of B
6. One soda $1.45
One slice pizza $2.25
7. a) 10 cm
b) 10π cm
c) 25π sq cm
8. a) 16 sq inches
b) 96 sq inches
c) 64 cubic inches
9. 8 feet
10. 64 cubic inches
H. Exponents, Powers and Roots
I. Exponents
A. definitions
a. Exponent
i. Def: a number that indicates how many times the base is to be used as a
factor
ii. Def: indicates repeated multiplication
b. Base
i. Def: number being multiplied
ii. Ex: 23 base is 2 exponent is 3
c. Power
i. Def: another word for exponent
ii. Ex: 23 can be read “2 raised to the third power”
B. Operations with exponents
a. To write a number using exponents
i. Ex: 5●5●5●5 = 54
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ii. Ex: x●x●x = x3
iii. Ex: (2y)(2y)(2y)(2y)(2y) = (2y)5
b. To evaluate a number containing an exponent
i. Ex: 34 = 3●3●3●3 = 81
ii. Ex: 25 = 2●2●2●2●2 = 32
iii. Ex: (5x)2 = (5x)(5x) = 25x2
c. Powers of ten
i. 103 (exponent 3 = 3 zeros)
ii. Ans: (10)(10)(10) = 1000
II.Expanded form vs Standard form
A. Remember Section 1.1 when we wrote whole numbers in expanded form:
4, 218 8 x 1
1 x 10
2 x 100
4 x 1000
= 8 x 10
= 1 x 101
= 2 x 102
= 4 x 103
So 8 x 10 + 1 x 101 + 2 x 102 + 4 x 103 = 4218
17,832,536
6x1
3 x 10
5 x 100
2 x 1000
3 x 10,000
8 x 100,000
7 x 1,000,000
1 x 10,000,000
= 6 x 100
= 3 x 101
= 5 x 102
= 2 x 103
= 3 x 104
= 8 x 105
= 7 x 106
= 1 x 107
So, 6 x 100 + 3 x 101 + 5 x 102 + 2 x 103 + 3 x 104 + 8 x 105 + 7 x 106 + 1 x 107 = 17,832.536
III. Roots
A. Definitions
a. Root: A number which, when multiplied by itself, results in the given number.
b. Index: the number used to tell which root you’re looking for
c. In square roots, the index is implied
B. Examples:
a. √16, because (16 = base, 2 = index, but the 2 is implied)
3
b. √8, because ( 8 = base, 3 = index)
4
c. √16, because (16 = base, 4 = index)
C. Not all roots are whole numbers
a. Can find others in tables or on calculator
b. Can estimate others:
i. Square root of 115 falls between 10 and 11
ii. Square root of 30 falls between 5 and 6
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IV. Comparing
1. Symbols
a. > greater than
b. < less than
c. = equal to
2. Examples
a. 16 + 4 ? 20
b. 8 + 7 ? 7 + 8
c. 12 ● 2 ? 18
d. 9 ? 4●3
e. 10 + 3 ? 2 + 10
f. (3 + 4) + 5 ? 3 + (4 + 5)
Answers:
a. =
b. =
c. >
d. <
e. >
f. =
V. Estimating
A. Way to see if answer makes sense, is logical
B. Round numbers being used and see if answer approximates actual answer.
C. The smaller the unit you round to, the closer your estimate will be to the actual answer
D. Examples:
a. Round to nearest whole number and estimate the answer.
i. (14.6)(3.2)
actual answer 46.72, rounded answer 45 (15 x 3), yes they are close
b. Round to the nearest tens and estimate the answer.
i. (42)(59)
Actual answer 2,478, rounded to 2400 (40 x 60), yes, they are close
c. Round to the nearest hundreds and estimate the answer.
i. (495)(98)
Actual answer 48,510, rounded to 50,000 (500 x 100), yes they are
close
E. Question on test could ask to estimate or approximate by rounding to a certain place
value.
a. Approximate by rounding to the nearest hundred
b. Approximate by rounding to the nearest thousand
Directions: Write using exponential notation:
1. 2●2●2●2●2●2 =
2. 7●7●7●7 =
3. 6●6●6 =
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4. 4=
5. 16 x 16 x 16 =
6. 23 x 23 x 23 x 23 =
7. Six cubed =
8. Three squared =
9. Eight squared =
10. Five to the sixth power =
11. Four to the fifth power =
12. 2●2●2 + 3●3●3●3 =
13. 3●3 + 9●9●9●9 =
14. 4●4 + 7●7●7 + 8
Direction: Write as repeated multiplication.
15. 45 =
16. 74 =
17. 32 =
18. 5353=
19. 61 =
20. 27 =
21. 32 + 54 =
22. (28)4 =
23. 83 =
24. (243)1 =
25. 65 + 21 =
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Directions: Calculate.
26. 35 =
27. (24)2 =
28. 24 =
29. 14 =
30. 05 =
31. 42 =
32. 73 =
33. 114 =
34. 64 =
Directions: Define or explain.
35. Exponent
36. Power
37. Base
38. Square root
Directions: Find the square root.
39. √64
40. √0
41. √49
42. √1
43. √144
44. √225
45. √169
46. √121
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47. √400
48. √100
49. √2500
50. √3600
Answer Key
1. 26
2. 74
3. 63
4. 22
5. 163
6. 234
7. 63
8. 32
9. 82
10. 56
11. 45
12. 23+34
13. 32+ 94
14. 42+ 73 + 8
15. 4 x 4 x 4 x 4 x 4
16. 7 x 7 x 7 x 7
17. 3 x 3
18. 5 x 5 x 5 x 5 x 5 x 5
19. 6
20. 2 x2 x 2 x 2 x 2 x 2 x 2
21. 3 x 3+5 x 5 x 5 x 5
22. 28 x 28 x 28 x 28
23. 8 x 8 x 8
24. 243
25. 6 x 6 x 6 x 6 x 6 + 2
26. 243
27. 576
28. 16
29. 1
30. 0
31. 16
32. 343
33. 1
34. 1296
35. a number that indicates how many times the base is to be used as a factor
36. another word for exponent
37. number being multiplied
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23
38. the index is implied
39. 8
40. 0
41. 7
42. 1
43. 12
44. 15
45. 13
46. 11
47. 20
48. 10
49. 50
50. 60
I. Simplifying radicals that are not perfect squares.
Example 1: recall √25 = 5 because 52 = 25
Example 2: √50 is not a perfect square so we simplify by factoring the 50.
√50 = √5𝑥5𝑥2
Now look for pairs √(5𝑥5)𝑥2
Answer: 5√2
Example 3: √80
√2𝑥2𝑥2𝑥2𝑥5
√(2𝑥2)(2𝑥2)𝑥5
(look for pairs)
2x2 √5
Answer: 4√5
Example 4: √27𝑥 3 𝑦 4
√3 ∗ 3 ∗ 3 ∗ 𝑥 ∗ 𝑥 ∗ 𝑥 ∗ 𝑦 ∗ 𝑦 ∗ 𝑦 ∗ 𝑦
(look for pairs)
√(3 ∗ 3)3(𝑥 ∗ 𝑥) ∗ 𝑥 ∗ (𝑦 ∗ 𝑦)(𝑦 ∗ 𝑦)
3xyy√3𝑥
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Answer: 3xy2 √3𝑥
Example 5: 2√3 + 4√3 is just like adding 2x + 4x: you add the coefficients.
2√3+ 4√3
(2+4)√3
6√3
Example 6: 9√2 – 7√2
(9 – 7) √2
Answer: 2√2
Example 7: 3√18 + 4√2 (simplify √18 to see if we can add
3√3 ∗ 3 ∗ 2 + 4√2
3√(3 ∗ 3)2 + 4√2
3x3√2 + 4√2
9√2 + 4√2
Answer: 13√2
Example 8: (√2)(√3) This can be done as long as they’re both square roots.
√2√3
√2●3
Answer: √6
Example 9: (3√5)(2√6)
(3●2)(√5√6)
Answer: 6√30
Example 10: (√2 + √3)(√2 + √3)
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Use foil
√2●2 + √2●3 + √2●3 + √3●3
2 + √6 + √6 + 3
Answer: 5 + 2√6
Example 11:
√12
√3
12
√
3
√4
√2 ∗ 2
Answer 2
Example 12:
√8
√32
8
√
32
1
√
4
√1
√4
√1∗1
√2∗2
1
Answer: 2
Example 13:
√2
√3
2
Writing √3 won’t help because there’s no cancellation, so we have to get
the radical out of the denominator. This is called rationalizing the
denominator.
√2
√3
●
(√3)
(√3)
√6
√3∗3
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Answer:
√6
3
Practice problems
1. √2√10
5
2. √3
12
15
3. √18 ● √40
4. 2√18 - 5√32 + 7√162
5. (3√2 + 2√5)(4√2 - 3√5)
6. √24𝑥 3 𝑦 6
Answers:
1. 2√5
2.
3.
√15
3
1
2
4. 49√2
5. -6 -√10
6. 2xy3√6𝑥
J. Rules of exponents
1. Multiplying like-base exponents
Example: x3 ● x4
Step 1: in expanded form
(x●x●x)(x●x●x●x)
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Step 2: remove parenthesis
x●x●x●x●x●x●x
Step 3: rewrite with a single exponent
x7
Step 4: shortcut-add original exponents
x3 ● x4
x3+4
x7
Rule: xa ● xb = xa+b (note: the bases must be the same)
Example: x24 ● x25
x24+25
x49
2. Raising a power to a power
Example: (x3)4
Step 1: in expanded form
x3 ● x3 ● x3 ● x3
(x● x ●x)(x● x ●x)(x● x● x)(x● x● x)
Step 2: remove parenthesis
x●x●x●x●x●x●x●x●x●x●x●x
Step 3: rewrite with a single exponent
x12
Step 4: Shortcut-multiply original exponents
(x3)4
x3●4
x12
Rule: (xa)b = xab
3. Dividing exponents
𝑥7
Example: 𝑥 3
Step 1: in expanded form
𝑥∗𝑥∗𝑥∗𝑥∗𝑥∗𝑥∗𝑥
𝑥∗𝑥∗𝑥
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Step 2: cancel
𝑥∗𝑥∗𝑥∗𝑥∗𝑥∗𝑥∗𝑥
𝑥∗𝑥∗𝑥
Step 3: rewrite with a single exponent
x●x●x●x
x4
Step 4: shortcut-subtract the original exponents
𝑥7
𝑥3
x7-3
x4
Rule:
𝒙𝒂
𝒙𝒃
𝒙𝒂
𝒙𝒃
= xa-b if a>b
𝟏
= 𝒙𝒃−𝒂 if b>a
4. Negative numbers as exponents
𝟏
Definition: x-a = 𝒙𝒂
1
Example: x-3 = 𝑥 3
1
1
Example: 2-4 = 24 = 16
1
Example: 𝑦 −5 =
1
1
𝑦5
1
1 𝑦5
= 1÷𝑦 5 = 1 ● 1 = 𝑦 5
Example: x4●x-3 = x4+(-3) = x1 = x
1
Example: x-5 ● x3 = x-5+3 = x-2 = 𝑥 2
In the above examples, notice that a negative exponent in the numerator ends up in the
denominator and a negative exponent in the denominator ends up in the numerator.
Example:
𝑥 −3 𝑦 −4
𝑥 −5 𝑦 6
𝑥5
can be written as 𝑥 3 𝑦 4 𝑦 6
𝑥2
Which then simplifies to 𝑦 10 (using the previous rules)
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1
Example
1
: (2x)-4 becomes 2-4x-4 which is 24 𝑥 4 or 16𝑥 4
Example: (4xyz)0 = 1 (any non-zero base to the zero power is 1)
Practice problems (write all answer with positive exponents only)
2𝑎2
1. (3𝑏4 )-3
2. (
𝑎−5 3
)
3
3. (2a2b-5c-6)3
4. (3a-5b4c3)-3
(2a3b-7c3)2
5. (5x0y5z6)(-3xy3z-3)
6. (-x4y5z6)4
7. (4x2y6z3)2(-x-3y4z5)6
8.
9.
40𝑥 4 −24𝑥 3 +8𝑥2
8𝑥 2
30𝑎2 𝑏 −6 𝑐 11
−4𝑎−6 𝑐 11
10. (x3 - 5x)(2x4 +7)
Answers:
1.
2.
3.
4.
27𝑏 12
8𝑎6
1
27𝑎15
8𝑎6
𝑏 15 𝑐 18
𝑎9 𝑏 2
108𝑐 15
5. -15xy8z3
6. x16y20z24
7.
16𝑦 36 𝑧 36
𝑥 14
8. 5x2-3x +1
9.
15𝑎8
−2𝑏 6
10. 2x7 -10x5+7x3-35x
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K. DEFINITION OF FACTORING
Factoring is the inverse (or reverse) of multiplying. To factor a quantity means
to write it as a product.
EXAMPLES:
Multiply
multiply
example 3:
56
example 2:
(2x)(3x2)
factor
Multiplying: (5xy)(2x + 3) = 10x2y + 15xy
example 4:
Factoring:
example 1:
8●7
=
=
6x3
factor
10x2y + 15xy = (5xy)(2x + 3)
II. METHODS OF FACTORING
A. Method 1: Factoring a Monomial
Remember, a monomial is a polynomial containing one term.
Example 1:
12x4 = (6x2)(2x2)
Example 2:
12x4 = (3x)(4x3)
EXERCISE II A: Complete the factoring of each monomial below.
1) 12x4 = (
)(3x3)
3) -8xy2 = (-2y)(
2) -8xy2 = (-8x)(
)
4)
20a2b2 = (5ab)(
4xy
4) 4ab
)
)
Answers:
1) 4x
2)
y2 or 1y2
3)
B. Method 2: Factoring a Polynomial by using a common monomial factor
Remember: a polynomial is one monomial OR the sum of several monomials.
Each monomial in the polynomial is called a term. A binomial is a polynomial
containing 2 terms. A trinomial is a polynomial containing 3 terms.
Example 1: 12x3 + 4x2 - 8x = 4x ( 3x2 + 1x - 2 )
note: 12x3 + 4x2 - 8x is a polynomial containing 3 terms. Each term contains
4x as a common monomial factor. So 4x is "factored out" of each term to obtain
the resulting
answer, which is the product of 4x and 3x2 + 1x - 2.
Example 2:
20x4 + 15x2 = 5x2 ( 4x2 + 3 )
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EXERCISE II B: Factor the common monomial factor out of each term in the polynomial.
1) 10n2 - 20n + 5 = 5 (
)
2)
16x2y - 8xy2 = (8xy)(
3)
20a2b2 - 10a2b + 15ab2 = (5ab)(
4)
14m2 - 28 mn - 7n2 =
5)
16x2 - 8x - 40 = ____ (
6)
5x3y2 - 10x2y3 = ____ (
)
7(
)
)
)
)
Answers:
1)
(2n2 - 4n + 1)
2)
(2x - y)
3)
(4ab - 2a + 3b)
4)
(2m2 - 4mn - n2)
5)
8(2x2 - x - 5)
6)
5x2y2 (x - 2y)
C. Method 3: Factoring a Polynomial by Grouping
This method should be attempted when there are four terms in the polynomial. When factoring
in this situation is possible, it is because the four terms happen to be the result of a multiplication
that was performed by the FOIL method. In other words, factoring by grouping is the reverse of
the FOIL method.
Review the FOIL method of multiplying:
F
O
I
L
(2x + 5)(3x - 4) = (2x)(3x) + (2x)(-4) + (5)(3x) + (5)(-4)
=
6x2
+
-8x
+ 15x
+ - 20
<----- notice that 4 terms result
=
6x2 + 7x - 20
Example of Factoring by Grouping (reverse of FOIL method):
Factoring:
Break the polynomial into 2 groups:
Factor each group separately:
6x2 - 8x + 15x - 20
6x2 - 8x
+
15x - 20
2x (3x - 4)
+
5 ( 3x - 4 )
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Factor out the common binomial:
(3x - 4) (2x + 5)
is the answer!
Note: the answer can be checked by FOIL. (see above)
EXERCISE II C: Factor by grouping.
1)
7x(2x - 3) + 1(2x - 3)
= (2x - 3) (
2)
x2( x + 5) + 3(x + 5) = (
3)
2x2 + 5x - 8x - 20 =
)
)(
4)
5) 3x2 + 12x - 4x + 8 =
)
14a2b2 + 12ab - 21ab - 18 =
6)
x2 + 10x + 10x + 100 =
Answers:
1) (7x + 1)
2)
(x2 +3)(x + 5)
5) not factorable
6)
(x + 10)(x + 10) or (x + 10) 2
3) (x - 4)(2x + 5)
4) (2ab - 3)(7ab + 6)
D. Method 4: Factoring a trinomial of the type 1x2 + bx + c
This is a guess and check method of factoring.
GUESS by looking at the numbers b and c, and CHECK by using FOIL.
example:
x2 - 7x + 10
is a trinomial of the type 1x2 + bx + c,
THINK:
x2 - 7x + 10 = (
+
where b = -7 and c = 10
) (
+
)
step 1: In the FOIL process, the product of the First terms is x2, so it is obvious
that the first terms must be x and x. Thus:
x2 - 7x + 10 = ( x ___
) (
x
___
)
step 2: In the FOIL process, the product of the Last terms is 10.
So the c value ( 10 ) is used to find two numbers whose product is 10.
GUESSING, some possibilities are: 1 and 10,
2 and 5,
-1 and -10, & -2 and -5
step3: In the FOIL process, the product of the Inner terms and the product of the
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Outer terms result in a sum of -7x. So the b value ( -7 ) is used to find
two numbers that add up to -7.
GUESSING, some possibilities are:
-3 and -4,
-6 and -1,
-10 and 3,
-5 and -2, .....
step 4: We have found that we need two numbers that add up to b ( -7 )
and at the same time have a product of c ( 10 ).
Thus, using our work in steps 2 and 3, we can complete the factoring
because -5 + -2 = b
and
(-5)(-2) = c
x2 - 7 x + 10 = (x-5) (x-2 )
step 5: CHECK your work by FOIL: x(x) + x(-2) + -5(x) + (-5)(-2) = x2 + -7x + 10
EXERCISE II D: Factor the given trinomial.
1)
Think:
____ + _____ = 12
x2 + 12x + 35
2)
Think:
= ( x
____ + _____ = -5
m2 - 5m - 24 = ( m
3)
Think:
and
+ ____
and
+
_____ x _____ = 35
____
) (
x
+ _____
_____ x _____ = -24
) (
____ + _____ = 3
and
a2 + 3a - 28 = (
a + ____
m
+
_____
)
_____ x _____ = -28
) (
a
-
_____
)
4)
x2 + 14x + 49
5)
n2 - n - 56 = (
6)
x2 + 8x - 20
= ( _____
+ _____
) (
_____
- _____
)
7)
a2 + 0a - 36
= ( _____
+ _____
) (
_____
- _____
)
= ( x
+ ____
n + ____
) (
)
) (
x
n
+ _____
- _____
)
)
Answers:
1) Use 7 and 5. Answer is (x + 5)(x + 7)
2) Use -8 and 3. Answer is (m - 8)(m + 3)
3) Use -4 and 7. Answer is (a + 7)(a - 4)
4) (x + 7)(x + 7)
5) (n - 8) (n + 7)
6) (x - 2)(x + 10)
7) (a + 6)(a - 6)
E. Method 5: Factoring a trinomial of the type ax2 + bx + c ,
( a ≠ 1)
(This method is sometimes referred to as the ac method.)
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Example: Factor 6x2 - 13x - 15
(note that the coefficient on the x2 term is not 1)
The strategy is to convert the trinomial into an equivalent polynomial having 4 terms,
and then to factor by grouping. This is not always possible, but when it is possible you
will be able to find two numbers which have:
a sum of b ( b is -13 in this case ) and
which also have a product of ac, which is (6)(-15) or -90 in this case.
Think: ____ + ____ = -13
and
at the same time ____ x ____ = -90
Making a factor tree for 90 may help you:
90
45 x 2
5x9x2
This tree should help you to see that (-18) x 5 = -90 and that (-18) + 5 = -13.
So use -18x + 5x to replace the middle term, which is -13x
Steps:
6x2 - 13x - 15
which is the original trinomial, can be written as
6x2 - 18x + 5x - 15
6x(x - 3) + 5(x - 3)
(x - 3)(6x + 5)
which now has 4 terms--so try factoring by grouping.
Factoring by grouping is possible in this case, so.....
is the final answer. Check it by FOIL.
EXERCISE II E: Factor the given trinomial. Remember to "factor by grouping".
1) 12x2 + x - 6
12x2 + ___x - ___ x - 6
Think: ac = -6 x 12 = -72 and b = 1
You need two numbers such that
___ x ___ = -72 and ___ + ___ = 1
2) 10x2 - 43x - 9
10x2 + ___x - ___ x
- 9
Think: ac = -9 x 10 = -90 and b = -43
You need two numbers such that
___ x ___ = -90 and ___ + ___ = -43
3) 16a2 + 8a - 3
16a2 + ___a - ___ a
- 3
Think: ac = -3 x 16 = -48 and b = 8
You need two numbers such that
___ x ___ = -48 and ___ + ___ = 8
+ 14
Think: ac = 8 x 14 = 112 and b = 23
8x2 + 23x + 14
4)
8x2 + ___x + ___ x
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You need two numbers such that
___ x ___ = 112 and ___ + ___ = 23
5)
18x2 - 13x
6)
12m2
- 5 = (
+ 35m
- 3
)(
= (
)
)(
)
Answers:
1) Use 9 and -8. Answer is (4x + 3)(3x - 2)
2) Use -45 and 2. Answer is (2x - 9)(5x + 1)
3) Use -4 and 12. Answer is (4a - 1)(4a + 3)
4) Use 7 and 16. Answer is (8x + 7)(x + 2)
5) Use -18 & 5. Answer is (18x + 5) (x - 1)
6) Use 36 & -1. Answer is (12m - 1)(m + 3)
F. Method 6: Factoring a Perfect Square Trinomial (Special Pattern)
Review:
36 is called a perfect square & 6 and -6 are both called square roots of 36
because 6 x 6 = 36 and -6 x -6 = 36. (Note: 6 x -6 is not equal to 36)
25n2 is a perfect square & 5n and -5n are both square roots of 25n2
because (5n)(5n) = 25n2 and (-5n)(-5n) = 25n2 . (Note: (5n)(-5n) is not equal to 25n2)
The first 12 Perfect Squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144
By FOIL method of multiplication, you should understand the PATTERN shown below:
(a + b)(a + b) = a2+2ab+ b2
This PATTERN shows what happens when a binomial, (a + b), has been squared.
The result, a2 + 2ab + b2, is called a perfect square trinomial.
By reversing the multiplication, the PATTERN can be used for factoring a perfect square
trinomial. You must KNOW (memorize) the PATTERN below and understand the
mathematical relationships that are indicated by the pattern.
The PATTERN:
The relationships:
a2 +2ab+ b2
= (a + b)(a + b)
b is the square root of the perfect square, b2
a is the square root of the perfect square, a2
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AND: the middle term, 2ab, is twice the product of the two square roots!!!
Example 1:
Given
x2 + 20x + 100, decide if it is a perfect square trinomial.
Solution: The first term, x2, is a perfect square and its square root is x. The third term, 100, is a
perfect square and its square root is 10 (or -10). Now, the middle term needs to be checked.
2 ( x) ( 10 ) = 20x , so the middle term does fit the pattern because it IS twice the product of the
two square roots. So, YES, the given trinomial is a perfect square trinomial and therefore CAN
be factored into the square of a binomial.
Factoring it:
Example 2:
x2 + 20x + 100 = (x + 10)(x + 10) or (x + 10)2
Given
n2 + 32n + 64, is it a perfect square trinomial ?
Solution:
The first term, n2, is a perfect square and its square root is n. The third
term, 64, is a perfect square and its square root is 8 (or -8). Now, the middle term needs to be
checked. 2 ( n ) ( 8 ) = 16n , so the middle term does NOT fit the pattern because it ISN'T twice
the product of the two square roots. So, NO, the given trinomial is not a perfect square trinomial
and therefore CANNOT be factored into the square of a binomial.
Example 3:
Given
9x2 + 30x + 25, is it a perfect square trinomial?
Solution: The first term, 9x2, is a perfect square and its square root is 3x. The third term, 25, is
a perfect square and its square root is 5 (or -5). Now, the middle term needs to be checked.
2(3x)(5) = 30x , so the middle term does fit the pattern because it IS twice the product of the
two square roots. So, YES, the given trinomial is a perfect square trinomial and therefore CAN
be factored into the square of a binomial.
Factoring it:
9x2 + 30x + 25 = ( 3x+ 5)(3x + 5) or (3x + 5)2
EXERCISE II F - Part I:
Is the trinomial a perfect square trinomial ?
If it is, factor it by use of the pattern.
1)
4x2 + 12x + 9
2)
m2 + 18m + 64
3)
25m2 + 70m + 49
4)
a2 + 12a + 36
5)
10w2 + 90w + 81
Answers:
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1) Yes. Answer is (2x + 3)(2x + 3) or (2x +3)2
2) No. The square roots are m and 8,
but 2(m)(8) is not equal to 18m.
3) Yes. Answer is (5m + 7)(5m + 7) or (5m + 7)2
4) Yes. (a + 6)(a + 6) or (a + 6)2
5) No. 10w2 is not a perfect square.
NOTICE that the pattern also works if the binomial being squared is a difference (not a sum).
= a2 - 2ab + b2
By FOIL, the multiplication pattern is:
(a - b)(a - b)
Reversing it, the FACTORING pattern is:
a2 - 2ab + b2 = (a - b)(a - b)
Again, you must MEMORIZE this factoring pattern and you must UNDERSTAND the
relationships contained within it. These relationships are: a is the square root of the first term in
the trinomial,-b is the square root of the third term in the trinomial, AND the middle term, -2ab,
is twice the product of the two square roots.
Example 1:
Given
x2 - 8x + 16, decide if it is a perfect square trinomial.
Solution: The first term, x2, is a perfect square and its square root is x. The third term, 16, is a
perfect square and its square root is -4 (or 4). Now, the middle term needs to be checked.
2(x)(-4) = -8x , so the middle term does fit the pattern because it IS twice the product of the two
square roots. So, YES, the given trinomial is a perfect square trinomial and therefore CAN be
factored into the square of a binomial.
Factoring it:
Example 2:
x2 - 8x + 16 = ( x -4 )( x -4 ) or ( x - 4 )2
Given
n2 - 25n + 50, decide if it is a perfect square trinomial.
Solution:
The first term, n2, is a perfect square and its square root is n.
The third term, 50, is a NOT perfect square. So, NO, the given trinomial is not a
perfect square trinomial and therefore CANNOT be factored into the "square of a binomial".
EXERCISE II F - Part 2:
Is the trinomial a perfect square trinomial ?
If it is, factor it by use of the pattern.
1)
n2 - 10n + 25
2)
x2 - 18x + 81
4)
4x2 - 12x + 9
5)
a2 - 18a - 49
3)
a2 - 14a + 49
Answers:
1) Yes. Answer is (n - 5)(n - 5) or (n - 5)2
2) Yes. (x - 9)( x - 9) or (x - 9)2
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3) Yes. Answer is (a - 7)(a - 7) or (a - 7)2
4) Yes. (2x - 3)(2x - 3) or (2x -3)2
5) No. -49 is not a perfect square.
G. Method 7: Factoring a Difference of Squares (Special Pattern)
REVIEW:
Multiplying by FOIL, (n - 5)(n + 5) = n2 - 5n + 5n - 25
(x - 3)(x + 3) = x2 - 3x + 3x - 9
=
n2 - 25
=
x2 - 9
(2x - 7)(2x + 7) = 4x2 - 14x + 14x - 49 = 4x2 - 49
(the SUM of 2 terms) x (the DIFFERENCE of those 2 terms) = difference of squares
The multiplication PATTERN:
(a + b)
In reverse, the factoring PATTERN:
a2 - b2
the difference of squares
(a - b)
=
a2 - b2
= (a + b) (a - b)
= (the SUM of 2 terms) x (the DIFFERENCE of those 2 terms)
You MUST know (memorize) this pattern and understand the relationships contained within the
pattern. The relationships are:
a2, the first term in the difference of squares, is a perfect square,
b2, the second term in the difference of squares, is a perfect square,
a is the square root of the first term ( a2 ),
and
b is the square root of the second term (b2).
EXERCISE II G:
Is the polynomial the difference of squares ?
If it is, factor it by use of the pattern.
1)
x2 - 100
2)
3x2 - 25
3)
4x2 - 49
4)
w2 - 16w - 64
5)
25a2 - 1/9
6)
x2 + 16
Answers:
1) Yes. Answer is (x + 10)(x - 10)
3) Yes. Answer is (2x - 7)(2x + 7)
2) No. 3x2 is not a perfect square.
4) No. difference of square pattern has
2 terms (not 3).
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5) Yes. (5a + 1/3)(5a - 1/3)
6) No. This is the sum of squares-- not the difference of squares.
III: MIXED PRACTICE
Since there are seven different factoring methods available, a GENERAL STRATEGY for
factoring is necessary. Deciding on the correct method to use is the key to success.
Step 1: ALWAYS begin by trying method 2 (Factoring out a common monomial factor).
Regardless of whether this works or not, you must continue to TRY factoring by another method.
Step 2: Count the number of terms in the polynomial that you are trying to factor. This is a big
HINT in choosing the next method which might work.
four terms:
three terms:
two terms:
try method 3
try method 4
try method 5
try method 6
try method 7
( factoring by grouping )
( factoring a trinomial of type 1x2 + bx + c ), OR
( factoring a trinomial of type ax2 + bx + c ), OR
( perfect square trinomial pattern )
( difference of squares pattern )
Step 3: Try to factor again, if possible. Although some polynomials cannot be factored at all,
others can be factored two, three, or even four times! You are not done factoring until the
polynomial has been COMPLETELY FACTORED.
EXERCISE III: Factor the polynomial completely.
1)
10x2 + 6x - 15x - 9
2)
3x2 - 48
3)
21a2 - 15a + 12
4)
x4 - 81
5)
15x2 - 20x - 75
6)
n2 + 22n + 121
7)
w2 + 25
8)
14x2 + 13x + 3
9)
3b2 + 75
10)
5n2 - 40n + 80
l2)
100n2 - 400n - 2100
11) x3 + 2x2 - 25x - 50
Answers:
1) By method 3:
(5x + 3)(2x - 3)
2) By method 1:
then by method 7:
3 (x2 - 16)
3(x + 4)(x - 4)
3) By method 1:
3 ( 7a2 - 5a + 4)
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(x2 - 9) (x2 + 9)
(x + 3)(x - 3)(x2 + 9)
4) By method 7:
then method 7 again:
5) By method 1:
then method 4:
5 (3x2 - 4x - 15)
5 (3x + 5)(x - 3)
6) By method 6:
(n + 11) (n + 11) or (n + 11)2
7) not factorable
8) By method 5:
(7x + 3)(2x + 1)
9) By method 1:
3(b2 + 25)
10) By method 1:
then method 6:
5 ( n2 - 8n + 16 )
5 (n - 4)(n - 4) or
11) By method 3:
then method 7:
(x2 - 25) (x + 2)
(x + 5)(x - 5)(x + 2)
12) By method 1:
then by method 3:
100 (n2 - 4n - 21)
100 (n - 7) (n + 3)
5(n - 4)2
L. Simplifying rational expressions using factoring
1. Multiplying rational expressions
Example:
𝑥 2 −3𝑥−10
𝑥 2 −4𝑥+4
●
2𝑥−4
4𝑥−20
Step 1: factor where possible
(𝑥−5)(𝑥+2)
(𝑥−2)(𝑥−2)
●
2(𝑥−2)
4(𝑥−5)
Step 2: Cancel
1
1
(𝑥−5)(𝑥+2)
2(𝑥−2)
(𝑥−2)(𝑥−2)
1
●
1
4(𝑥−5)
2
1
Step 3: multiply
𝑥+2
Step 4: Answer 2(𝑥−2)
2. Dividing rational expressions
𝑥 2 +𝑥−20
Example: 𝑥 2 −7𝑥+12 ÷
𝑥 2 +10𝑥+25
𝑥 2 −6𝑥+9
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Step 1: rewrite as a multiplication (this is a rule for fractions)
𝑥 2 +𝑥−20
●
𝑥 2 −7𝑥+12
𝑥 2 −6𝑥+9
𝑥 2 +10𝑥+25
Step 2: factor where possible
(𝑥+5)(𝑥−4)
(𝑥−3)(𝑥−4)
●
(𝑥−3)(𝑥−3)
(𝑥+5)(𝑥+5)
Step 3: cancel
1
1
1
(𝑥+5)(𝑥−4)
(𝑥−3)(𝑥−4)
1
●
(𝑥−3)(𝑥−3)
(𝑥+5)(𝑥+5)
1
1
𝑥−3
Step 4: answer
𝑥+5
3. Adding/subtracting rational expressions
3
6
Example: 2𝑦 2 + 𝑦
Step 1: get a common denominator of 2y2
Step 2: multiply second fraction top and bottom by 2y
3
6(2𝑦)
2𝑦 2
+ 𝑦(2𝑦)
Step 3: multiply the second numerator
3
12𝑦
2𝑦 2
+ 2𝑦 2
Step 4: write a single fraction over the common denominator
Answer:
2
3+12𝑦
2𝑦 2
1
Example: 9𝑎3 + 6𝑎2
Step 1: get a common denominator of 18a3
Step 2: multiply 1st fraction top and bottom by 2 and the second fraction top and
bottom by 3a.
2(2)
1(3𝑎)
+ 2
3
9𝑎 (2)
6𝑎 (3𝑎)
4
3𝑎
Step 3: 18𝑎3 + 18𝑎3
Answer:
Example:
𝑥−2
3𝑥
-
4+3𝑎
18𝑎3
2𝑥−1
5𝑥
Step 1: get a common denominator of 15x
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Step 2: multiply 1st fraction top and bottom by 5 and the second fraction top and
bottom by 3.
(𝑥−2)5 (2𝑥−1)3
- (5𝑥)3
(3𝑥)5
Step 3:
Step 4:
Step 5:
(5𝑥−10)
15𝑥
(6𝑥−3)
-
15𝑥
(5𝑥−10)−(6𝑥−3)
15𝑥
5𝑥−10−6𝑥+3
15𝑥
Step 6: Answer
−𝑥−7
15𝑥
𝑎
5
Example: 𝑎2 +11𝑎+30 - 𝑎2 +9𝑎+20
Step 1: factor both denominators
𝑎
𝑎
(𝑎+6)(𝑎+5)
- (𝑎+4)(𝑎+5)
Step 2: get a common denominator of
(a+b)(a+5)(a+4)
Step 3: multiply the first fraction top and bottom by (a+4) and multiply the second
fraction top and bottom by (a+6).
𝑎(𝑎+4)
5(𝑎+6)
(𝑎+6)(𝑎+5)(𝑎+4)
- (𝑎+4)(𝑎+5)(𝑎+6)
(𝑎2 +4𝑎)
(5𝑎+30)
Step 4: (𝑎+6)(𝑎+5)(𝑎+4) - (𝑎+4)(𝑎+5)(𝑎+6)
𝑎2 +4𝑎−5𝑎−30
Step 5: (𝑎+6)(𝑎+5)(𝑎+4)
𝑎2 −𝑎−30
Step 6: (𝑎+6)(𝑎+5)(𝑎+4)
Step 7: factor the numerator to see if they fraction will simplify
(𝑎−6)(𝑎+5)
(𝑎+6)(𝑎+5)(𝑎+4)
(𝑎−6)
Answer: (𝑎+6)(𝑎+4)
3
2
Example: 𝑥−1 - 1−𝑥
Step 1: notice the denominators are opposites of each other. This means all we
have to do is multiply one fraction top and bottom by -1.
3
2(−1)
Step 2: 𝑥−1 - (1−𝑥)(−1)
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3
Step 3: 𝑥−1 Step 4:
(−2)
𝑥−1
3−(−2)
𝑥−1
5
Step 5: 𝑥−1
Practice problems
1.
2.
3.
4.
5.
6.
𝑎2 −9
●
𝑎2
𝑎2 +4𝑎
𝑎2 +𝑎−12
12𝑥−36
3𝑥 2 +2𝑥−8
12𝑥−16
●
8𝑥−24
5𝑎2 +5𝑎−30
2𝑎2 −8
÷ 6𝑎2 +36𝑎+54
10𝑎+30
4−𝑥 2
𝑥−2
÷ 𝑥+2
𝑥
𝑥+3
𝑥−5
+
2𝑥−1
5−𝑥
𝑥
6
𝑥 2 +15𝑥+56
- 𝑥 2 +13𝑥+42
Answers
𝑎+3
1. 𝑎
2.
3.
4.
5.
6.
6
𝑥+2
3(𝑎+3)(𝑎+3)
2(𝑎+2)
−1(𝑥+2)2
𝑥
−𝑥+4
𝑥−5
𝑥 2 −48
(𝑥+6)(𝑥+7)(𝑥+8)
M. Solving quadratic equations
Example: Solve x2+5x +6 = 0
Step 1: factor
(x+3)(x+2) =0
Step 2: set each factor = to 0 and solve
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x+3=0 or x+2=0
Answer: x= -3 or x = -2
Example: 2x2 -3x = 5
Step 1: get the equation = to 0 by subtracting 5 from both sides
2x2 – 3x – 5 = 0
Step 2: factor
(2x -5)(x + 1) = 0
Step 3: set each factor = to 0 and solve
2x -5 = 0 or x + 1 = 0
2x =5
5
Answer: x = 2 or x= -1
Example: x2 –x – 4 = 0
Step 1: This is not factorable so we use the quadratic formula 𝑥 =
−𝑏±√𝑏 2 −4𝑎𝑐
2𝑎
Step 2: identify a, b and c
a = 1 (the coefficient on x2)
b = -1 (the coefficient on x)
c = -4 (the constant)
Step 3: plug into the formula and do the arithmetic
𝑥=
𝑥=
1±√1−4(1)(−4)
2(1)
1±√1+16
2
Answer: 𝑥 =
1±√17 = 1+√17 or 1−√17
2
2
2
Example: 4x2 + 9x + 2 = 0
Step 1: Even though this is factorable, we can use the quadratic formula
Step 2: a = 4
b=9
c=2
Step 3: 𝑥 =
−9±√81−4(4)(2)
2(4)
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𝑥=
𝑥=
𝑥=
x=
−9±√81−32
8
−9±√49
8
−9±7
8
−2
8
or
−16
8
answer: x =
−1
4
or -2
Practice problems
1. x2 – 6x -16 = 0
2. x2 -4x – 11 = 0
3. x2 +22x + 21 = 0
4. 4x2 + 12x = 7
5. 3x2 – 4x = 3
6. (3x+2)2 =15
7. 9x2 -25 = 0
8. 2x2 = 5x + 12
Answers
1. -2, 8
2. 2±√15
3. -21, -1
4.
5.
6.
7.
−7 1
2
,2
2±√13
3
−2±√15
3
5 −5
3
,
3
3
8. − 2, 4
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