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Unit #2: Powers and Polynomials
Unit Outline:
1
Date Lesson Title Assignment Completed
2.1 Introduction to Algebra
2.2 Discovering the Exponent Laws – Part 1
2.3 Discovering the Exponent Laws – Part 2
2.4 Multiplying and Dividing Polynomials
Mid Unit Test Lesson 2.1 to 2.4
2.5 Collecting Like Terms
2.6 Adding and Subtracting Polynomials
2.7 Multiplying a Polynomial by a Monomial
Mid Unit Test Lessons 2.5 to 2.5
2.8 Simplifying Algebraic Expressions – Part 1
2.9 Simplifying Algebraic Expressions – Part 2***
Final Algebra Test
2.1 Introduction to Algebra
Warm-up: A room is constructed with the following dimensions.
1. CalculateShow your calculations and show units in all of your calculations
a) The perimeter of the floor b) The area of the floor c) The volume of the room
Reflect:i) Can you add the perimeter and the area?
ii) Can you add the perimeter and the volume?
iii) Can you add the area and the volume?
7m
5m
4m
6m5m
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Learning Goals: By the end of this lesson, you will be able to:
Understanding Algebraa) 5m + 6m b) 1m + 1m c) x + x d) 4x + 7x
e) (3m) (4m) f) (1m) (1m) g) (x)(x) h) (5x)(2x)
What is algebra?
Learning algebra is a little like learning another language. By learning the simple language of algebra, mathematical models of real-world situations can be created and solved.
In algebra, letters are often used to represent numbers. Algebra also uses the same symbols as arithmetic for adding, subtracting, multiplying and dividing.
A visual of algebra:
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Visualizing Algebra:Example 1) Draw the following algebraic terms:
x 2x 3x 4x -2x 4
Vocabulary Term:
Variable:
Coefficient:
Constant:
Example 2) Draw the following algebraic terms:
2x 2x2 4x 3x2 2x3 4
Vocabulary Like Terms:
Example 3) Terms can be added and/or subtracted together to form polynomials.
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Example 4) We can simplify like terms in polynomials to form simpler but equivalent polynomials.
Important: Polynomials are in simplest form when they contain no like terms.
Vocabulary Polynomial:
Trinomial:
Binomial:
Monomial:
Example 5) Polynomials are in standard form when they are written with the exponents in descending order.
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Practice Problems:
1. State the coefficient and variable(s) in each terma) 15x2 b) -3y c) -w3
Coefficient: Coefficient: Coefficient:Variable: Variable: Variable:
2. Circle the constant term in each polynomial.a) 3x2 + 2x + 4 b) 3y 5 c) 4a2 + 5a
3. Draw out each of the following.
a) 3x b) 4x2 c) 2x3
d) 4x + 2x e) 4x + 2x2 f) 3x + 2x2 + x3 + 6x + x2
g) 4x2 + 6x + 3 h) 3x2 + 4x – 2x2 – 5x i) x – 2x2 + 2x3 + 4x + 7x2
In your own words Like terms are….
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Writing Algebraic Expression
Writing algebraic expressions from word problems is a very important skill in mathematics.
Example 1) Seven times a number plus three
Example 2) Seven times the square of a number plus three
Example 3) Three quarters of a number
Example 4) A gym membership costs $35 upfront, then $10 per month
Example 5) Write each of the following
a) The sum of a number and seven
b) Seven minus a number
c) Seven less than a number
d) A number less seven
e) The difference between a number and seven
f) The difference between seven and a number
g) 2 times a number plus seven
h) 2 times, a number plus seven
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Assignment 2.1: Introduction to Algebra
1. Sketch models to represent each of the following algebraic expressions. The variables x and y do not represent the same number.a) x2 b) x3 c) 2y2 d) 5x3
2. For each term, identify the coefficient and the variable.a) 4x b) –5p4 c) 3m2n d) g3h2
e) –2y5 f) –p4q5 g) ab h) 0.6r4s2
3. Classify each polynomial by type.a) 2x + 1 b) 3p2 – p + 4 c) 4b2d3 d) 6 + gh5
e) 2 – y5 – y2 + 4y f) x2 – y2 + 4 g) ab – b h) 6p3q3
4. We read 42 as “4 squared” and 43 as “4 cubed”. Sketch models that show why this makes sense.
5. Write an algebraic expression for each phrase.a) double a number b) triple a number c) quadruple a numberd) one half of a number e) one third of a number f) one quarter of a number g) 6 more than a number h) a number increased by 3 h) 2 increased by a numberj) 5 decreased by a number k) 7 less than a number l) a number decreased by 6
6. Write an algebraic expression for each phrase.a) 4 more than triple a number b) half a number, less 5c) quadruple a number decreased by 1 d) 2 less than double a number
7. State the problems that are in standard form. If it is not in standard form, re-write in standard form.
a. 23 11xx b. c.
24 2173 xxx d.
8. Given: 12252 23 xxx
a) How many terms are there? b) What is the coefficient of the 3rd term? c) What is the constant?
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2.2 Discovering the Exponent Laws – Part 1
Warm Up:Evaluate:
a)
4
21
b)
2
32
b)
2
73
Identify the base, and write each expression in expanded form.
a) 32
b) 32
c) 42
d) 42
Evaluate:
a) 32 2154231
b)
Write the following as repeated multiplication:
a) 35
b) 4x
c) 4)2( y
d) 42y
Learning Goals: By the end of this lesson, you will be able to:
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Investigation #1: Multiplication of Powers
Complete this statement: When multiplying powers with the same base….
Challenge:
2 3 4 5c d c d c c d d d c c c c d d d d d 6 8c d
3 4 5 2k j k j
4 2 2 3p q p q
5 3 3 4v z v z
2 5 4 3g t g t
4 2 5 3 4 4a b c a b c
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Product Expanded Form Single Power4 3x x x x x x x x x 7x
5 6y y y y y y y y y y y y y 11y
3 2m m 4 5a a 6t t
7 2w w 2 4h h 8 5p p a bx x
Investigation #2: Division of Powers
Complete this statement: When dividing powers with the same base…
Challenge:
7 3 4 2c d c d c c c c c c c d d dc c c c d d
3c d
8 4 5 2k j k j
4 6 2 3p q p q
5 8 3 4v z v z
6 5 4 3g t g t
4 7 6 3 4 4a b c a b c
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Product Expanded Form Single Power
7 3x x x x x x x x xx x x
4x
5 2y y y y y y yy y
3y
6 2m m
4 3a a
6t t
a bx x
Let’s Review:
a1=
a0=
a5⋅a2=
a5
a3 =
x4⋅y3=
x4
y3 =
a5⋅b6
a4⋅b3 =
b5
b5 =
Consolidation:
a) b) b)
Confirm your answers by writing the expressions in standard form. (When in doubt, write it out)
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Assignment 2.2 Discovering the Exponent Laws – Part 1
1. Write each expression as a single power.a) 72 74 b) 35 33 c) 5 52 d) 32 34 33
e) (–2)2 (–2)3 f) (–1)3 (–1)2 (–1) g) 0.53 0.52 h)
31 12 2
2. Evaluate each expression in question 1.
3. Write each expression as a single power.a) 86 84 b) 55 53 c) 77 72 d) 48 45 4
e) (–9)7 (–9)6 f) 0.16 0.14 g)(–0.3)4 (–0.3) h)
5 32 23 3
4. Evaluate each expression in question 3.
5. Simplify.
a) b5 b3 b) p4 p c) w5 w2 d) x8 x4
e) a4b5 ab3 f) m2n4 m3n3 g) p6q5 p3q2 h) xy2 y
6. Simplify each of the following, write as a single power if possible.
a) b) c)
d) e) f)
g) h) i)
j) k) L)
7. Explain why it is necessary for the bases to be the same in order to apply the multiplication and division principles for exponents, use examples in your explanations.
Enrichment: Write 125 625 5 as a single power
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2.3 Discovering the Exponent Laws – Part 2
Warm Up:Evaluate:
27
26
25
24
23
21
20
Use the patterns to explain what 20 might mean
Evaluate by writing the expanding form
Simplify using the exponent principles.
Use the exponent principals to explain what 30 might mean
Evaluate:(convert all decimals to fraction form)
23
22
21
20
2-1
2-2
2-3
Use the patterns to explain what a negative exponent might mean.
Evaluate by writing the expanding form
Simplify using the exponent principles.
Use the exponent principals to explain what a negative exponent might mean.
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Investigation 3: Power of a Power
Product Expanded Form Single Power
24x x x x x x x x x 8x
43y y y y y y y y y y y y y 12y
23m
52g
34t
45w
26k
33r
bax
Do these using your shortcut, without the middle step
53m
68q
410x
87n
53 4a b
(x2 y4 z3 )6
( a2
b3 )4
Complete this statement: When simplifying powers of powers….
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Let’s Review:
There are still two more:
Example: (xy)3 = Example
3
ba
=
Power of a ProductThe power of a product is the __________________________________
i.e.
Practice Problems: Simplifya) (x3y2)2 b) (5xy)3 c) (2x2y4)3
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Power of a QuotientThe power of the quotient is the _________________________________
i.e.
Practice Problems:
a)
b) c)
Remember that exponent principles can be combined as long as the ____________ are the ____________.
Practice Problems: Simplify
a) b) c)
Extension: Determine the value that makes each statement true
a) b) c)
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Assignment 2.3 Discovering the Exponent Laws – Part 2
1. Simplify each expression using exponent laws.
a) 2935 444 b) 245 c) 753
d) 547 e)6410 777 f) 7532 333
2. Simplify each expression using exponent laws.
a) 52x b) 245 c) 753
d) 547 e) 37a f) 553 x
g) 322x h) 243y i)
j) 2532 xxx k) 553ts l) 652 yx
m)
4
3
2
ba
n)
2
56
812
xyyx
o)
3
43
75
baba
3. Simplify each expression using exponent laws.
a) (m5)2 b) (k2)3 k2 c) g5 g5 g7 d) (a6)3 (a5)2
e) (gh4)3 f) 2k2m3 (2k2)2 g) (2g5h3)2 2gh6 h*)
2 2 2
2
6 33
b d b dbd
4. Show that 310 is the same as 95 using your understanding of exponent.
5. Determine the value of the exponent that makes each expression true.
a) 274 = 3? b) (-125)7 = (-5)? c) 6? = 2162
6. Evaluate for a = 5 and b = 3. Are the expressions equal? If not, which expression has the lesservalue?a) a2 + b3 or a3 + b2 b) a2b2 or (ab)2
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2.4 Multiplying and Dividing Monomials
Warm Up:
Explain why 08 and 08 have different values.
Is each statement true or false? Explain.
a) 33 = 81
b) 6(–2)3 = 48
c) y2 × y4 = y6
d) ((–2)3)3 = –512
e) (–a)4 ÷ (–a)2 = a2
f) (–5)3 ÷ (–5)2 = 5
After marking a quiz, a teacher recorded the most common errors made by the students. In each case, identify the error made by the student and provide the correct simplification.
a) 1243 222 b) 523 44
c) 954 933 d) 3632 55 baba
Summarize each of the exponent principles that you have learned in the past lessons.
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Multiplying Monomials Recall: You can multiply powers with the SAME BASE by _____________ their exponents! But what happens when we multiply something like this….
2x2y • 3xy2 “When in doubt, write it out”
How-to Multiply Monomials
(6x2y6z)(2xy4)
Practice Problems: Multiply the following monomials.
a) xx 32 b) c) yxz 236
d) 42 27 mnmn e) 72 15
32 xx
f) nmnp 82
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How-to Divide Monomials
Practice Problems: Divide the following monomials.
a) xx
510 2
b) mnnm
216 2
c) 2
53
16
48
xy
yx
d) pmp 515 e)
xyzyzx
51
54 24
f) 22 4xx
Challenge Questions:
a)
2
4
35
86
yyx
b)
42
252
396
yx
yxyx
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Assignment 2.4 Multiplying and Dividing Monomials
1. Multiplying Monomials: Simplify each expression using exponent laws.
a) bb 42 b) b63 c) 532 bb
d) xx 53 e) 522 32 bb f) 234 xx
g) 52 32 bb h) 522 635 yxyx i) 35 2 x
j) 62 47 aa k) xxx 235 l) xxx 432
m) 232 32 abba n) 253 yxy o) 34 523 aaa
p) nmnm 647 543 q) yyxx 832 233 r) 422433 38 yxyxyx
2. Dividing Monomials: Simplify each expression using exponent laws. Fractions must be in lowest terms – no decimals!
a) 39 2x
b) yy
428 5
c)5
6
1812nn
d)
7
545zz
e)4
6
927tt
f)5
7
3624xx
g)127
83
205
yxyx
h)810
97
4914
baba
i)
1142
333
81
9
yx
yx
j) k)
43
25
557nmnm
l)
343
32
5
55
yx
yx
3. Write an expression to represent the area of the following shapes.
a) xy3 b)
22x m12
22
4. Determine the volume of the cube shown.
5. Determine the volume of the following shapes.
6. Write an expression to represent the volume of the following shape.
7. The volume of a box is 64a5b4. The length of the box is 4a3 and the width of the box is 8b4. Determine the height of the box.
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3 102x x
2. 5 Collecting Like Terms
Warm Up:Draw a model to represent x, x2, and x3 Define the following words
Term
Variable
Coefficient
Exponent
Illustrate and explain the difference between
a) 2x and x2.
b) 3x2 and 2x3
Classify the expression as a monomial, binomial, trinomial or neither, then list the coefficients and state the constant term.
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Representing Polynomials
Algebra Tiles are often used to represent polynomials. Algebra tiles can help you visualize equivalent algebraic expressions and/or equations. Many of the tiles have a positive and a negative side (normally represented with different colours).
Represent the following polynomials with algebra tiles
a) 3x + 1 b) 4x2 – 3x c) –2x2 – 2x + 4 d) –x2 – 5
Like and Unlike Terms
Like terms have the same set of variable bases and corresponding exponents.
Examples: Represent each of the following sets with algebra tiles
a) 4x2 and -5x2 b) -6x and 9x c) 3xy2 and 5xy2
Did you notice how algebra tiles with the same size and shape represent like terms? And even though some of the coefficients are negative, as long as they still have the same variable base and exponent combination as the other terms, we consider them like terms.
Main Idea: Like terms can have ____________________ coefficients as long as the variable base and the exponent are the ____________________.
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Practice Problem: Define and give an example of Unlike Terms
Collecting Like TermsAlgebraic expressions that contain like terms can be simplified by combining each group of like terms into a single term.
Examples:3x + 4x 9x2 – 6x2 12x3y2 - 5x3y2
Why can’t you simplify?4x2 + 4x x2 – 7 6x3y + 5xy3
Practice Problems: Simplify
1) 7x + 5 – 3x 2) 6w2 + 11w + 8w2 – 15w 3) 6x + 4 – 5 – (-7x)
4) (-12x) – 5 – 7x – 11 5) 2x2 – (-3x) + 7 – (-3x2) + 4x – 7 6) 11a2b – 12ab2
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Assignment 2. 5: Collecting Like Terms
1. Are the terms in each pair like or unlike?
a) 5a and –2a b) 3x2 and x3 c) 2p3 and –p3 d) 4ab and ab
e) –3b4 and –4b3 f) 6a2b and 3a2b g) 9pq3 and –p3q h) 2x2y and 3x2y2
2. Simplify.
a) 4 + v + 5v – 10 b) 7a – 2b – a – 3b c) 8k + 1 + 3k – 5k + 4 + k
d) 2x2 – 4x + 8x2 + 5x e) 12 – 4m2 – 8 – m2 + 2m2 f) –6y + 4y + 10 – 2y – 6 – y
3. Simplify.
a) 2a + 6b – 2 + b – 4 + a b) 4x + 3xy + y + 5x – 2xy – 3y
c) m4 – m2 + 1 + 3 – 2m2 + m4 d) x2 + 3xy + 2y2 – x2 + 2xy – y2
4. Simplify:
a) b)
5. Find the perimeter of each figure below.a) b)
4 y−6
2 y+43n+2
n
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2.6 Adding and Subtracting Polynomials
Warm Up:Define the following terms
a) Opposite
b) Reciprocal
Give an example of …
A monomial
A binomial
A trinomial
A constant term
Evaluate the following
a) 10 (4)
b) 2 ( 3)
c) 11 (5 2)
d) 10 ( 1 3 8)
Your friend forgets how to properly subtract integers. Give an explanation and two examples of how to subtract integers.
Determine a simplified expression the perimeter of the following rectangle.
If d = 7, what is the perimeter of the rectangle on the left?
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Adding Polynomials
In order to distinguish one polynomial from another polynomial in an algebraic expression, the polynomials are often placed in separate pairs of brackets.
Consider the algebraic expression 2 2(3 5 1) (4 2 )x x x x .
Practice Problems: Simplify.
a. (4 7) ( 3 2)y y b.2 2( 2 ) (6 10 ) (5 1)a a a a a
Problem Solving: The measures of two sides of a triangle are given. P is the perimeter. Find the measure of the third side.
P = 4x2 + 5x + 5 x2 + 3x – 5
2x2 + 3x + 6
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Subtracting Polynomials
Suppose you were asked to evaluate the following expression: 4 ( 2) . Explain the procedure for subtracting integers.
This same idea can be used to subtract polynomials.
Consider the following algebraic expression: 2 2(7 2 13) (4 5 6)b b b b .
Practice Problems: Simplify.
a. (3 2) (4 9)x x b. 2(5 12 ) ( 4 8)z z z
c. 212 (4 5 1)x x d.
2 2(8 7 ) (4 3) ( 5 9)x x x x
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Assignment 2.6 Adding and Subtracting Polynomials
1. Add.a) (y2 + 6y – 5) + (–7y2 + 2y – 2) b) (–2n + 2n2 + 2) + (–1 – 7n2 + n)c) (3m2 + m) + (–10m2 – m – 2) d) (–3d2 + 2) + (–2 – 7d2 + d)
2. For each shape below, write the perimeter as a sum of polynomials and in simplest form.i) ii)
iii) iv)
3. Subtract. a) (2x + 3) – (5x + 4) b) (4 – 8w) – (7w + 1)c) (x2 + 2x – 4) – (4x2 + 2x – 2) d) (–9z2 – z – 2) – (3z2 – z – 3)
4. A student subtracted (3y2 + 5y + 2) – (4y2 + 3y + 2) like this:= 3y2 – 5y – 2 – 4y2 – 3y – 2= 3y2 – 4y2 – 5y – 3y – 2 – 2= –y2 – 8y – 4a) Explain why the student’s solution is incorrect.b) What is the correct answer? Show your work.
5. The difference between two polynomials is (5x + 3). One of the two polynomials is (4x + 1 – 3x2). What is the other polynomial? Explain how you found your answer.
6. Subtract.a) (mn – 5m – 7) – (–6n + 2m + 1) b) (2a + 3b – 3a2 + b2) – (–a2 + 8b2 + 3a – b)c) (xy – x – 5y + 4y2) – (6y2 + 9y – xy)
7. The sum of the perimeters of two shapes is represented by 13x + 4y. The perimeter of one shape is represented by 4x – 2y. Determine an expression for the perimeter of the other shape. Show your work.
8. A rectangular field has a perimeter of 10a – 6 meters. The width is 2a meters. Determine an expression for the length of this field.
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2.7 Multiplying a Polynomial by a Monomial
Warm Up:Simplify
a) 2(3b)
b) –2(6h)
c) 4(2b2)
d) –2(2x2)
e) –2(–y2)
f) –3(–2f)
Simplify.a) (6k – 4) + (2k + 4)
b) (2a + 1) – (4a + 2)
c) (b – 6) – (2 – 5b) + (b + 4)
d) (2m2 + m + 12) – (3m2 + 4m – 6)
Simplify
a) 5 34 2a a
b)
c)
d) 5 2(9 )y
e)
3 2
2
8 42 2b d bd
bd
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The Distributive Property
Minds On: A rectangle has an unknown length and a width of 4 units. If the length is increased by 7 units to create a larger rectangle, write a simplified algebraic expression for the area of the new, larger rectangle.
4
xoriginal
7
7
4
x
This property is known as the distributive property. This is also known as expanding.
Distributive Property: ( )a b c ab ac
--------------------------------------------------------------------------------------------------------------------Practice Problems: Expand
a) 3(g + 4) b) –7(q + 3) c) –(2t – 1) d) –4(–w – 5)
What is the width of the new rectangle?
What is the length of the new rectangle?
What is the area of the new rectangle?
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Example 1) Simplify: (2 5)x x
Practice Problems: Expand.a) b(b + 1) b) 3p(p + 4) c) –r(–5r + 2) d) –3w(2w – 1)
----------------------------------------------------------------------------------------------------------------------------------------
Example 2) Simplify: 23 (9 4 )x x x .
Practice Problems: Simplify the expressions.
a) 6 33 ( 4 2 )y y y b)
25 (3 1)z z
c) 4 22 ( 3 9)x x x d) ( 1) 11a a
----------------------------------------------------------------------------------------------------------------------------------------Example 3) Simplify 2(b2 – 4b + 3) + 5b(b + 4)
Practice Problems: Expand and simplify.a) -3(p – 2) + 6(p + 1) b) 3[–2(6 – t) + 5t] c)–5m(m + 5) – 2(3m2 – 4m – 7)
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Assignment 2.7 Multiplying a Polynomial by a monomial
1. Determine each product.a) 4(3a + 2) b) (d2 + 2d)(–3) c) 2(4c2 – 2c + 3)d) (–2n2 + n – 1)(6) e) –3(–5m2 + 6m + 7)
2. Here is a student’s solution for a multiplication question.(–5k2 – k – 3)(–2)= –2(5k2) – 2(k) –2(3)= –10k2 – 2k – 6a) Explain why the student’s solution is incorrect.b) What is the correct answer? Show your work.
3. Expand.a) 3(x2 + x – 4) b) 2(m2 – 3m + 5) c) –4(b2 – 2b – 3) d) 5c(c2 – 6c – 1)e) –3h(4 – h2) f) (n2 + 4n + 3)(–2) g) (5t2 – 2t)(–t) h) (w2 + 2w – 5)(4w)
4. Write a simplified expression for the area of the following rectanglesa) b)
5. Expanda) 4x2(3x + 2) b) 2n(2n – 3) c) 5m(4m2 + 3m) d) (y4 – y2)y3 e) pq(3p + 2q) f) 3d(2d2 – 4d + 1) g) (t2 + t – 4)(-2t) h) 3a2b(3a2 – b2)
6. Simplifya) 2(x + 4) – 4(2x + 3) b) 3a(2a + 4b – 3) – 2b(3a + 2ab) c) 2p(p – 4) + 6(p2 + 4p – 3) d) 4x(4x – 4y – 4) + 2y(6x + 3y)
Enrichment: Do some research to learn how to multiply a binomial by a binomial (or a polynomial)a) (x + 1)(x – 3) b) (3x – 5)(x – 2) c) (4x – 3y)(5x + y)
35
2.8 Simplifying Algebraic Expressions
Simplify 1) 3p – 4q + 2p + 3 + 5q – 21 2) 5x2y + 12xy2 – 8x2 – 12xy2
3) 5 – 2y2 + 3y – 8y – 9y2 – 12 4) 2x2 + 3x – 7x – (-5x2)
5) 5(a – 3b) + 2(-b – 4a) 6) 3(x + y) – 5(-2y + 3x)
7) -3b(5a – 3b) + 4(-3ab – 5b2) 8) 3x(x - 2y) – 4(-3x2 - 2xy)
9) -3(x2 + 3y) + 5(-6y – x2) 10) -3(7xy - 11y2) – 2y(-2x + 3y)
11) 4(2 – x) - 3(-5 – 12x) 12) 7(3 - x) – 6(8 - 13x)
13) 14)
15) 17)
18) 19) 2(x – 2y) – [3 – 2(x – y)]
20) 21)
22) 23)
24) 25) 2 x ( x−3 )+4 (x2+5 )
26) A triangle has sides of length 2a centimeters, 7b centimeters, and 5a + 3 centimeters. What is the perimeter of the triangle?
27) A rectangle has sides of length 7x – 2 meters and 3x + 4 meters. What is the perimeter of the rectangle?
28) A square has a side of length 9x – 2 inches. Each side is shortened by 3 inches. What is the perimeter of the new smaller square?
29) A triangle has sides of length 4a – 5 feet, 3a + 8 feet, and 9a + 2 feet. Each side is doubled in length. What is the perimeter of the new enlarged triangle?
30) Subtract four times the quantity 2x + 5 from three times the quantity 7x – 8.
36
37
Unit #2 Answers
Assignment 2.11. Answers will be taken up in class 2. a) coefficient: 4; variable: x b) coefficient: 5; variable: p4 c) coefficient: 3; variable: m2n
d) coefficient: 1; variable: g3h2 e) coefficient: 2; variable: y5 f ) coefficient: 1; variable: p4q5
g) coefficient:
34 ; variable: ab h) coefficient: 0.6; variable: r4s2
3. a) binomial b) trinomial c) monomial d) binomiale) four-term polynomial f) trinomial g) binomial h) monomial
5. a) 2x b) 3x c) 4x d)12x
e)
13x
f )
14x
g) x + 6 h) x + 3i) 2 + x j) 5 x k) x 7 l ) x 6
6. a) 3x + 4 b)
1 52x
c) 4x 1 d) 2x 2
7. a) In standard form b) c) c) 8. a) 4 terms b) -2 c) 12
Assignment 2.2
1. a) 76 b) 38 c) 53 d) 39 e) (2)5 f ) (1)6 g) 0.55 h) 4
12
2. a) 117 649 b) 6561 c) 125 d) 19 683 e) 32 f) 1 g) 0.031 25 h)
116
3. a) 82 b) 52 c) 75 d) 42 e) (9)1 f ) 0.12 g) (0.3)3 h) 2
23
4. a) 64 b) 25 c) 16 807 d) 16 e) 9 f) 0.01 g) 0.027 h)
49
5. a) b8 b) p5 c) w3 d) x4
6. e) a5b8 f) m5n7 g) p3q3 h) xy7. a) 1047 b) a19 c)c 11 d) d16 e) x10 f) w1133 g) a11 h) 10a + b
i) g42 j) 104 k) 43 l) 93
8. 76 and 74
9. answers may vary.
Assignment 2.3
1. a) 224 b)
85 c) 353 d)
207 e) 127 f)
43
2. a) 10x b)
85 c) 353 d)
207 e) 21a f)
253x g) 68x h)
89y i) i) 16y8x8 j) 10 x
k) 2515ts l)
3012 yx m) 12
8
ba
n) 414 yx o)
96ba3. a )m10 b) k8c) g3 d) a8 e) g3h12 f ) 8k6m3 g) 2g9 h) 2b2d4. 95 = (32) 5 = 310
5. a) 12 b) 21 c) 66. a) 52 and 134, a2 + b3 has the lesser value b) 225 and 225, they are equal
Assignment 2.4
38
1. a) 28b b) b18 c)
66b d) 215x e)
912b f) 312x g)
76b
h) 7390 yx i)
215x j) 28a8 k) 330x l)
412x m) 536 ba n)
315xy
o) 830a p) -168m8n5 q)
3648 yx r) 9924 yx
2. a) 23x b)
47y c) n
32
d) 69z e)
23t f)
2
32 x
g)
44
41 yx
h) ba 3
72
i)
211
91 yx
j) 6x2y k) 227 nm l)x -7 y -3
3. a) 6x3y b) 30m2 4. 8y3 5.a) 24xy2 b) 24x2y c) 512xy2 d) 324x2y 6. 6x3 y5
7. 2a2
Assignment 2.51. a) like b) unlike c) like d) like e) unlike f) like g) unlike h) unlike2. a) 6v 6 b) 6a 5b c) 7k + 5 d) 10x2 + x e) 4 3m2 f ) 5y + 43. a) 3a + 7b 6 b) 9x + xy 2y c) 2m4 3m2 + 4 d) 5xy + y2
4. a) -w2 b) 5. a) 8n + 4 b) 8y + 2
Assignment 2.61. a) –6y2 + 8y – 7 b) –n – 5n2 + 1 c) –7m2 – 2 d) –10d2 + d 2. i) (2n + 2) + (n + 1) + (2n + 2) + (n + 1) = 6n + 6 ii) (3p + 4) + (3p + 4) + (3p + 4) = 9p + 12
iii) (4y + 1) + (4y + 1) + (4y + 1) + (4y + 1) = 16y + 4 iv) (a + 8) + (a + 3) + (12) = 2a + 233. a) –3x – 1 b)3 – 15w c) –3x2 – 2 d) –12z2 + 14. a) The student is incorrect because he changed the signs in the first polynomial.
b) (3y2 + 5y + 2) – (4y2 + 3y + 2) = 3y2 + 5y + 2 – 4y2 – 3y – 2 = 3y2 – 4y2 + 5y – 3y + 2 – 2 = –y2 + 2y 5. (4x + 1 – 3x2) – (5x + 3) = –3x2 – x – 2, or (5x + 3) + (4x + 1 – 3x2) = –3x2 + 9x + 46. a) mn – 7m – 8 + 6n b) –a + 4b – 2a2 – 7b2 c) 2xy – x – 14y – 2y2
7. 9x + 6y : (4x – 2y) + (9x + 6y) = 13x + 4y *answers may vary8. length = 3a-3 : (2a) + (2a) + (3a – 3) + (3a – 3) = 10a -6
Assignment 2.71. a) 12a + 8 b) –3d2 – 6d c) 8c2 – 4c + 6 d) –12n2 + 6n – 6 e) 15m2 – 18m – 212. a) The negative signs were omitted on the first polynomial when (–2) was distributed;
(–5k2)(–2) + (–k)(–2) + (–3)(–2) = 10k2 + 2k + 63. a) 3x2 + 3x 12 b) 2m2 6m + 10 c) 4b2 + 8b + 12 d) 5c3 30c2 5c
e) 12h + 3h3 f ) 2n2 8n 6 g) 5t3 + 2t2 h) 4w3 + 8w2 20w4. a) 2d(3d + 4) = 6d2 + 8d
b) y(4y + 6) = 4y2 + 6y5. a) 12x3 + 8x2 b) 4n2 – 6n c) 20m3 + 15m2 d) y7 – y5
e) 3p2q + 2pq2 f) 6d3 – 12d2 + 3d g) -2t3 – 2t2 + 8t h) 9a4b – 3a2b3 6. a) -6x – 4 b) 6a2 + 6ab – 9a – 4ab2 c) 8p2 + 16p – 18 d) 16x2 – 4xy – 16x + 6y2
Assignment 2.8 1. a) 4(x + 5) b) 5x(1 + 6x) c) 12x(x 4) d) 7x(3x 7)
e) 3(6x + 11) f) 10x(2 5x) g) 3x(16x + 21) h) 36x2(x + 2) 2. a) 4(x2 + 3x + 2) b) 3(x2 + 2x 3) c) 5x(x2 + 2x 24) d) 3x2(x2 12x + 35)
3. a)2 22(9 25 )x y b)
5 425 (4 2 3)z z z c) 236 (1 3 )rs rs d)
7 3( )a b a
e)
3 2 4(2 3 4)c c d c f) 73 ( 1)g g g)
4 3 22 (9 24 28 43)x x x x h)2( 1)c c c
4. 3x by 7x + 25. a) ab2c (ab – c + ac) b) (x+y)(3x + 2y) c) (2x – 3)(5x – 1)
39
40
Possible Test Questions1. If x = 3, what is the value of 2x2 + 5x?
a 21 b 27 c 33 d 51
2. Simplify the following expression: 3x(2x + 3) – 5xa 6x 2 – 5x + 3 b 6x 2 – 6x c 15x 2 – 5x d 6x 2 + 4x
3. Simplify the following algebraic expression:
a6 b4
a2b
aa3
b3b a4
b3c a
3b3d a
4b3
4. Tim shows the steps he took in simplifying the following algebraic expression: In which step did Tim make an error?
a Step 1
b Step 2
c Step 3
d Step 4
5. Sabeeta expands and simplifies the expression below. 2(3x 2 – 5x) + 4x(7 + x) Which expression is equivalent to the one above?
a 6x 2 + 22x b 10x 2 + 18x c 10x 2 – 38x d 28x 2
1) 4x + (–1) – (+5x) – (–6) 2) (–5a + 6) + (8a – 7) 3) (b – 4) – (3b – 8) 4) 2y(3y – 5)
5) 3(x + 6) – 4(x + 3) 6)
12
(4 x−3 )+ 23
(9 x+1 )
7) 2m(3m – 5) – 4(m2 – 2m – 1) 8) –3[2(p + 2) – 3p]
9) 10)
(a2 )3
a2a3
=
a5
a2a3
Step 1
=
a5
a2+3
Step 2
=
a5
a5
Step 3
41
