Download Algebra Tiles

Algebra Tiles
Activity 1: Adding Integers
We are going to use positive (yellow) and negative (red) tiles to discover the rules for
adding and subtracting integers. Each tile has a value of 1 or –1.
+
3=
+
+
–
–2 =
–
A.
Try the following additions with the tiles. Draw the tiles and find the integer answer.
1.
3+1
2.
3+5
3.
2+4
4.
In every example above, what type of numbers were you adding?
5.
What operation did you perform with the number parts?
6.
What happened with the sign?
7.
What rule can you develop for adding positive integers?
B.
Try the following additions with the tiles. Draw the tiles and find the integer answer.
Model
Model
1.
–2 + –1
2.
–4 + –3
3.
–2 + –5
4.
In every example above, what type of numbers were you adding?
5.
What operation did you perform with the number parts?
6.
What happened with the sign?
7.
What rule can you develop for adding negative integers?
Answer
Answer
The Zero Pair
When two tiles of unlike signs are combined, they form a zero pair (+1 + –1 = 0)
+
-
=0
When you combine positive and negative tiles, find the zero pairs and remove them. The
tiles that remain show the answer.
C.
Try the following additions with the tiles. Draw the tiles, remove the zero pairs (cross them
out) and find the integer answer.
Model
1.
–3 + 1
2.
–4 + 5
3.
–2 + 6
4.
7 + –4
5.
2 + –5
6.
In every example above, what type of numbers were you adding?
7.
What operation did you perform with the number parts?
8.
What happened with the sign?
9.
What rule can you develop for adding positive and negative integers?
Answer
Conclusion:
When adding integers with the same sign, you _______ the numbers and ________ the sign.
When adding integers with different signs, you ___________ the numbers and use the sign of the
__________ absolute value.
Algebra Tiles
Activity 2: Subtracting Integers
We are going to use positive (yellow) and negative (red) tiles to discover the rules for
adding and subtracting integers. Each tile has a value of 1 or –1.
When you subtract with tiles, you remove the tiles that are being subtracted. What remains
is your answer.
+
5–3=
A.
+
+
+
+
=2
Try the following subtractions with the tiles. Draw the tiles and find the integer answer.
Model
1.
3–1
2.
–5 – (–2)
3.
6–4
4.
Using tiles, can you do the following?
Answer
4–7
At first it appears not, but you can subtract more than you have by introducing zero pairs until
you have enough.
Step 1: Start with 4 tiles.
+
+
+
Step 2: Introduce 3 zero pairs
+
=
+
+
+
+
+
+
+
–
–
–
Step 3: Remove the 7 positive tiles. This leaves 3 negative tiles, which is your answer.
+
+
+
+
+
+
+
–
–
–
B.
Try the following subtractions with the tiles. Draw the tiles and find the integer answer.
Model
1.
–2 – (–4)
2.
–4 – (–6)
3.
2–5
4.
–2 – 4
5.
5 – (–2)
6.
What did you notice about subtracting integers?
7.
What rule can you develop for subtracting integers?
Answer
The rule for subtracting integers is often stated as “ to subtract integers you add the
opposite.”
C.
Try the following subtractions with the tiles by adding the opposite. Draw the tiles and find
the integer answer.
Model
1.
–3 – 4
2.
4 – (–4)
3.
–2 – 6
4.
–3 – (–4)
5.
2–5
Answer
Write a paragraph explaining to a friend how to subtract integers.
EXPLORING THE DISTRIBUTIVE PROPERTY
–x2 =
x2 =
blue
x=
red
green
–x =
red
+1 =
–1 =
yellow
red
The Distributive Property combines multiplication with addition or subtraction.
Example: 2(x + 3)
2(x + 3) = 2x + 6
Model each of the following problems to simplify the variable expression.
1.
2(x + 4)
2.
3(x + 2)
3.
2(2x + 1)
4.
3(x – 1)
5.
4(x – 2)
6.
2(3x + 1)
7.
2(-x + 1)
8.
3(x – 4)
9.
5(-2x + 3)
10.
In your own words explain how the distributive property works or write an algebraic
expression to represent the concept.
EXPLORING SIMPLIFYING VARIABLE EXPRESSIONS
–x2 =
x2 =
blue
–x =
x=
red
green
+1 =
red
yellow
–1 =
red
You simplify variable expressions by combining like terms. Use zero pairs, where needed,
and perform as many operations as possible within the expression.
Example. 3a + (–a) =
+
=
= 2a
0
2y + 7 + y – 3 =
+
+
= 3y + 4
–3
Use models to simplify these expressions.
1.
4x + 7x =
2.
5b – 2b =
3.
3a + (-2a) =
4.
5y – (–2y) =
5.
3c2 – 8 + 2c2 =
6.
8 – 5x + 3x2 – 2 + 3x =
Remember 2(x + 3) means
+
= 2x + 6
Use models to simplify these expressions.
7.
2(y2 – 4) =
8.
3(a + 3) =
9.
2(z – 4) + 9 =
10.
2(2b2 – 1) + 7 =
11.
10c2 + 4(c2 + 3c) – 5c =
12.
8y + 2(3 – y) – 4 =
EXPLORING SOLVING EQUATIONS
x=
green
–x =
–1 =
+1 =
red
yellow
red
IMPORTANT: Equations are like a scale in balance: to keep it balanced you
must always do the operations on both sides of the equal sign at the same time.
Ex. x + 5 = 8
=
Remove 5 yellow squares from each side. Your answer x = 3.
=
Now you try a few using your algebra tiles.
1. x + 4 = 11
2.
x + 3 = 10
3.
x+6=9
4.
x+5=9
5.
2x + 2 = 12 + x
6.
4 + 3x = 5 + 2x
Sometimes you need to group the tiles to find the value of the variable.
Think about these examples. Use your tiles and group them to find the values.
7. 2x = 8
8. 3x = 12
9.
4x = 8
10. 5x = 20
Combine the two methods to solve these equations.
11. 3x + 4 = 10
12. 4x + 5 = 2x + 11
13. 8 + 5x = 3x + 12
14. 7x + 1 = 4x + 10
EXPLORING S0LVING EQUATIONS 2
x=
green
–x =
–1 =
+1 =
red
yellow
red
IMPORTANT: Equations are like a scale in balance: to keep it balanced you
must always do the operations on both sides of the equal sign at the same time.
REMEMBER: When subtracting integers you add the opposite.
Ex. x – 3 = 6
becomes
x + –3 = 6
=
You must add positive 3 to both sides then remove the zero pairs. Your answer
x = 9.
=
Now you try a few using your algebra tiles. Write the steps you used below the
problem.
1.
x – 4 = 10
2.
x–3=7
3.
x–6 = 4
4.
x–5 =6
5.
2x – 3 = 1
6.
3x – 2 = 10
7.
4x – 5 = 7
8.
2x – 8 = 14
EXPLORING ADDING AND SUBTRACTING POLYNOMIALS
–x2 =
x2 =
blue
x=
red
green
–x =
–1 =
+1 =
red
yellow
red
Add and subtract by combining like terms and by using “adding zero” where needed.
Ex. (2x2 + x + 3) + (x2 – 3x + 1)
+
Form a zero pair with +x and –x.
= 3x2 – 2x + 4
0
Try these:
1.
(2x2 + 3x + 3) + (x2 + 2x – 2)
2.
(3x2 + 4x – 3) + (4x2 – 2x – 4)
3.
(x2 + 3x – 4) – (2x2 + 2x – 3)
4.
(5x2 – 2x + 1) – (x2 +2x – 4)
5.
(2x2 + x + 3) – (x2 + 2x – 2)
6.
(-3x2 + 4x + 3) + (3x2 – 3x – 5)
EXPLORING MULTIPLYING POLYNOMIALS
–x2 =
x2 =
blue
red
ex. 1. x(x + 1)
x
–x =
x=
green
red
x
Try these examples:
2.
x(x + 2)
3.
–2x(x + 2)
4.
x(2x – 1)
5.
3x(2x + 1)
6.
2x(–x + 4)
x
x
x
= –2x2 + 6x
= x2 + x
–x2
2(3x + 1)
red
+ 1
x2
1.
yellow
2. 2x(–x + 3)
–x2
x
–1 =
+1 =
x
x
x
EXPLORING MULTIPLYING BINOMIALS
–x2 =
x2 =
blue
ex. 1.
–x =
x=
red
green
red
(x + 1)(x + 1)
x
–1 =
+1 =
yellow
red
2. (x + 1)(–x + 3)
+ 1
x
x
x2
x
+
1
x
1
1.
(x + 2)(x + 1)
2.
(x + 2)(x + 2)
3.
(–2x – 1)(x + 2)
4.
(x –3)(x - 2)
5.
(-x + 3)(2x + 1)
6.
(x – 4)(x + 4)
x
x
x
= –x2 + 2x + 3
= x2 + 2x + 1
1
Try these examples:
–x2
–x
1
1
1
EXPLORING DIVIDING POLYNOMIALS
In order to divide polynomials, put your divisor on the left side and arrange your algebra tiles
into a rectangular array so that the left edge matches your divisor. You may need to add zero
pairs to complete your model.
Ex. Divide x2 + 5x + 6 by x + 2
What is the
factor
across the
top?
Rearrange
x2 + 5x + 6 divided by x + 2 = x + 3, or (x + 2)(x + 3) = x2 + 5x + 6
Divide x2 – 3x + 2 by x – 1
Rearrange
x2 – 3x + 2 divided by x – 1 = x – 2, or
(x – 1)(x – 2) = x2 – 3x + 2
Divide x2 – 2x – 3 by x + 1
Rearrange
Notice I had to
add a zero
-pair (x & –x)
in order to match the +1 in the divisor.
x2 – 2x – 3 divided by x + 1 = x – 3, or (x + 1)(x – 3) = x2 – 2x – 3
Divide each problem.
1.
x2 – 5x + 6 divided by x – 2
2.
x2 – 4 divided by x + 2
3.
x2 – 6x + 8 divided by x – 2
4.
a2 – 4a + 4 divided by x – 2
5.
x2 + 8x + 15 divided by x + 3
6.
2y2 – 5y – 3 divided by 2x + 1
7.
6m2 + m – 1 divided by 2x + 1
8.
x2 – 7x + 12 divided by x – 3
EXPLORING FACTORING POLYNOMIALS
In order to factor polynomials, put your algebra tiles into a rectangular array. You may need to
add zero pairs to complete your model.
Ex. Factor x2 + 4x + 3
Rearrange
What should be on the
left hand side?
above?
x2 + 4x + 3 = (x + 1)(x + 3)
Factor 2x2 + 3x
Rearrange
2x2 + 3x = x(2x + 3)
Factor –x2 + 3x + 4
Rearrange
–x2 + 3x + 4 = (x + 1)(–x + 4)
Notice I had to
add a zero
-pair (x & –x).
Factor each problem.
1. –3x2 + 2x
2. 5x2 –15x + 10
3. x2 + 6x + 8
4. a2 + 4a + 4
5. x2 – 6x + 9
6. y2 – 2y – 15
7. m2 + m – 12
8. 6x2 + 7x + 2
EXPLORING SQUARE TRINOMIALS
In order to factor square trinomials, put your algebra tiles into a square array.
Ex. Factor x2 + 4x + 4
Rearrange
x2 + 4x + 4 = (x + 2)(x + 2)
Factor 4x2 – 4x + 1
Rearrange
4x2 – 4x + 1 = (2x – 1)(2x – 1)
Factor p2 – 2p + 1
Rearrange
p2 – 2p + 1 = (p – 1)(p – 1)
What are the factors
on the left hand side?
above? (They should
be the same.)
Determine if any of the following are square trinomials. Explain why or why not. Can you tell
without using the algebra tiles?
1.
x2 + 6x + 9
2.
x2 + 6x + 36
3.
p2 – 8p + 16
4.
a2 + 4a + 64
5.
q2 – 10q – 25
6.
9y2 + 12y + 4
7.
16m2 + 40m + 25
8.
4x2 – 6x + 9
COMPLETING THE SQUARE
Completing the square enables you to solve equations that are not easily factorable. It will lead
to the quadratic formula for solving quadratic equations.
Method 1: In the equation ax2 + bx +c = 0, if a is a square and b is evenly divisible by 2 a ,
then arrange ax2 + bx into the beginning of a square. Add as many zero pairs as needed to make
a square. Combine the extra +1's and –1's together.

Ex. Solve x2 + 4x + 3 = 0
Rearrange
x2 + 4x + 3 = (x + 2)2 – 1 = 0, or
(x + 2)2 = 1,
x + 2 = ±1,
x = –1, –3
Solve 4x2 – 4x – 3 = 0
Rearrange
4x2 – 4x – 3 = (2x – 1)2 – 4 = 0, or
(2x – 1)2 = 4, 2x – 1 = ±2,

x=
3 1
,
2 2
Method 2: In the equation ax2 + bx +c = 0, if a is not a square, then divide the equation by a.
b
b
Divide the in half and arrange the x2  x into the beginning of a square. Add as many zero
a
a
pairs as needed to make a square. Combine the extra +1's and –1's together.
2
Solve 3p2 – 6p – 2 = 0  p2 – 2p – = 0
3


–

Rearrange
–
2
3

3p2 – 6p – 2 = 0  p2 – 2p –
2
= 0,
3
(p – 1)2 –

5
3
Solve: 2x2 – 5x = 3  x 2  x 
2
2

5
4
 5
4
2
3
5
= 0,
3

5
3

5
4
=
p = 1
3
2
=
+
25
16
3
2


5
4
25
16




25
25
Notice that in this example we did not add a zero pair of +
and –
to the left side of the
16
16


25
equation, but instead added +
to both sides of the equation.
16

 5 2 49
x    ,
 4  16
5
3
2x2 – 5x = 3  x 2  x  ,
2
 2

x
5
7
1
  , x = 3,  .
4
4
2
Method 3: In the equation ax2 + bx +c = 0, if a is not a square, then divide the equation into a
separate squares and complete the squares.




 5 2 25 
 5 2
 2 5 
25 49
2

Solve: 2x – 5x = 3  2x  x 3  2
 3  2x    3 
x    


 4 
8
8

2 
 4  16 
which gives the same solution as Method 2.



Solve the following equations by completing the square.
1.
x2 2x – 3 = 0
2.
x2 + 4x = – 2
3.
p2 – 6p + 3 = 0
4.
a2 + 4a = 14
5.
q2 – 10q – 5 = 0
6.
9y2 + 12y = – 4
7.
m2 – 7m + 3 = 0
8.
4x2 – 6x = 1
The Quadratic Formula
b
c
b
c
Solve: a  x 2  b  x  c  0  x2  x   0  x2  x  
a
a
a
a
b
2a



b
=
2a


c
a

b
2a

b

2a

2

b  b 2  4ac
a  x 2  b  x  c  0  x   
,
 2a 
4a 2


2
b  b  4ac
,
x

2a
2a



=

c
a
+
b2

4a 2

or x =
b  b 2  4ac
.
2a
b2
4a 2