Answer :
Sure! Let's go through the steps to factor the given quadratic expressions using algebra tiles and then verifying the factors.
### Step 1: Factor [tex]\( x^2 - 3x + 2 \)[/tex]
#### a. Building and Sketching a Rectangular Figure
1. Represent [tex]\( x^2 \)[/tex]: First, we have a single [tex]\( x^2 \)[/tex] tile.
2. Represent [tex]\( -3x \)[/tex]: Next, we have 3 negative [tex]\( x \)[/tex] tiles.
3. Represent +2: Finally, we have 2 positive unit tiles.
To factor the expression, we arrange these tiles in a rectangular shape:
- Place the [tex]\( x^2 \)[/tex] tile in the top-left corner.
- Arrange -3x tiles in such a way to form the sides of a rectangle.
- The positive unit tiles will fit in the remaining spaces to complete the rectangle.
This setup should help us identify the length and width of the rectangle formed by the algebra tiles.
#### b. Identifying the Side Lengths
By arranging the tiles correctly:
- The [tex]\( x^2 \)[/tex] tile suggests the dimensions start with [tex]\( x \)[/tex].
- The factors [tex]\( (x-2) \)[/tex] and [tex]\( (x-1) \)[/tex] complete the rectangle since:
- The length [tex]\( x - 1 \)[/tex] represents 1 [tex]\( x \)[/tex] tile and 1 unit tile.
- The width [tex]\( x - 2 \)[/tex] represents 1 [tex]\( x \)[/tex] tile and 2 unit tiles.
#### c. Writing an Equation
We have:
[tex]\[ (x - 2) \times (x - 1) = x^2 - 3x + 2 \][/tex]
### Step 2: Factor [tex]\( 2x^2 + x - 6 \)[/tex]
#### a. Building and Sketching a Rectangular Figure
1. Represent [tex]\( 2x^2 \)[/tex]: We need 2 [tex]\( x^2 \)[/tex] tiles.
2. Represent [tex]\( x \)[/tex]: We need 1 positive [tex]\( x \)[/tex] tile.
3. Represent [tex]\( -6 \)[/tex]: We need 6 negative unit tiles.
We might need to add zero pairs (an [tex]\( x \)[/tex] tile and a [tex]\( -x \)[/tex] tile) to form a complete rectangle.
#### b. Identifying the Side Lengths
Arrange the tiles to form a rectangle:
- Place [tex]\( 2x^2 \)[/tex] tiles to start with.
- Arrange the remaining [tex]\( x \)[/tex] tile and [tex]\(-6\)[/tex] unit tiles around these.
- Zero pairs help fill any gaps.
The correct arrangement will help us identify the factors [tex]\( (x + 2) \)[/tex] and [tex]\( (2x - 3) \)[/tex]:
- The length [tex]\( 2x - 3 \)[/tex]: Two [tex]\( x \)[/tex] tiles and subtract 3 unit tiles.
- The width [tex]\( x + 2 \)[/tex]: One [tex]\( x \)[/tex] tile and add 2 unit tiles.
#### c. Writing an Equation
We have:
[tex]\[ (x + 2) \times (2x - 3) = 2x^2 + x - 6 \][/tex]
### Verification
1. For [tex]\( x^2 - 3x + 2 \)[/tex]:
[tex]\[ (x - 2)(x - 1) = x^2 - x - 2x + 2 = x^2 - 3x + 2 \][/tex]
2. For [tex]\( 2x^2 + x - 6 \)[/tex]:
[tex]\[ (x + 2)(2x - 3) = 2x^2 - 3x + 4x - 6 = 2x^2 + x - 6 \][/tex]
Both factorizations are verified correct.
### Step 1: Factor [tex]\( x^2 - 3x + 2 \)[/tex]
#### a. Building and Sketching a Rectangular Figure
1. Represent [tex]\( x^2 \)[/tex]: First, we have a single [tex]\( x^2 \)[/tex] tile.
2. Represent [tex]\( -3x \)[/tex]: Next, we have 3 negative [tex]\( x \)[/tex] tiles.
3. Represent +2: Finally, we have 2 positive unit tiles.
To factor the expression, we arrange these tiles in a rectangular shape:
- Place the [tex]\( x^2 \)[/tex] tile in the top-left corner.
- Arrange -3x tiles in such a way to form the sides of a rectangle.
- The positive unit tiles will fit in the remaining spaces to complete the rectangle.
This setup should help us identify the length and width of the rectangle formed by the algebra tiles.
#### b. Identifying the Side Lengths
By arranging the tiles correctly:
- The [tex]\( x^2 \)[/tex] tile suggests the dimensions start with [tex]\( x \)[/tex].
- The factors [tex]\( (x-2) \)[/tex] and [tex]\( (x-1) \)[/tex] complete the rectangle since:
- The length [tex]\( x - 1 \)[/tex] represents 1 [tex]\( x \)[/tex] tile and 1 unit tile.
- The width [tex]\( x - 2 \)[/tex] represents 1 [tex]\( x \)[/tex] tile and 2 unit tiles.
#### c. Writing an Equation
We have:
[tex]\[ (x - 2) \times (x - 1) = x^2 - 3x + 2 \][/tex]
### Step 2: Factor [tex]\( 2x^2 + x - 6 \)[/tex]
#### a. Building and Sketching a Rectangular Figure
1. Represent [tex]\( 2x^2 \)[/tex]: We need 2 [tex]\( x^2 \)[/tex] tiles.
2. Represent [tex]\( x \)[/tex]: We need 1 positive [tex]\( x \)[/tex] tile.
3. Represent [tex]\( -6 \)[/tex]: We need 6 negative unit tiles.
We might need to add zero pairs (an [tex]\( x \)[/tex] tile and a [tex]\( -x \)[/tex] tile) to form a complete rectangle.
#### b. Identifying the Side Lengths
Arrange the tiles to form a rectangle:
- Place [tex]\( 2x^2 \)[/tex] tiles to start with.
- Arrange the remaining [tex]\( x \)[/tex] tile and [tex]\(-6\)[/tex] unit tiles around these.
- Zero pairs help fill any gaps.
The correct arrangement will help us identify the factors [tex]\( (x + 2) \)[/tex] and [tex]\( (2x - 3) \)[/tex]:
- The length [tex]\( 2x - 3 \)[/tex]: Two [tex]\( x \)[/tex] tiles and subtract 3 unit tiles.
- The width [tex]\( x + 2 \)[/tex]: One [tex]\( x \)[/tex] tile and add 2 unit tiles.
#### c. Writing an Equation
We have:
[tex]\[ (x + 2) \times (2x - 3) = 2x^2 + x - 6 \][/tex]
### Verification
1. For [tex]\( x^2 - 3x + 2 \)[/tex]:
[tex]\[ (x - 2)(x - 1) = x^2 - x - 2x + 2 = x^2 - 3x + 2 \][/tex]
2. For [tex]\( 2x^2 + x - 6 \)[/tex]:
[tex]\[ (x + 2)(2x - 3) = 2x^2 - 3x + 4x - 6 = 2x^2 + x - 6 \][/tex]
Both factorizations are verified correct.