Answer :
Sure, let's break down the solution step-by-step for each part of the problem.
### Part A: Factoring the equations
1. Equation: [tex]\( x^2 - 36 \)[/tex]
This is a difference of squares. The formula for the difference of squares is [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex]. Here, [tex]\( a = x \)[/tex] and [tex]\( b = 6 \)[/tex], so:
[tex]\[ x^2 - 36 = (x - 6)(x + 6) \][/tex]
2. Equation: [tex]\( 9x^2 - 1 \)[/tex]
This is also a difference of squares where [tex]\( a = 3x \)[/tex] and [tex]\( b = 1 \)[/tex]. Applying the formula:
[tex]\[ 9x^2 - 1 = (3x - 1)(3x + 1) \][/tex]
3. Equation: [tex]\( 4x^2 - 16 \)[/tex]
First, notice that 4 is a common factor in both terms:
[tex]\[ 4x^2 - 16 = 4(x^2 - 4) \][/tex]
Then [tex]\( x^2 - 4 \)[/tex] is a difference of squares where [tex]\( a = x \)[/tex] and [tex]\( b = 2 \)[/tex]:
[tex]\[ x^2 - 4 = (x - 2)(x + 2) \][/tex]
Therefore:
[tex]\[ 4x^2 - 16 = 4(x - 2)(x + 2) \][/tex]
Thus, the factored forms of the equations are:
- [tex]\( x^2 - 36 \)[/tex] factors into [tex]\( (x-6)(x+6) \)[/tex].
- [tex]\( 9x^2 - 1 \)[/tex] factors into [tex]\( (3x-1)(3x+1) \)[/tex].
- [tex]\( 4x^2 - 16 \)[/tex] factors into [tex]\( 4(x-2)(x+2) \)[/tex].
### Part B: Non-equivalent Factored Forms
All the provided equations were factorable using the difference of squares method. There are no binomials in part A that did not have an equivalent factored form.
### Part C: Products using Properties of Complex Numbers
The products can be derived by simply expanding the factored forms back:
1. Product of [tex]\( (x - 6)(x + 6) \)[/tex]:
[tex]\[ (x - 6)(x + 6) = x^2 - 36 \][/tex]
2. Product of [tex]\( (3x - 1)(3x + 1) \)[/tex]:
[tex]\[ (3x - 1)(3x + 1) = 9x^2 - 1 \][/tex]
3. Product of [tex]\( 4(x - 2)(x + 2) \)[/tex]:
[tex]\[ 4(x - 2)(x + 2) = 4(x^2 - 4) = 4x^2 - 16 \][/tex]
Thus, the products of each expression, reaffirming the given factored forms, are the original expressions:
- [tex]\( (x-6)(x+6) = x^2 - 36 \)[/tex]
- [tex]\( (3x-1)(3x+1) = 9x^2 - 1 \)[/tex]
- [tex]\( 4(x-2)(x+2) = 4x^2 - 16 \)[/tex]
These steps confirm the accuracy and correctness of the original information, ensuring that the factored forms are appropriate and that all of the provided binomials are factorable.
### Part A: Factoring the equations
1. Equation: [tex]\( x^2 - 36 \)[/tex]
This is a difference of squares. The formula for the difference of squares is [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex]. Here, [tex]\( a = x \)[/tex] and [tex]\( b = 6 \)[/tex], so:
[tex]\[ x^2 - 36 = (x - 6)(x + 6) \][/tex]
2. Equation: [tex]\( 9x^2 - 1 \)[/tex]
This is also a difference of squares where [tex]\( a = 3x \)[/tex] and [tex]\( b = 1 \)[/tex]. Applying the formula:
[tex]\[ 9x^2 - 1 = (3x - 1)(3x + 1) \][/tex]
3. Equation: [tex]\( 4x^2 - 16 \)[/tex]
First, notice that 4 is a common factor in both terms:
[tex]\[ 4x^2 - 16 = 4(x^2 - 4) \][/tex]
Then [tex]\( x^2 - 4 \)[/tex] is a difference of squares where [tex]\( a = x \)[/tex] and [tex]\( b = 2 \)[/tex]:
[tex]\[ x^2 - 4 = (x - 2)(x + 2) \][/tex]
Therefore:
[tex]\[ 4x^2 - 16 = 4(x - 2)(x + 2) \][/tex]
Thus, the factored forms of the equations are:
- [tex]\( x^2 - 36 \)[/tex] factors into [tex]\( (x-6)(x+6) \)[/tex].
- [tex]\( 9x^2 - 1 \)[/tex] factors into [tex]\( (3x-1)(3x+1) \)[/tex].
- [tex]\( 4x^2 - 16 \)[/tex] factors into [tex]\( 4(x-2)(x+2) \)[/tex].
### Part B: Non-equivalent Factored Forms
All the provided equations were factorable using the difference of squares method. There are no binomials in part A that did not have an equivalent factored form.
### Part C: Products using Properties of Complex Numbers
The products can be derived by simply expanding the factored forms back:
1. Product of [tex]\( (x - 6)(x + 6) \)[/tex]:
[tex]\[ (x - 6)(x + 6) = x^2 - 36 \][/tex]
2. Product of [tex]\( (3x - 1)(3x + 1) \)[/tex]:
[tex]\[ (3x - 1)(3x + 1) = 9x^2 - 1 \][/tex]
3. Product of [tex]\( 4(x - 2)(x + 2) \)[/tex]:
[tex]\[ 4(x - 2)(x + 2) = 4(x^2 - 4) = 4x^2 - 16 \][/tex]
Thus, the products of each expression, reaffirming the given factored forms, are the original expressions:
- [tex]\( (x-6)(x+6) = x^2 - 36 \)[/tex]
- [tex]\( (3x-1)(3x+1) = 9x^2 - 1 \)[/tex]
- [tex]\( 4(x-2)(x+2) = 4x^2 - 16 \)[/tex]
These steps confirm the accuracy and correctness of the original information, ensuring that the factored forms are appropriate and that all of the provided binomials are factorable.
