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Explanation:
The equation [tex]x^2 - 36[/tex] can be factored into [tex](x - 6)(x + 6)[/tex] using the difference of two squares.
The equation [tex]9x^2 - 1[/tex] is the difference of two squares, [tex](3x)^2 - 1^2[/tex], so it factors into [tex](3x - 1)(3x + 1)[/tex].
We can factor 4 out of the expression [tex]4x^2 - 16[/tex] to get [tex]4(x - 2)(x + 2)[/tex].

Part B:
Which two binomials in part A didn't have an equivalent factored form? What made these expressions different from the three that you could factor?

Part C:
Find the product of each expression using properties of complex numbers.

Sagot :

GinnyAnswer
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.
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