Problem 31 - Algebra 2 and Trigonometry

Question

In \(27-39,\) factor each polynomial completely. $$ 12 c^{2}-3 $$

Step-by-Step Solution

Answer

The polynomial \(12c^2 - 3\) factors completely as \(3(2c - 1)(2c + 1)\).

Step 1: Identify the Common Factor

Look at the polynomial \( 12c^2 - 3 \) and identify the common factor in both terms. Both terms have a common factor of 3.

Step 2: Factor Out the Common Factor

Factor out the greatest common factor, 3, from each term in the polynomial: \[ 12c^2 - 3 = 3(4c^2 - 1). \]

Step 3: Recognize the Difference of Squares

Notice that \( 4c^2 - 1 \) is a difference of squares, which can be factored further. Recall that \( a^2 - b^2 = (a - b)(a + b) \), so here \( a^2 = 4c^2 \) and \( b^2 = 1 \).

Step 4: Factor the Difference of Squares

Apply the difference of squares formula: \[ 4c^2 - 1 = (2c - 1)(2c + 1). \]

Step 5: Write the Completely Factored Form

Combine all the factored parts to get the completely factored form of the original polynomial: \[ 12c^2 - 3 = 3(2c - 1)(2c + 1). \]

Key Concepts

Common Factor

When you're faced with a polynomial and need to factor it, finding a common factor is often the first step. A common factor is a number or variable that divides each term of the polynomial without leaving a remainder. This step simplifies the polynomial, making it easier to work with in subsequent factorization steps.
For the polynomial given, \(12c^2 - 3\), we can identify that both terms, 12 and -3, share a common factor of 3.

Here's how to factor out the common factor:

• Divide each term of the polynomial by the common factor, 3.
• You'll get: \(12c^2 \div 3 = 4c^2\) and \(-3 \div 3 = -1\), leading to \(3(4c^2 - 1)\).

Factoring out the common factor is key because it reduces complexity and allows us to see further factorization opportunities, such as the case with the difference of squares.

Difference of Squares

Polynomial factorization often uncovers special forms, like the difference of squares. The difference of squares is specific to polynomials structured as \(a^2 - b^2\), which can be rewritten utilizing the formula: \((a-b)(a+b)\).

In our polynomial example, after extracting the common factor, we arrive at \(4c^2 - 1\).
This expression is a perfect candidate for the difference of squares:

• Recognize that \(4c^2\) equates to \((2c)^2\) and \(1\) is \(1^2\).
• Implement the formula by identifying \(a = 2c\) and \(b = 1\), then apply \((a-b)(a+b)\).

This gives us \((2c-1)(2c+1)\) as the factors of \(4c^2 - 1\), illustrating a textbook case of the difference of squares.

Polynomial Factorization

Polynomial factorization is a method of breaking down a polynomial into simpler "factor" polynomials, whose product is the original polynomial. This is like finding the building blocks of the polynomial expression.

For our polynomial, \(12c^2 - 3\), we pursued factorization by:

• First, identifying and extracting the common factor, giving us \(3(4c^2 - 1)\).
• Next, recognizing and utilizing the difference of squares on \(4c^2 - 1\), further simplifying it to \((2c - 1)(2c + 1)\).

Ultimately, the fully factored form of the polynomial is \(3(2c - 1)(2c + 1)\).
Factorization helps break down complex expressions into simpler components that are easier to manage and use in solving equations or other mathematical operations. It's valuable for simplifying computations and gaining insights into polynomial behavior.