Problem 39 - Elementary Algebra

Question

Factor completely. Identify any prime polynomials. $$ 6 x^{2}-6 y^{2} $$

Step-by-Step Solution

Answer

The factored form is 6(x - y)(x + y). The polynomials x - y and x + y are prime.

Step 1: Identify the common factor

Both terms in the polynomial, 6x^{2} and -6y^{2}, have a common factor of 6. Factor the 6 out of the polynomial.

Step 2: Apply Difference of Squares Formula

The remaining polynomial after factoring out the 6 is x^{2} - y^{2}. This fits the difference of squares formula: a^{2} - b^{2} = (a - b)(a + b), where a = x and b = y.

Step 3: Write the Factorized Form

Apply the formula to get (x - y)(x + y). Thus, 6x^{2} - 6y^{2} factors completely to 6(x - y)(x + y).

Step 4: Identify Any Prime Polynomials

The factors x - y and x + y are both polynomials of degree 1 and are therefore considered prime polynomials.

Key Concepts

Common Factor

To solve the polynomial factoring problem, the first key concept is identifying the common factor.
In the given polynomial, we have:

• 6x^{2}
• -6y^{2}

Both terms share a common factor of 6.
This means we can divide each term by 6 and factor it outside the polynomial.
The polynomial then simplifies to:

• 6(x^{2} - y^{2})

Identifying and factoring out a common factor simplifies the polynomial and makes the next steps more manageable.

Always check for common factors as your first step when factoring polynomials.

Difference of Squares

The next concept involves recognizing and applying the difference of squares formula.
The simplified polynomial from identifying the common factor is:

• x^{2} - y^{2}

This matches the form of the difference of squares:

• a^{2} - b^{2}

The formula for factoring a difference of squares is:

• a^{2} - b^{2} = (a - b)(a + b)

Here, we can identify:

• a = x
• b = y

Applying the formula, we get:

• (x - y)(x + y)

Factoring using the difference of squares formula breaks the polynomial down into two binomial terms.

Prime Polynomials

In the final step, it’s essential to check whether the resulting factors can be simplified further or are prime polynomials.
From our factored form, we have:

• 6(x - y)(x + y)

We need to verify if

• (x - y)
• (x + y)

can be factored further. Both terms are polynomials of degree 1.
Degree 1 polynomials are defined as prime because they cannot be factored any further.
Hence,

• (x - y)
• (x + y)

are both prime polynomials. In summary, 6x^{2} - 6y^{2} completely factors to:

• 6(x - y)(x + y)

with both (x - y) and (x + y) being prime polynomials.