Problem 41 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises, factor the polynomial. If the polynomial is prime, state it. $$ u^{2} v^{6}-8 u^{2} $$

Step-by-Step Solution

Answer

The factored form of the given polynomial is \(u^2(v^3 - 2)(v^3 + 2)\).

Step 1: Identify common factors of the expression

In this expression, \(u^{2}v^{6} - 8u^{2}\), we can observe that both terms have a common factor, \(u^2\). Keeping this in mind, we'll move to the next step of extracting common factors.

Step 2: Extract the common factor

Now, let's extract the common factor \(u^2\) from both terms. Doing so, we can rewrite the expression as \(u^2(v^6 - 8)\).

Step 3: Factor the remaining expression

Now, let's focus on the expression inside the parentheses, \(v^6 - 8\). This is a difference of two squares, with \(v^6\) a perfect square, and \(8\) being written as \(2^3\). We have the following formula to factor a difference of squares: \(a^2 - b^2 = (a - b)(a + b)\) In our case, \(a = v^3\) and \(b = 2\). Therefore, we can factor the remaining expression as: \(v^6 - 8 = (v^3 - 2)(v^3 + 2)\)

Step 4: Combine the extracted factor with the factored expression

Now, we will combine the extracted factor, \(u^2\), to the factored expression inside the parentheses. Doing so, we get \(u^2(v^6 - 8) = u^2(v^3 - 2)(v^3 + 2)\) So the factored form of the given polynomial is \(u^2(v^3 - 2)(v^3 + 2)\).

Key Concepts

Common Factors

When dealing with polynomial expressions, finding common factors is one of the first steps in the factoring process. Essentially, a common factor is a factor that is shared by all terms of a given expression. In our polynomial, \(u^2v^6 - 8u^2\), the common factor is \(u^2\), as both terms have this factor in common.

• Examine each term of the polynomial.
• Identify any terms that are multiplied consistently across all components of the polynomial.
• Extract the common factor as it will simplify the expression making further factorization easier.

Factoring out the common factor simplifies the expression and sets you up for further steps. This helps in solving polynomial equations more efficiently.

Difference of Squares

A difference of squares is a specific algebraic expression that takes the form \(a^2 - b^2\). It results in a structure that can be factored using the identity \((a - b)(a + b)\). In the problem, after extracting the common factor, we find ourselves with \(v^6 - 8\). Even though it isn't a perfect difference of squares directly, it demonstrates a similar pattern.

• Recognize that \(v^6\) is indeed \((v^3)^2\).
• Notice that \(8\) can be seen as \(2^3\), allowing manipulation into our formula \((a^2 - (b)^2)\).
• Apply the difference of squares formula to factor further.

This technique is useful in simplifying polynomials even if they don't immediately appear to fall into a classic pattern.

Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of simpler polynomials. The goal is to break down the polynomial expression into factors that are easier to work with. In our case, the expression \(u^2(v^3 - 2)(v^3 + 2)\) is the result of factoring by common factors and applying a difference of squares.

• Look for common factors first to simplify the expression.
• Check if special algebraic identities, such as difference of squares, can be applied.
• Each step of factoring simplifies further calculations and leads the way to solving polynomial equations.

Factoring is a cornerstone technique in algebra that opens the door to numerous problem-solving methods, including solving equations and simplifying complex expressions.

Perfect Squares

Perfect squares refer to numbers or expressions that are the product of a term multiplied by itself. Recognizing and utilizing perfect squares is fundamental, especially when exploring techniques like factoring.For instance, in the expression \(v^6\), it's important to see it as \((v^3)^2\), which is a perfect square. Perfect squares form a basic building block in factorization patterns and can simplify complex polynomial expressions.

• Identify and express terms as squares whenever possible.
• Recognizing patterns in squares aids in applying formulas such as the difference of squares.
• This identification simplifies the polynomial, converting it to a more factorable form.

Understanding and identifying perfect squares allows for efficient use of various algebraic techniques, streamlining simplification and solution processes.