Problem 36 - Introductory Algebra for College Students

Question

Factor completely, or state that the polynomial is prime. $$32-2 y^{2}$$

Step-by-Step Solution

Answer

The factored form of the given equation \(32 - 2y^2\) is \((4 - \sqrt{2}y)(4 + \sqrt{2}y)\).

Step 1: Identify the Square Terms

In this equation, the two terms are 32 and \(2y^2\). We can rewrite 32 as \(4^2\) and \(2y^2\) as \((\sqrt{2}y)^2\). So our equation can be rewritten as: \(4^2 - (\sqrt{2}y)^2\).

Step 2: Apply the Difference of Squares

As per the difference of squares formula \((a^2 - b^2) = (a - b)(a + b)\), we can apply it to the equation \(4^2 - (\sqrt{2}y)^2\). This gives us: \((4 - \sqrt{2}y)(4 + \sqrt{2}y)\).

Step 3: Final Answer

Hence, the factored form of the given equation \(32 - 2y^2\) is \((4 - \sqrt{2}y)(4 + \sqrt{2}y)\).

Key Concepts

Difference of Squares

The 'difference of squares' is a fundamental concept in algebra which refers to an expression of the form \(a^2 - b^2\). This expression can be factored into the product of two binomials that are conjugates of each other, that is, \((a - b)(a + b)\). This factorization is possible because when the two binomials are multiplied, the middle terms cancel each other out, leaving only the square terms.

In practice, when you encounter a polynomial subtraction where both terms are perfect squares, this method allows for quick and straightforward factorization. Understanding and identifying the difference of squares is an important skill, as it makes simplifying expressions much easier and is often a step towards solving more complex algebraic equations.

Polynomial Prime Factorization

When we talk about 'polynomial prime factorization', we're treating a polynomial like a number and breaking it down into factors that cannot be factored further. These 'prime' factors are analogous to prime numbers in the realm of integers.

In algebra, a polynomial is considered 'prime' if it cannot be factored into the product of two non-constant polynomials. The process of factorizing a polynomial completely involves breaking it down as far as possible until all that remains are its prime factors or, if it's already in its simplest form, recognizing that the polynomial itself is prime. This is critical in solving equations, simplifying expressions, and can also be applied to finding the least common denominators for adding and subtracting rational expressions.

Identifying Square Terms in Polynomials

To successfully factor polynomials, one needs to be adept at 'identifying square terms in polynomials'. A square term is simply an expression raised to the second power. Recognizing square terms is particularly useful because they play a major role in several factoring techniques, including the difference of squares.

In our exercise, we transformed the number 32 into \(4^2\) and the term \(2y^2\) into \((\radsqrt{2}y)^2\), showing their square nature which is not immediately obvious. This transformation is a crucial step because it allows us to use specific factoring rules that simplify the polynomial significantly. It's important to remember that not all terms have square roots that are neat integers or rational numbers; sometimes they involve irrational numbers, as seen with \radsqrt{2}.

Square Root Transformations

Working with 'square root transformations' is an integral part of manipulating algebraic expressions, especially when factoring. This involves taking the square root of a term to find another term which, when squared, will give the original term. It's a reverse-engineering process of sorts.

These transformations are used in identifying the square terms we talked about in the previous section. Recognizing that \(32\) can be seen as \(4^2\) or that \(2y^2\) can be envisaged as \((\radsqrt{2}y)^2\) leverages the power of square roots in simplifying expressions. This technique is particularly useful when dealing with quadratics or any polynomials where the identification of squares can lead to easier factorization through methods such as the difference of squares.