Problem 83 - College Algebra

Question

Factor completely, or state that the polynomial is prime. $$ 20 y^{4}-45 y^{2} $$

Step-by-Step Solution

Answer

The completely factored form of the polynomial \(20y^{4} - 45y^{2}\) is \(5y^{2}(2y - 3)(2y + 3)\).

Step 1: Identify common factors

All terms in the polynomial \(20y^{4}-45y^{2}\) have \(y^{2}\) in them and this is the most power of y that every term has. Additionally, the numerical coefficients 20 and 45 have a common factor, which is 5.

Step 2: Factor out the common factors

Factoring out the common factors yields: \(5y^{2}(4y^{2}-9)\)

Step 3: Apply the difference of squares rule

The quadratic expression in the brackets is a difference of two squares which follows the pattern \(a^{2}- b^{2} = (a+b)(a-b)\). Applying this method of factoring to \(4y^{2}-9\) yields \((2y-3)(2y+3)\).

Step 4: Write the final factorised form

Substitute the factorization of the quadratic expression back into the original polynomial to yield the final factorised form: \(5y^{2}(2y - 3)(2y + 3)\).

Key Concepts

Understanding Common Factors

When dealing with the factorization of polynomials, the first step often involves identifying the common factors among the terms. Common factors are elements that are present in each term of the polynomial.
For instance, in the polynomial \(20y^{4} - 45y^{2}\), both terms

• have a variable factor of \(y^{2}\),
• and numerical coefficients 20 and 45, which share the numerical factor of 5.

By recognizing these common factors, we can simplify the expression by factoring them out.
This results in the expression \(5y^{2}(4y^{2} - 9)\).
Factoring out reduces the complexity of the polynomial and prepares it for further simplification, such as recognizing patterns like the difference of squares.

Difference of Squares Simplification

The difference of squares technique is a powerful method for factoring specific types of polynomials. This method applies when you have a polynomial that can be expressed in the form \(a^{2} - b^{2}\).
The beauty of this pattern is that it can be factored into \((a+b)(a-b)\). These are the linear factors.
For the problem at hand, inside the parentheses we have \(4y^{2} - 9\), which can be rewritten as \((2y)^{2} - 3^{2}\). This clearly matches the difference of squares pattern.
So we apply the rule:

• \(a = 2y\)
• \(b = 3\)

Hence, \(4y^{2} - 9\) becomes \((2y + 3)(2y - 3)\). This technique is essential when you encounter expressions that appear cumbersome, but reveal their simplicity through careful analysis.

Defining a Prime Polynomial

A polynomial is considered prime if it cannot be factored into a product of lower-degree polynomials using integer coefficients. It is similar to a prime number, which cannot be divided evenly by anything other than itself and one.

When working with polynomials like \(20y^{4} - 45y^{2}\), after factoring out common factors and recognizing special patterns (like the difference of squares), you reach a factorized form. In this case, \(5y^{2}(2y - 3)(2y + 3)\).

If no further factorization is possible, and all factors are linear or irreducible quadratics (in case of polynomials with real coefficients), then the polynomial cannot be simplified further and is not prime. In our example:

• The polynomial was factored successfully into products of lower-degree polynomials.
• Therefore, it's not prime since it can be broken down.

Recognizing when a polynomial is irreducible is key in ensuring a polynomial is, indeed, in its simplest form.