Problem 58 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$y^{5}-16 y$$

Step-by-Step Solution

Answer

The fully factorized form of \(y^{5}-16 y\) is \(y(y-2)(y+2)(y^{2} + 4)\).

Step 1: Identify Common Factors

The first step is to observe and identify any common factors. In the given polynomial \(y^{5} - 16 y\), the common factor is \(y\).

Step 2: Factor out the Common Factor

Take out the common factor which results into \(y(y^{4} - 16)\).

Step 3: Recognize Difference of Squares

Recognize the expression \(y^{4}-16\) as a difference of squares. Note that this can be rewritten as \((y^{2})^{2} - (4)^{2}\).

Step 4: Apply Difference of Squares

Apply the formula for difference of squares, \(a^{2}-b^{2} = (a - b)(a + b)\) to get the expression \(y^{2} - 4)(y^{2} + 4)\).

Step 5: Further Factorize

Factorize \(y^{2} - 4\) further since it is a difference of squares into \((y-2)(y+2)\). The expression \(y^{2} + 4\) cannot be factored over the real numbers, as it does not follow the difference of squares format.

Step 6: Final Factorized Expression

Combining all the factored elements, the final expression will be \(y(y-2)(y+2)(y^{2} + 4)\).

Key Concepts

Difference of Squares

Breaking down complex polynomials can sometimes be made simpler by recognizing patterns, such as the difference of squares. This particular algebraic identity states that for any two perfect squares, the expression \(a^2 - b^2\) can be factored into \(a - b\) and \(a + b\).

Think of it like this: you have two squares, \(a\) and \(b\), and you take the smaller square out from the larger square; what’s left can be neatly split along the middle, creating factors that mirror each other, one with a minus and one with a plus.

Applying this to our exercise, we have \(y^4 - 16\), which is indeed a difference of squares since both terms are squares: \(y^2)^2\) and \(4^2\). Factoring based off of our identity, we tear apart the difference and get \(y^2 - 4\) and \(y^2 + 4\), unveiling the factors hidden within the expression.

Common Factors in Polynomials

Polynomials may look daunting due to their multiple terms, but they can often share a common thread that attributes to their simplification. Identifying common factors across the terms is key to reducing the polynomial's complexity.

This act—much like finding what multiple children have in common so you can unify them to work together—is the essence of factoring polynomials. In our exercise, we notice that each term of the polynomial \(y^5 - 16y\) contains a \(y\). This \(y\) acts as a common factor, acting much like a team leader, bringing the terms together by being factored out.

Once factored out, we’re left with \(y(y^4 - 16)\), a simpler expression. This reveals a leftover expresson where we can then apply further factorization techniques, demonstrating the power of removing common factors as an initial step in polynomial factorization.

Polynomial Factorization Techniques

To completely tame the wild polynomials means to delve into the arsenal of polynomial factorization techniques. Each technique is like a unique key, unlocking a specific type of polynomial expression. Understanding and identifying which key fits which lock is part of the beauty of algebra.

In the exercise provided, we utilized a combination of factoring by common factors and recognizing a difference of squares. These are two critical techniques amongst others, like factoring by grouping, using the sum or difference of cubes, or applying the quadratic formula for higher-degree polynomials.

After extracting the common factor \(y\), and splitting the difference of squares, our polynomial transformed into \(y(y-2)(y+2)(y^2 + 4)\). It's evident that being well-versed in various polynomial factorization techniques can greatly simplify an intimidating expression into a product of factors that are much more manageable.