Problem 37 - Introductory Algebra for College Students

Question

Factor completely, or state that the polynomial is prime. $$3 y^{3}-48 y$$

Step-by-Step Solution

Answer

The fully factored form of the given polynomial is \(3y(y+4)(y-4)\).

Step 1:Identify common factors

Start by looking at the polynomial and identifying any common factors in each term. The polynomial is \(3 y^{3}-48 y\). Both terms in the polynomial have a common factor of \(3y\).

Step 2: Factor out the common factors

Factor out the common factor \(3y\) from both terms. After factoring out, the polynomial simplifies to: \(3y(y^2 - 16)\).

Step 3: Apply difference of squares

Recognize that the second factor, \(y^2 - 16\), is a difference of squares. A difference of squares can be factored into \((a+b)(a-b)\). Applying this rule here gives us \((y+4)(y-4)\). So, the completely factored polynomial is: \(3y(y+4)(y-4)\)

Key Concepts

Common factors

In polynomial expressions, identifying common factors is one of the first steps in simplification or factorization. When you look at a polynomial like \(3y^3 - 48y\), you need to check each term to see what they all have in common.

• Common factors are elements that evenly divide each term in the polynomial. In the example given, each term in \(3y^3 - 48y\) can be divided by \(3y\).
• To factor out the common factor, simply divide each term by \(3y\) and write it outside of a parenthesis. This step simplifies our polynomial to \(3y(y^2 - 16)\).

Identifying and factoring out common factors simplifies complicated expressions, making it easier to apply other factoring techniques. This foundational step prevents mistakes and streamlines the factorization process.

Difference of squares

The difference of squares is a special pattern used in factoring. It applies when you have a polynomial in the form \(a^2 - b^2\).

• The pattern can be rewritten as \((a + b)(a - b)\). This method takes advantage of a specific algebraic identity.
• In the simplified polynomial \(3y(y^2 - 16)\), the \(y^2 - 16\) part is a difference of squares, because \(y^2\) is \((y)^2\) and \(16\) is \((4)^2\).
• Applying the difference of squares, \(y^2 - 16\) becomes \((y + 4)(y - 4)\), leading us to the further factored expression \(3y(y + 4)(y - 4)\).

Recognizing these patterns is key to simplifying complex expressions and quickly arriving at fully factored form. It demonstrates the power of algebraic identities to make factoring more manageable.

Factoring polynomials

Factoring polynomials is about breaking down a polynomial into its simplest components or factors. This process can appear challenging at first, but breaking it into steps makes it easier.

• The first step usually involves looking for common factors across all terms, such as in \(3y^3 - 48y\), where the common factor \(3y\) simplifies the expression.
• Next, check if the simplified expression includes patterns like the difference of squares or other notable factorization identities, as we've seen with \(y^2 - 16\).
• Combining these methods, you can rewrite polynomials concisely. For the given polynomial, we reach \(3y(y + 4)(y - 4)\) after applying common factors and then the difference of squares.

Factoring polynomials is crucial for solving equations and understanding algebra on a deeper level. It opens doors to solving not only polynomials but also more advanced mathematical concepts. With practice, recognizing patterns like common factors and difference of squares becomes second nature.