Problem 15 - College Algebra

Question

For the following exercises, solve the following polynomial equations by grouping and factoring. $$ 4 y^{3}-9 y=0 $$

Step-by-Step Solution

Answer

The solutions are \(y = 0\), \(y = \frac{3}{2}\), and \(y = -\frac{3}{2}\).

Step 1: Set Equation to Zero

The given equation is already set to zero: \[4y^3 - 9y = 0\]

Step 2: Factor Out Common Terms

Identify and factor out the greatest common factor from the terms. The greatest common factor here is \(y\):\[y(4y^2 - 9) = 0\]

Step 3: Recognize Patterns

Notice that \(4y^2 - 9\) can be factored further as a difference of squares. Recall that \(a^2 - b^2 = (a-b)(a+b)\).

Step 4: Factor the Difference of Squares

Apply the difference of squares: \[4y^2 - 9 = (2y)^2 - 3^2 = (2y - 3)(2y + 3)\]So the equation becomes:\[y(2y - 3)(2y + 3) = 0\]

Step 5: Solve the Factors Separately

Set each factor equal to zero and solve for \(y\):1. \(y = 0\)2. \(2y - 3 = 0\) leads to \(y = \frac{3}{2}\)3. \(2y + 3 = 0\) leads to \(y = -\frac{3}{2}\)

Step 6: Compile the Solutions

The solutions for the equation \(4y^3 - 9y = 0\) are \(y = 0\), \(y = \frac{3}{2}\), and \(y = -\frac{3}{2}\).

Key Concepts

Factoring

Factoring is a method used to simplify polynomial expressions by breaking them down into simpler components. It transforms a complex expression into the product of its simpler factors. In this exercise, the task is to factor the polynomial equation \(4y^3 - 9y = 0\).

To begin, it's important to identify terms that can be grouped together by recognizing patterns or common factors. By factoring out the greatest common factor, we reduce the polynomial into smaller components.

• Initially, factor by identifying the largest expression common to all terms. In this case, it's \(y\).
• This gives us: \[y(4y^2 - 9) = 0\].

This transformation simplifies the expression, making it easier to recognize additional factoring patterns like the difference of squares.

Difference of Squares

The difference of squares is a special algebraic pattern recognized as \(a^2 - b^2 = (a - b)(a + b)\). It's a pivotal concept for simplifying expressions further. In our equation, after factoring out \(y\), we have \(4y^2 - 9\). This expression can be viewed as a difference of squares:

• We recognize \(4y^2\) as \((2y)^2\) and \(9\) as \(3^2\).
• This gives us: \[4y^2 - 9 = (2y)^2 - 3^2 = (2y - 3)(2y + 3)\].

By factoring this way, we efficiently break down the polynomial into simpler binomial factors. This makes solving the equation more straightforward, as each factor can be set independently to zero to find solutions.

Solving Equations

To solve polynomial equations, it's crucial to first factor the expression fully. Once factored, each component can be set to zero independently. This is because of the zero-product property, which states that if a product equals zero, at least one of the factors must be zero. For \[y(2y - 3)(2y + 3) = 0\], the solution process involves:

• Setting each factor equal to zero: \((1)\, y = 0, \ \ (2)\, 2y - 3 = 0, \ \, \text{and}\, (3)\, 2y + 3 = 0\).
• Solving each equation gives us: \(y = 0\), \(2y = 3 \Rightarrow y = \frac{3}{2}\), and \(2y = -3 \Rightarrow y = -\frac{3}{2}\).

These solutions satisfy the original equation. Thus, factoring helps simplify the solving process by isolating each viable solution.

Common Factors

Identifying common factors is the first crucial step in the factoring process. It simplifies a polynomial expression by pulling out the largest term common to each part of the equation. This step is key to setting the stage for further simplification. In our example, the polynomial \(4y^3 - 9y\) has the common factor \(y\):

• This common factor simplifies the polynomial from \(4y^3 - 9y\) to \[y(4y^2 - 9)\].
• This reduction makes it easier to apply additional factoring techniques, such as the difference of squares.

By recognizing and factoring out common factors, you lay the groundwork for further simplifications and easily solving the equation.