Problem 38 - Algebra A Combined Function

Question

Solve each equation. $$ 4 y^{3}-36 y=0 $$

Step-by-Step Solution

Answer

The solutions are \( y = 0 \), \( y = 3 \), and \( y = -3 \).

Step 1: Recognize the structure of the equation

First, notice that the equation is in the format of a polynomial equation, particularly a cubic polynomial, which can often be solved by factoring. The equation is given by \( 4y^3 - 36y = 0 \).

Step 2: Factor out the greatest common factor

Identify and factor out the greatest common factor from the terms in the polynomial. Here, the greatest common factor of \( 4y^3 \) and \(-36y \) is \( 4y \). So, we factor \( 4y \) out: \( 4y(y^2 - 9) = 0 \).

Step 3: Further factor the inner expression

Notice that \( y^2 - 9 \) is a difference of squares, which can be factored as: \( (y - 3)(y + 3) \). So, the equation becomes \( 4y(y - 3)(y + 3) = 0 \).

Step 4: Set each factor equal to zero

Using the zero product property, set each individual factor equal to zero to find potential solutions: 1. \( 4y = 0 \)2. \( y - 3 = 0 \)3. \( y + 3 = 0 \)

Step 5: Solve each resulting simple equation

Solve each of the equations from Step 4: 1. From \( 4y = 0 \), divide both sides by 4 to get \( y = 0 \).2. From \( y - 3 = 0 \), add 3 to both sides to get \( y = 3 \).3. From \( y + 3 = 0 \), subtract 3 from both sides to get \( y = -3 \).

Key Concepts

Factoring Polynomials

Factoring polynomials is an invaluable skill in solving algebraic equations. In essence, it involves expressing a polynomial as a product of its factors, which are polynomials of lower degrees. This is particularly useful because it simplifies the equation and often turns a complex equation into a simpler one.
For example, let's look at the polynomial from the original problem:

• We start with the equation: \( 4y^3 - 36y = 0 \).
• Recognizing that both terms share a common factor allows us to simplify the equation by factoring. This process breaks down polynomials into smaller, more manageable pieces.

The goal of factoring is to find these smaller polynomials that, when multiplied together, give us the original polynomial. This greatly simplifies the process of solving the equation because once factored, we can apply the zero product property to each term. Factoring is analogous to "breaking apart" a problem to make it more digestible.

Greatest Common Factor

The greatest common factor (GCF) is the largest factor that divides each term in a polynomial. Finding the GCF is the first crucial step in factoring polynomials, as it helps in simplifying the polynomial to a more workable form.
In the given problem, we have the polynomial \( 4y^3 - 36y \). Let's find the GCF for this expression:

• Look at each coefficient and the variables. The coefficients are 4 and -36.
• The greatest number that divides both 4 and 36 is 4.
• Now, consider the variables: both terms have at least one \( y \). So, include \( y \) as part of the GCF.
• Thus, the GCF is \( 4y \).

By factoring out \( 4y \), we get \( 4y(y^2 - 9) \). This process simplifies the polynomial and sets the stage for further factoring.

Zero Product Property

The zero product property is a fundamental principle in algebra that states: if the product of two factors is zero, then at least one of the factors must be zero. This property is crucial for solving factored polynomial equations.
In the example we are examining, once we have factored the polynomial, it looks like this: \( 4y(y - 3)(y + 3) = 0 \). According to the zero product property:

• If \( 4y \) is zero, then \( y \) must be zero.
• If \( y - 3 \) is zero, then \( y \) must be 3.
• If \( y + 3 \) is zero, then \( y \) must be -3.

This property allows us to break a complicated equation into simple linear ones, making it much easier to find all the possible solutions for \( y \). By dealing with each factor separately, we can efficiently find all roots of the polynomial equation.

Difference of Squares

The difference of squares is a special case of polynomial factoring. It occurs when you have an expression in the form of \( a^2 - b^2 \), which can be factored into \( (a - b)(a + b) \).
In our problem, after factoring out the GCF, we encountered \( y^2 - 9 \), which is a difference of squares:

• Recognize that \( y^2 \) is \( (y)^2 \).
• Realize that 9 is \( (3)^2 \).
• This means the expression \( y^2 - 9 \) can be rewritten as \( (y - 3)(y + 3) \).

The difference of squares technique is fast and straightforward. It’s a powerful tool that helps simplify equations quickly, transforming what looks like a complicated polynomial into a multiplication problem. Once factored, it becomes much easier to apply the zero product property and find the solutions. Using the difference of squares can often save time and reduce errors in solving polynomial equations.