Question

What is the solution set to the equation \(0=(3 a+1)^2(a-4)\) ? A) \(\left\\{\frac{1}{3},-4\right\\}\) ?) \(\left\\{-\frac{1}{3}, 4\right\\}\) C) \(\left\\{-\frac{1}{3}, \frac{1}{3},-4\right\\}\) D) \(\left\\{-\frac{1}{3}, \frac{1}{3}, 4\right\\}\)

Short answer

The solution set to the equation \(0=(3 a+1)^2(a-4)\) is B) \(\left\{-\frac{1}{3}, 4\right\}\).

Step-by-step solution

Identify the equation

The given equation is: \(0 = (3a + 1)^2(a - 4)\)

Utilize the zero-product property

According to the zero-product property, if the product of two or more factors is zero, then at least one of the factors must be zero. In this case, we have two factors: \((3a + 1)^2\) and \((a - 4)\). So, either \((3a + 1)^2 = 0\) or \((a - 4) = 0\).

Solve for \(a\) in the first factor

To solve the equation \((3a + 1)^2 = 0\), take the square root of both sides: \[\sqrt{(3a + 1)^2} = \sqrt{0}\] \[3a + 1 = 0\] Now, solve for \(a\): \[3a = -1\] \[a = -\frac{1}{3}\]

Solve for \(a\) in the second factor

To solve the equation \((a - 4) = 0\), simply add 4 to both sides: \[a = 4\]

Write the solution set

The values of \(a\) that satisfy the equation are \(-\frac{1}{3}\) and \(4\). So, the solution set is: \(\left\\{-\frac{1}{3}, 4\right\\}\) The correct answer is option B) \(\left\\{-\frac{1}{3}, 4\right\\}\).

Key concepts

Zero-Product Property

The zero-product property is a fundamental principle in algebra that helps solve equations efficiently. When you have an expression set to zero, like in the equation \(0 = (3a + 1)^2(a - 4)\), the zero-product property comes into play. This property states that if the product of several factors equals zero, then at least one of those factors must be zero.

In practical terms, it means you can take each factor of a multiplication that results in zero and set it equal to zero individually. For our example, you set \((3a + 1)^2 = 0\) and \((a - 4) = 0\). Solving these individual equations helps you find the potential solutions to the original equation.

This property can significantly simplify the solving process for quadratic and higher-order polynomial equations, turning a complex problem into a series of simpler linear equations.

Solution Set

A solution set is the complete collection of all possible values that satisfy a given algebraic equation. When dealing with equations like \(0 = (3a + 1)^2(a - 4)\), your task is to identify all the values of the variable that make the equation true. These values are collected into what we call a solution set.

In our original problem, by applying the zero-product property, we found two factors: \(a = -\frac{1}{3}\) and \(a = 4\). These are the numbers that make each factor of the equation equal to zero. Therefore, the solution set for the equation is \(\left\{-\frac{1}{3}, 4\right\}\).

It's essential to ensure the solution set includes all possible solutions, as leaving out any could mean missing a crucial answer in real-world applications or academic exercises.

Algebraic Equations

Algebraic equations are mathematical statements composed of variables, constants, and operators that define a particular relationship. In simplest terms, they say, "these two expressions are equal." They can be linear, quadratic, or higher order in form.

The equation \(0 = (3a + 1)^2(a - 4)\) is a polynomial equation common in algebra, involving multiplication and exponentiation of expressions in the variable \(a\). Such equations are typically solved by isolating the variable on one side or using properties like factoring and the zero-product property.

Understanding how to manipulate and solve algebraic equations is crucial not only for mathematics but also for fields like physics, engineering, and economics, where modeling relationships with variables is common.

Factoring in Algebra

Factoring is a technique in algebra used to break down complex expressions into simpler products that are easier to solve or study. It involves rewriting a polynomial as a product of its factors. For instance, in the given equation \(0 = (3a + 1)^2(a - 4)\), the expression is already factored.

Factoring becomes essential especially in solving quadratic equations. By expressing a quadratic equation as a product of binomials, you can easily use the zero-product property to find solutions.

Here are some important steps in factoring:

• Look for common factors in all terms.
• Identify special binomial products, like difference of squares.
• Use techniques like grouping when applicable.

Mastering factoring techniques can transform complicated algebraic problems into much simpler tasks, paving the way for more advanced mathematical operations.