Problem 27 - Elementary Algebra

Question

For the following problems, solve the equations, if possible. $$ y(5 y+2)(2 y-1)=0 $$

Step-by-Step Solution

Answer

Answer: The values of y that make the equation true are \(y = 0\), \(y = \frac{-2}{5}\), and \(y = \frac{1}{2}\).

Step 1: Identify the factors

In the given equation, \(y(5y + 2)(2y - 1) = 0\), we have three factors: \(y\), \((5y+2)\), and \((2y-1)\). We will set each of these factors equal to zero and solve for y in each case.

Step 2: Set first factor equal to zero and solve

Set the first factor equal to zero: \(y=0\). We can see that the first solution for y is \(y=0\).

Step 3: Set second factor equal to zero and solve

Set the second factor equal to zero: \(5y + 2 = 0\). To solve this equation for y, we first subtract 2 from both sides of the equation: \(5y = -2\). Then, divide both sides by 5: \(y = \frac{-2}{5}\).

Step 4: Set third factor equal to zero and solve

Set the third factor equal to zero: \(2y - 1 = 0\). To solve this equation for y, we first add 1 to both sides of the equation: \(2y = 1\). Then, divide both sides by 2: \(y = \frac{1}{2}\). In conclusion, the solutions to the given equation are \(y = 0\), \(y = \frac{-2}{5}\), and \(y = \frac{1}{2}\).

Key Concepts

Factoring

Factoring is a fundamental concept in algebra that involves expressing a polynomial as a product of simpler terms, known as factors. In the exercise, the equation given is already factored as \(y(5y + 2)(2y - 1) = 0\). Recognizing these factors is crucial because it simplifies the process of finding the solution to the equation.

When you factor an equation, you break it down into components that, when multiplied together, give you the original equation. This process helps simplify solving because each factor can be treated as an individual equation equal to zero. Once factored completely, you can use the zero product property to find the solution.

• Factoring can help reduce complex equations into manageable parts.
• Always look for common factors before branching out into grouping terms.
• Factored equations are essential for leveraging the zero product property.

Understanding how to factor correctly can make solving polynomial equations much more straightforward.

Zero Product Property

The zero product property is a key principle in algebra that makes solving factored equations much easier. It states that if the product of two or more factors is zero, then at least one of the factors must be equal to zero. This is because multiplying any number by zero results in zero.

In the exercise, we have three factors: \(y\), \((5y + 2)\), and \((2y - 1)\). According to the zero product property, you set each factor equal to zero to find the possible solutions. This gives you the equations \(y = 0\), \(5y + 2 = 0\), and \(2y - 1 = 0\). Solving each of these will give you the values of \(y\) that satisfy the original equation.

• The zero product property helps isolate solutions quickly.
• This property turns complex polynomial equations into single-variable linear equations.
• Every separate equation derived from setting each factor to zero contributes to finding all solutions.

Applying the zero product property reduces the complexity of polynomial equations by exploiting the power of zero.

Solving Equations

Solving algebraic equations involves finding all possible values of the variable that make the equation true. With polynomial equations like the one in the exercise, the equation's factored form can tell you a lot about potential solutions.

Once you have set each factor to zero using the zero product property, the next step is solving those simpler equations. From the exercise, the factors were \(y = 0\), \(5y + 2 = 0\), and \(2y - 1 = 0\). Each of these is a linear equation that can be solved simply:

• To solve \(y = 0\), the solution is evident as \(y = 0\).
• For \(5y + 2 = 0\), you isolate \(y\) by subtracting 2 and then dividing by 5, resulting in \(y = \frac{-2}{5}\).
• For \(2y - 1 = 0\), you add 1 and divide by 2, finding \(y = \frac{1}{2}\).

Solving each as outlined brings you to the conclusion that the solutions are \(y = 0\), \(y = \frac{-2}{5}\), and \(y = \frac{1}{2}\). These steps are a crucial aspect of solving algebraic equations and are vital for understanding how to handle these types of problems effectively.