Problem 12 - Algebra 1: Concepts and Skills

Question

Use the zero-product property to solve the equation. \((y+9)(y-2)(y-5)=0\)

Step-by-Step Solution

Answer

The solutions to the equation are \(y=-9\), \(y=2\), and \(y=5\).

Step 1: Identify the Factors

Identify the factors in the equation \((y+9)(y-2)(y-5)=0\). Here, the factors are \(y+9\), \(y-2\), and \(y-5\).

Step 2: Apply the Zero Product Property

Apply the zero product property which states that if a product of multiple factors is zero, then at least one of the factors must be zero. Therefore, set each factor equal to zero.

Step 3: Solve for 'y' in each equation

Once each factor is set to zero, we are left with three equations to solve for y: \n1) \(y+9=0\) solving for y gives \(y=-9\)\n2) \(y-2=0\) solving for y gives \(y=2\)\n3) \(y-5=0\) solving for y gives \(y=5\)

Key Concepts

Quadratic Equations

Quadratic equations are a special group of polynomials where the highest exponent of the variable is 2. These equations typically take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. In our exercise, the equation is already neatly factored as \((y+9)(y-2)(y-5)=0\), showcasing another way quadratics can appear through their factors directly.

Quadratic equations often involve factoring, which is the process of breaking down the equation into simpler "factors" that, when multiplied, return the original equation. The beauty of quadratics is their versatility and the variety of methods available for solving them, such as completing the square, using the quadratic formula, or factoring. The choice of method depends on the specific equation and its complexity.

Factoring

Factoring is a crucial skill when working with quadratic equations, as it simplifies them and makes solving possible. It involves expressing the equation as a product of its factors. In this case, the equation \((y+9)(y-2)(y-5)=0\) is already factored.

To factor a quadratic equation, you look for two expressions that multiply to give the original quadratic. For example, you might factor \(x^2 + 5x + 6\) into \((x+2)(x+3)\). These expressions are the components that, when multiplied together, recreate the quadratic's original form.

One goal of factoring is to transform the equation into a more workable state where the zero-product property can be applied, assisting in solving the equation. Successful factoring leads directly to simple equations that are easily solved by setting each factor equal to zero.

Solving Equations

Once quadratic equations are factored, solving them becomes straightforward using the zero-product property. This property is key here — it states that if the product of several factors equals zero, at least one of these factors must be zero.

In practice, this means you can set each factor from the equation \((y+9)(y-2)(y-5)=0\) equal to zero to find the solutions for \(y\). Here’s how it looks:

• For \(y+9=0\), solve to get \(y=-9\).
• For \(y-2=0\), solve to get \(y=2\).
• For \(y-5=0\), solve to get \(y=5\).

Each solution represents a point where the product of the entire equation equals zero, making it the solution of the quadratic equation. Solving these equations is akin to identifying where the function intersects the x-axis, representing its roots.