Problem 73 - Algebra 2

Question

Find each product. $$ (y-2)(y+5) $$

Step-by-Step Solution

Answer

The product of the binomials is \(y^2 + 3y - 10\).

Step 1: Identify the Expression

We need to find the product of two binomials: \((y-2)\) and \((y+5)\). This is typically done using the distributive property, often referred to as the FOIL method when dealing with binomials.

Step 2: Apply the FOIL Method

FOIL stands for First, Outer, Inner, and Last, referring to the terms we will multiply. For the binomials \((y-2)(y+5)\), we calculate:- **First:** Multiply the first terms of each binomial: \(y imes y = y^2\).- **Outer:** Multiply the outer terms: \(y imes 5 = 5y\).- **Inner:** Multiply the inner terms: \(-2 imes y = -2y\).- **Last:** Multiply the last terms: \(-2 imes 5 = -10\).

Step 3: Combine Like Terms

Now, we add all the products we have from applying the FOIL method:\[ y^2 + 5y - 2y - 10 \]Combine the like terms, which are the middle terms in this case \(5y - 2y\). So the expression becomes:\[ y^2 + 3y - 10 \]

Step 4: Write the Final Expression

The product of the binomials \((y-2)(y+5)\) simplifies to the quadratic expression \(y^2 + 3y - 10\).

Key Concepts

Distributive Property

The distributive property is a foundational concept in algebra that allows us to simplify expressions by distributing a single value across terms within parentheses. When we talk about the distributive property, it essentially states that for any numbers or expressions \( a \), \( b \), and \( c \), the equality \( a(b+c) = ab + ac \) holds true. This property is particularly useful when dealing with binomials, such as in the expression \((y-2)(y+5)\).
In the context of multiplying binomials, we can extend the distributive property to systematically multiply each term from one binomial with each term in the other binomial, which is exactly what the FOIL method accomplishes.

• First: Multiply the first terms from each binomial.
• Outer: Multiply the outer terms.
• Inner: Multiply the inner terms.
• Last: Multiply the last terms from each binomial.

By applying the distributive property through these steps, we ensure that all combinations of terms are multiplied, leading us to the next step of simplifying the expression.

Combining Like Terms

Once we have used the distributive property to multiply the terms in our binomials, we are often left with a longer expression that can be further simplified. To simplify it, we need to combine like terms. Like terms are terms within an expression that have the same variable raised to the same power.
For example, in the expression we derived from our binomials multiplication: \[ y^2 + 5y - 2y - 10 \] we can identify the like terms:

• \( 5y \) and \( -2y \) are like terms because they both contain the variable \( y \) raised to the first power. By combining them, we perform the operation \( 5y - 2y = 3y \), simplifying our expression further.

This simplification is crucial as it condenses the expression to its simplest form, making it easier to work with and understand. Combining like terms is a critical step when dealing with any polynomial expressions, especially in order to prepare it or solve it for different operations.

Quadratic Expression

A quadratic expression is a type of polynomial expression where the highest power of the variable is 2. The general form of a quadratic expression is \( ax^2 + bx + c \) where \( a \), \( b \), and \( c \) are constants. In our simplified expression coming from the product \((y-2)(y+5)\), we have the quadratic expression:\[ y^2 + 3y - 10 \] Here, \( a = 1 \), \( b = 3 \), and \( c = -10 \).
Quadratic expressions are commonly encountered in algebra and have a range of applications from calculating areas to solving for values in physics problems. They often lead to quadratic equations when set equal to zero or another value, which can be solved using different methods like factoring, completing the square, or using the quadratic formula.
Understanding how to recognize and simplify quadratic expressions is a crucial skill in math, helping students get a better grasp on a variety of mathematical problems and equations they may encounter.