Problem 26 - Algebra 1: Concepts and Skills

Question

Factor the expression. \(x^{2}+2 x+x y+2 y\)

Step-by-Step Solution

Answer

The factored form of the expression is \((x + 2)(x + y)\).

Step 1: Identify Common Factors

First, look at the terms in the expression and identify any common factors. Here, we can separate the expression into two groups: \(x^{2} + 2x\) and \(xy + 2y\) which both have common factors of x and y, respectively.

Step 2: Apply the Distributive Property in Reverse

Apply the distributive property in reverse to factor out the common factors from each group. For the first group, factor out x so we have \(x(x + 2)\). For the second group, factor out y so we have \(y(x + 2)\).

Step 3: Factor Out the Common Binomial

Lastly, notice that \(x + 2\) is a common factor in both expressions. Factor this out to get \((x + 2)(x + y)\).

Key Concepts

Algebraic Expressions

Algebraic expressions are combinations of variables and constants joined by mathematical operators such as addition, subtraction, multiplication, and division. In simpler terms, these are like mathematical phrases that can be simplified or manipulated. For example, in the exercise provided, the expression is \(x^{2} + 2x + xy + 2y\).

Every term in this expression is an individual part, separated by addition or subtraction signs. Here, \(x^{2}\), \(2x\), \(xy\), and \(2y\) are separate terms:

• The first term, \(x^{2}\), consists of just one variable raised to the second power.
• The second term, \(2x\), has a coefficient (2) and a variable (x).
• The third term, \(xy\), includes two variables multiplied together.
• The last term, \(2y\), like \(2x\), includes a coefficient and a variable.

Understanding the parts of algebraic expressions helps in simplifying and factoring them effectively.

Common Factor

Finding the common factor is a crucial initial step in the process of factoring algebraic expressions. A common factor is a number or expression that divides each term of the expression completely.

In the given problem, we broke down the expression into two groups: \(x^{2} + 2x\) and \(xy + 2y\). From these:

• The first group, \(x^{2} + 2x\), has a common factor of \(x\) because both terms contain the variable \(x\).
• The second group, \(xy + 2y\), exhibits a common factor of \(y\) because both terms include the variable \(y\).

By identifying and factoring out the common factor from each group, we simplify the expression and prepare it for further factoring. This step significantly simplifies the algebraic manipulation in the following stages.

Distributive Property

The distributive property is a key operation in algebra that allows us to multiply a single number by each of the terms inside a parenthesis. The reverse application of this property is used to factor expressions.

In the expression \(x^{2} + 2x + xy + 2y\), we apply the distributive property in reverse:

• From the group \(x^{2} + 2x\), we factor out \(x\) by using the property in reverse, resulting in \(x(x + 2)\).
• Similarly, in the group \(xy + 2y\), we factor out \(y\), yielding \(y(x + 2)\).

This property shows that multiplication can be distributed over addition, and by reversing it, we gather similar terms together to simplify the expression. This makes it easier to spot additional factors.

Binomial Factoring

Binomial factoring involves expressing a polynomial as the product of two simpler binomial expressions.

After factoring out common factors as seen in the exercise, we use binomial factoring. In the expression \(x(x + 2) + y(x + 2)\), the binomial \(x + 2\) appears in both terms.

This allows us to factor \(x + 2\) out completely:

• We take \(x + 2\) as the common binomial factor, and the remaining factors are \(x\) and \(y\), forming the expression \((x + 2)(x + y)\).

By doing this, we fully factor the original expression into a product of simpler two binomials. This final step condenses the expression into a more manageable form, illustrating the power and utility of factoring in algebra.