Problem 55 - Beginning and Intermediate Algebra

Question

Factor completely, if possible. Begin by asking yourself, "Can I factor out a GCF?" $$2 r^{2}+8 r+6$$

Step-by-Step Solution

Answer

The completely factored expression is \(2(r + 1)(r + 3)\).

Step 1: Determine GCF

Identify the Greatest Common Factor among the coefficients. The coefficients are 2, 8, and 6. The GCF of these numbers is 2.

Step 2: Factor out the GCF

Divide each term of the expression by the GCF, 2, and rewrite the expression: \(2r^2 + 8r + 6 = 2(r^2 + 4r + 3)\)

Step 3: Factor the quadratic expression

Factor the quadratic expression in the parentheses using the trial and error method or factoring by grouping. In this case, we will use trial and error to find two binomials that multiply to the quadratic expression: \((r^2 + 4r + 3) = (r + 1)(r + 3)\)

Step 4: Write the completely factored expression

Combine the GCF from Step 2 and the factored quadratic expression from Step 3 to write the completely factored expression: \(2r^2 + 8r + 6 = 2(r + 1)(r + 3)\)

Key Concepts

Greatest Common Factor

To begin factoring a quadratic, it's crucial to check if there is a Greatest Common Factor (GCF) that you can factor out. The GCF is the largest number that divides all the coefficients of the terms in the polynomial without leaving a remainder. By factoring out the GCF, you simplify the quadratic, making further factoring steps easier. For example, in the expression \(2r^2 + 8r + 6\), the coefficients are 2, 8, and 6. The highest number that can divide all these coefficients is 2. Factoring the GCF of 2 from the expression results in the simpler quadratic \(2(r^2 + 4r + 3)\). Always begin with this step to simplify your expressions.

Trial and Error Method

After factoring out the GCF, the expression \(r^2 + 4r + 3\) remains. A common approach to factor such expressions is the trial and error method. This technique involves guessing pairs of numbers that multiply to give the constant term and checking if their sum equals the middle coefficient. - For the expression \(r^2 + 4r + 3\), the numbers should multiply to 3 (the constant term) and add up to 4 (the coefficient of the middle term).- You can try pairs like (1,3) and find that \((r + 1)(r + 3)\) works.This method relies on a bit of practice and intuition, as you test possible combinations.

Factoring by Grouping

Sometimes, the trial and error method might not be readily applicable, and factoring by grouping offers a viable alternative. Although it's not used in our specific example, understanding it enriches your toolkit for different scenarios. Factoring by grouping works well for four-term polynomials. - You begin by splitting the middle term into two terms so the expression has four terms. - Then, you group pairs of terms and factor them separately, looking for a common factor in each group. - Finally, if done correctly, there should be a common binomial factor that emerges. This method provides a structured approach for more complex quadratics that are not easily guessed with trial and error.

Binomial Expressions

Binomial expressions are mathematical expressions containing two terms. For example, each factor in the fully factored quadratic like \((r + 1)(r + 3)\) is a binomial expression. When factoring quadratics, you often aim to rewrite the expression as a product of binomials. This rewriting simplifies solving the expression because setting each binomial equal to zero allows you to find the roots of the original quadratic equation. Understanding how to form binomials from factored terms is key in algebra, laying the groundwork for solving equations and deeper algebraic manipulations.