Problem 33 - Elementary Algebra

Question

For the following problems, factor the trinomials when possible. $$ b^{2}+15 b+56 $$

Step-by-Step Solution

Answer

Answer: \((b + 7)(b + 8)\)

Step 1: Identify the factors of the last term

List all possible pairs of factors of 56 (the constant term). Remember to include both positive and negative factors since the signs could be different. The pairs are: 1. \((1, 56)\) 2. \((-1, -56)\) 3. \((2, 28)\) 4. \((-2, -28)\) 5. \((4, 14)\) 6. \((-4, -14)\) 7. \((7, 8)\) 8. \((-7, -8)\)

Step 2: Determine which pair of factors adds up to the middle term

Out of the pairs listed above, find the pair that adds up to the middle term, which is 15 in this case. We can see that only \((7, 8)\) adds up to 15. So, we rewrite the middle term, \(15b\), as the sum of these two factors multiplied by b: \(7b + 8b\)

Step 3: Rewrite the trinomial with the factored middle term

Now, rewrite the given trinomial replacing the middle term with the two terms found in Step 2: $$ b^2+15b+56 \Rightarrow b^2 + 7b + 8b + 56 $$

Step 4: Factor by grouping

To factor by grouping, split the terms into two groups and factor out the greatest common factor (GCF) from each group: $$ (b^2 + 7b) + (8b + 56) $$ Now, factor out the GCF from each group: $$ b(b + 7) + 8(b + 7) $$

Step 5: Factor out common binomial

We can see that both terms have a common binomial factor of \((b + 7)\). Factor this out and write the factored form of the given trinomial: $$ (b + 7)(b + 8) $$ The given trinomial, \(b^2 + 15b + 56\), factors into \((b + 7)(b + 8)\).

Key Concepts

Polynomial Factoring

When we talk about polynomial factoring, we refer to the process of breaking down a polynomial into simpler components, or "factors," which when multiplied together give back the original polynomial. This is a key concept in algebra, as it allows us to simplify and solve polynomial equations more easily. For instance, a trinomial is a polynomial with three terms, and a common task is to factor them. In our exercise, the polynomial is a trinomial of the form \(b^2 + 15b + 56\).

Factoring trinomials requires identifying pairs of factors for the constant term that can combine to replace the middle term. This involves both multiplication to match the constant term and addition to match the linear coefficient. This systematic approach to factoring makes it easier to manage complex expressions by reducing them into products of simpler polynomials, which can be directly multiplied to check the correctness of the factorization.

Factoring by Grouping

Factoring by grouping is a powerful method used in algebra to simplify polynomial expressions, especially useful for trinomials that can be divided into two binomials. In the current problem, once we have rewritten the middle term through suitable factor pairs like \(7b + 8b\), the expression becomes a four-term polynomial: \(b^2 + 7b + 8b + 56\).

The core idea here is to split the polynomial into two groups of terms: \((b^2 + 7b)\) and \((8b + 56)\). By factoring out the greatest common factor (GCF) from each group, we can find common terms that further simplify the polynomial. This results in an expression like \(b(b + 7) + 8(b + 7)\), where \(b + 7\) is the shared binomial factor.

• First, group terms to highlight common factors.
• Next, factor out the GCF from each group.
• Finally, factor out the common binomial factor from the grouped expression.

This method not only aids in simplifying polynomials but also in resolving complex algebraic expressions in a structured way.

Algebraic Expressions

Algebraic expressions are at the heart of algebra, representing a combination of numbers, variables, and operations. They can take many forms, such as polynomials, which include terms with variables raised to different powers. Understanding the structure of algebraic expressions is essential when performing operations such as factoring.

In expressions like \(b^2 + 15b + 56\), recognizing the relationship between coefficients and terms allows us to perform operations like factoring efficiently. The structure of algebraic expressions dictates how we manipulate them:

• Identify terms: Determine what constitutes a term and how each term relates to the other through addition or subtraction.
• Understand coefficients: The numbers in front of variables that determine the magnitude of each term.
• Recognize patterns: In the context of trinomials, recognizing standard forms and patterns helps streamline the factoring process.

Mastering how to read and interpret these expressions is fundamental in solving equations and understanding algebra, allowing students to apply this knowledge to more complex mathematical problems in the future.