Problem 71 - Beginning and Intermediate Algebra

Question

Factor by grouping. $$5 t u+6 t-5 u-6$$

Step-by-Step Solution

Answer

The given expression \(5tu + 6t - 5u - 6\) can be factored by grouping as follows: 1. Group the terms: \((5tu + 6t) - (5u + 6)\) 2. Find common factors in each group: \(t(5u + 6) - (5u + 6)\) 3. Factor by grouping: \((5u + 6)(t - 1)\) So, the factored expression is \((5u + 6)(t - 1)\).

Step 1: Group the terms

Identify the terms that can be grouped together in a reasonable way. In our case, it is appropriate to group the first two terms and the last two terms, like this: $$5tu + 6t - 5u - 6$$ becomes $$(5tu + 6t) - (5u + 6)$$

Step 2: Find common factors in each group

Now, find the common factors in each group. For the first group (5tu + 6t): The common factor is \(t\), so we can factor it out: $$t(5u + 6)$$ For the second group (5u + 6): There are no common factors, so we can rewrite it as: $$-(5u + 6)$$

Step 3: Write the expression with factored groups

Now that we have factored each group, let's write the entire expression: $$t(5u + 6) - (5u + 6)$$

Step 4: Factor by grouping

Finally, we can see that the expression contains the common factor (5u + 6) in both terms. Let's factor this term out: $$(5u + 6)(t - 1)$$ Now, the given expression has been factorized by grouping. The final factored form is: $$(5u + 6)(t - 1)$$

Key Concepts

Algebraic Expressions

An algebraic expression is a combination of variables, numbers, and operations (such as addition, subtraction, multiplication, and division). It does not have an equality sign like equations do. These expressions can represent real-world quantities and make problem-solving easier by establishing a formula you can work with. For example, in the expression \(5tu + 6t - 5u - 6\), each term consists of numbers and variables like \(t\) and \(u\), combined using operations:

• \(5tu\): represents 5 times \(t\) times \(u\)
• \(+6t\): represents 6 times \(t\)
• \(-5u\): represents -5 times \(u\)
• \(-6\): a constant term

Understanding algebraic expressions is a key component in identifying patterns, creating formulas, and solving algebra problems.

Polynomial Factorization

Polynomial factorization is the process of expressing a polynomial as a product of its factors. Factors are simpler polynomials that, when multiplied together, give the original polynomial. This technique simplifies algebraic expressions and solves polynomial equations by finding roots more easily. For the expression \(5tu + 6t - 5u - 6\), factorization helps us break it down into the form of \((5u + 6)(t - 1)\), making it simpler to analyze or solve:

• Identify terms that can group together and form factors, optimizing the process.
• Use the distributive property in reverse to find common factors and simplify the expression.

Ultimately, factorization is crucial for algebra, enabling deeper insights into algebraic structures and equations.

Common Factors

Common factors are terms that are shared by different parts of an expression. Identifying common factors simplifies algebraic expressions, making them easier to work with. In our problem, we identify a common factor in the grouped terms of the polynomial. Here’s how we identify and use them:

• For the group \(5tu + 6t\), the common factor is \(t\).
• By factoring \(t\) out, the group simplifies to \(t(5u + 6)\).
• The expression \( -(5u + 6) \) already contains the structure of a common factor.

Factoring out the common \((5u + 6)\) from the entire polynomial brings it to \((5u + 6)(t - 1)\), which clearly shows the simplification process. Listening to the "music" of common factors helps us harmonize algebraic expressions and streamline calculations.

Algebraic Manipulation

Algebraic manipulation involves rearranging, simplifying, or solving expressions using algebraic rules and operations. This essential skill allows for transforming complex expressions into simpler forms. For factorization by grouping, algebraic manipulation is crucial:

• Start with grouping terms: In the expression \(5tu + 6t - 5u - 6\), we create groups \((5tu + 6t)\) and \(-(5u + 6)\).
• Next, factor common elements within each group, finding \(t(5u + 6)\) in one group.
• Lastly, recognize the shared factor \((5u + 6)\) and factor it out of the full expression, resulting in \((5u + 6)(t - 1)\).

Algebraic manipulation enables us to rearrange expressions for ease of computation, bringing clarity and simplicity to what might initially seem complex.