Problem 73 - Elementary and Intermediate Algebra

Question

Factor. If an expression is prime, so indicate. $$ 10 u^{2}-13 u-6 $$

Step-by-Step Solution

Answer

\((5u + 2)(2u - 3)\) is the factored form.

Step 1: Identify the Quadratic Expression

Given the quadratic expression is \(10u^2 - 13u - 6\). This is in the form \(ax^2 + bx + c\), where \(a = 10\), \(b = -13\), and \(c = -6\).

Step 2: Use the AC Method

Multiply \(a\) and \(c\): \(10 \times -6 = -60\). We need two numbers that multiply to \(-60\) and add to \(-13\).

Step 3: Find the Pair of Numbers

The numbers \(-15\) and \(2\) multiply to \(-60\) and add to \(-13\).

Step 4: Rewrite the Expression

Rewrite \(-13u\) using the pair found: \(10u^2 - 15u + 2u - 6\).

Step 5: Factor by Grouping

Group terms: \((10u^2 - 15u) + (2u - 6)\). Factor each group: \(5u(2u - 3) + 2(2u - 3)\).

Step 6: Factor Out the Common Factor

Factor out the common binomial \((2u - 3)\): \((5u + 2)(2u - 3)\).

Step 7: Verify the Solution

Expand \((5u + 2)(2u - 3)\) to check: \(5u \cdot 2u - 5u \cdot 3 + 2 \cdot 2u - 2 \cdot 3\) which simplifies to \(10u^2 - 15u + 4u - 6 = 10u^2 - 13u - 6\).

Key Concepts

AC method

The AC method is a popular technique used to factor quadratic expressions. It is especially useful when dealing with a trinomial where the coefficient of the quadratic term, denoted as \( a \), is not equal to 1. This method involves several important steps:

• First, multiply the coefficient \( a \) of the squared term and the constant term \( c \). In the exercise example, it would be \( 10 \times -6 = -60 \).
• Next, look for a pair of numbers that will multiply to this product (\(-60\)) and add up to the linear coefficient \( b \). Here, you need numbers that both multiply to \(-60\) and sum to \(-13\).

Through practice, you will learn to quickly spot such pairs. In this case, those numbers are \(-15\) and \(2\). This step is crucial because finding the correct pair allows you to decompose the middle term efficiently.

factoring by grouping

Factoring by grouping is an effective technique to handle expressions once they have been suitably rearranged. It's more like organizing the expression into parts that are easier to factor further. Here's a breakdown:

• After determining the pair of numbers through the AC method, rewrite the middle term using these numbers. So, for \(10u^2 - 13u - 6\), rewrite \(-13u\) as \(-15u + 2u\).
• This changes our expression to \(10u^2 - 15u + 2u - 6\).
• Next, group the expression into two pairs: \((10u^2 - 15u) + (2u - 6)\).

Now we can factor each group separately. For \(10u^2 - 15u\), the common factor is \(5u\), leaving us with \(5u(2u - 3)\). For \(2u - 6\), the factor is \(2\), resulting in \(2(2u - 3)\). Since \(2u - 3\) is common in both groups, you can factor it out to get \((5u + 2)(2u - 3)\). This process effectively reduces the complexity of the initial equation, making it easier to solve or graph.

verify factorization

Verification of factorization is critical to ensure that the steps taken actually reconstruct the original expression. Essentially, you're performing a check to catch any potential mistakes. Here’s how:

• After obtaining the factors \((5u + 2)(2u - 3)\), expand them to see if they simplify back to the original expression.
• Distribute each term: \(5u \cdot 2u = 10u^2\), \(5u \cdot -3 = -15u\), \(2 \cdot 2u = 4u\), and \(2 \cdot -3 = -6\).

Now, combine these terms: \(10u^2 - 15u + 4u - 6\). Simplifying, \(10u^2 - 11u - 6\) adds \(-15u\) and \(4u\) to yield \(-13u\), returning you to the original equation, \(10u^2 - 13u - 6\).This verification step is your safety net. It reveals whether the factorization is correct or whether an error sneaked into the process. A mistake here and uncovering it means you can fix it, building confidence in your algebra skills.