Problem 30 - Elementary and Intermediate Algebra

Question

Factor. See Example 2 or Example 6. $$ 15 a^{2}-28 a+5 $$

Step-by-Step Solution

Answer

The expression factors to \((5a - 1)(3a - 5)\).

Step 1: Identify the Quadratic Form

The expression is a quadratic in the form of \( ax^2 + bx + c \), where \( a = 15 \), \( b = -28 \), and \( c = 5 \).

Step 2: Multiply and Identify the Middle Term

Multiply \( a \) and \( c \): \( 15 \times 5 = 75 \). We need two numbers whose product is 75 and whose sum is \( b = -28 \).

Step 3: Choose Suitable Numbers

The numbers that satisfy these conditions are \(-25\) and \(-3\), since \((-25) \times (-3) = 75\) and \((-25) + (-3) = -28\).

Step 4: Rewrite the Middle Term

Rewrite the expression as: \( 15a^2 - 25a - 3a + 5 \).

Step 5: Factor by Grouping

Group the terms as: \((15a^2 - 25a) + (-3a + 5)\). Factor out the common factors: \( 5a(3a - 5) - 1(3a - 5) \).

Step 6: Factor the Common Binomial

Extract the common binomial factor \((3a - 5)\): \( (5a - 1)(3a - 5) \).

Key Concepts

Quadratic Forms

Quadratic forms are fundamental when dealing with polynomial expressions, especially in factoring problems. A quadratic expression typically follows the format \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. In the exercise, for the expression \( 15a^2 - 28a + 5 \), we identify that it's in quadratic form because it fits perfectly into this format:

• \( a = 15 \)
• \( b = -28 \)
• \( c = 5 \)

Recognizing a quadratic form is essential because it determines the subsequent steps for factoring. Quadratics are often set to zero for solving equations or manipulated for simplification, depending on the desired outcome.

Factor by Grouping

Factor by grouping is one of the intuitive techniques used to simplify quadratic expressions. This method involves rearranging and regrouping terms to make it easier to factor out common elements. Once you've rewritten the middle term based on a chosen pair of numbers, you'll rearrange the expression to facilitate grouping. Consider the exercise where we expanded the middle term of \( 15a^2 - 28a + 5 \) into \( 15a^2 - 25a - 3a + 5 \).
Here's the process broken down:

• Group terms to identify common elements, \((15a^2 - 25a) + (-3a + 5)\).
• Factor out the greatest common factor from each group: \(5a(3a - 5) - 1(3a - 5)\).

After grouping and factoring, you arrive at a point where you can further simplify by extracting the common binomial \((3a - 5)\), resulting in \((5a - 1)(3a - 5)\). This technique is efficient and often the go-to strategy for trinomials that can be reorganized into simpler products of binomials.

Polynomial Expressions

Polynomial expressions, which include quadratics, play a crucial role in algebra. A polynomial is essentially a sum of terms with varying degrees, each involving powers of a variable. The focus is often on simplifying these expressions or solving associated equations through factoring methods.
The original polynomial expression \( 15a^2 - 28a + 5 \) presents a challenge that many students face. The key is understanding how different parts of the polynomial can interact based on their coefficients and terms.
The entire process of factoring addresses the nature of polynomials:

• Identify and arrange terms properly.
• Apply suitable factoring methods such as grouping.
• Simplify the polynomial by reducing it to lower-degree factors wherever possible.

Handling polynomial expressions becomes straightforward by applying these organized steps. It allows students to transform complex expressions into more manageable ones, providing a deeper comprehension of the inner workings of algebra.