Problem 5 - Elementary Algebra

Question

In the expression \(8 x^{2}-5 x+6\), list the factors of the first term: second term: third term:

Step-by-Step Solution

Answer

Answer: Factors of the first term \(8x^2\): \((1 \cdot x)(8 \cdot x)\) and \((2 \cdot x)(4 \cdot x)\). Factors of the second term \(-5x\): \((1)(-5 \cdot x)\) and \((-1)(5 \cdot x)\). Factors of the third term \(6\): \((1)(6)\), \((2)(3)\), \((-1)(-6)\), and \((-2)(-3)\).

Step 1: Identify the terms in the expression

In the quadratic expression \(8 x^{2}-5 x+6\), there are three terms: 1) The first term is \(8 x^{2}\), 2) the second term is \(-5 x\), and 3) the third term is \(6\).

Step 2: Factor the first term \(8x^2\)

To find the factors of the first term \(8 x^{2}\), we need to find the numbers that can be multiplied together to get 8 and the expressions that can be multiplied together to get \(x^2\). The factors of 8 are 1 and 8, 2 and 4. The factors of \(x^2\) are x and x. Therefore, the factors of \(8 x^{2}\) are: $$(1 \cdot x)(8 \cdot x) \quad \text{and} \quad (2 \cdot x)(4 \cdot x).$$

Step 3: Factor the second term \(-5x\)

To find the factors of the term \(-5x\), we need to find the numbers that can be multiplied together to get -5 and combined with x. The factors of -5 are 1 and -5, -1 and 5. Therefore, the factors of \(-5x\) are: $$(1)(-5 \cdot x) \quad \text{and} \quad (-1)(5 \cdot x).$$

Step 4: Factor the third term \(6\)

Lastly, we need to find the factors of the constant term 6. The factors of 6 are 1 and 6, 2 and 3, -1 and -6, -2 and -3. Therefore, the factors of 6 are: $$(1)(6), (2)(3), (-1)(-6), \quad \text{and} \quad (-2)(-3).$$ To summarize, the factors of the given expression are: First term (\(8 x^{2}\)): \((1 \cdot x)(8 \cdot x)\) and \((2 \cdot x)(4 \cdot x)\). Second term (\(-5 x\)): \((1)(-5 \cdot x)\) and \((-1)(5 \cdot x)\). Third term (\(6\)): \((1)(6)\), \((2)(3)\), \((-1)(-6)\), and \((-2)(-3)\).

Key Concepts

Algebraic Terms

Algebraic expressions are made up of parts called terms, which are separated by plus (+) or minus (−) signs. In algebra, a term is a product of numbers and variables. For example, in the expression \(8 x^{2}-5 x+6\), there are three distinct algebraic terms: \(8 x^{2}\), \(5 x\), and \(6\).

Breaking down these terms, \(8 x^{2}\) includes the coefficient 8 and the variable \(x\) raised to the second power, \(5 x\) consists of the coefficient 5 and the variable \(x\), and \(6\) is a constant term with no variable attached. It's crucial to understand each term's components—coefficients, variables, and exponents—as they provide the foundation for further operations such as factoring.

Factors of a Term

A factor is a number or algebraic expression that divides another number or expression without leaving a remainder. When factoring terms in an expression, the goal is to break down the term into its simplest building blocks—its factors.

Let's consider the term \(8 x^{2}\). The numerical coefficient 8 can be factored into the pairs of numbers (1 and 8) and (2 and 4). Additionally, the variable part, \(x^{2}\), can be separated into two \(x\)'s because \(x \times x = x^{2}\). Therefore, the complete set of factors for the term \(8 x^{2}\) is \(1 \times 8 \times x \times x\) or \(2 \times 4 \times x \times x\). By identifying the factors of each term in an algebraic expression, students build essential skills for more advanced algebraic manipulations, such as finding the greatest common factor or factoring polynomial expressions.

Polynomial Factors

Polynomials are expressions consisting of multiple terms, and factoring them is about finding the polynomial's divisor that yields other polynomials or algebraic expressions when divided. Factoring quadratic expressions—second-degree polynomials with the form \(ax^{2} + bx + c\)—is a foundational skill.

To factor quadratic expressions, it is crucial to find two numbers whose product equals the coefficient of the constant term \(c\) and whose sum equals the coefficient of the \(x\)-term \(b\). In practice, factoring opens the doors to solving quadratic equations, comparing different forms of expressions, and understanding the properties such as roots and zeros of polynomials.

With careful consideration of these concepts, students can deeply understand and solve a wide range of problems, from simple factoring to applying the Zero Product Property for finding the solutions to polynomial equations.