Problem 43 - College Algebra with Modeling and Visualization

Question

Factor the expression completely. \(6 x^{2}-x-2\)

Step-by-Step Solution

Answer

\(6x^2 - x - 2 = (2x + 1)(3x - 2)\).

Step 1: Identify the Form of the Quadratic

The expression \(6x^2 - x - 2\) is a quadratic trinomial of the form \(ax^2 + bx + c\). Here, \(a = 6\), \(b = -1\), and \(c = -2\). Our goal is to express this quadratic in the form \((px + q)(rx + s)\).

Step 2: Multiply a and c

Multiply the coefficient \(a\) and the constant term \(c\). In this case, \(a = 6\) and \(c = -2\), so we calculate \(6 \times (-2) = -12\). We need to find two numbers that multiply to \(-12\) and add to \(b = -1\).

Step 3: Find the Numbers

Look for two numbers that multiply to \(-12\) and add up to \(-1\). These numbers are \(-4\) and \(3\) because \(-4 \times 3 = -12\) and \(-4 + 3 = -1\).

Step 4: Rewrite the Middle Term

Rewrite the middle term \(-x\) using the numbers found: \(-4x + 3x\). The expression becomes \(6x^2 - 4x + 3x - 2\).

Step 5: Group and Factor by Grouping

Group the terms: \((6x^2 - 4x) + (3x - 2)\). Factor out the greatest common factor in each group:- For \(6x^2 - 4x\), take out \(2x\) to get \(2x(3x - 2)\).- For \(3x - 2\), there is no common factor, so it stays \(1(3x - 2)\).This can be rewritten as \( (2x + 1)(3x - 2) \) by factoring out the common binomial \((3x - 2)\).

Step 6: Verify the Factoring

Multiply \((2x + 1)\) and \((3x - 2)\) to ensure that it equals the original expression:\((2x + 1)(3x - 2) = 2x(3x) + 2x(-2) + 1(3x) + 1(-2) = 6x^2 - 4x + 3x - 2 = 6x^2 - x - 2\). The original expression is correctly factored.

Key Concepts

Polynomial Factoring

Factoring polynomials can be thought of as finding a product of simpler expressions that together equal the original polynomial. This is a useful tool in algebra as it simplifies complex expressions, enabling us to solve equations more easily. When we factor polynomials, we are essentially breaking them down into their building blocks. For instance, a quadratic polynomial like the one in the exercise, which is expressed in the standard form \(ax^2 + bx + c\), can often be decomposed into the product of two binomials like \((px + q)(rx + s)\).

Breaking down the expression involves recognizing the numbers \(p, q, r, \) and \(s\), such that their product and sum correspond to certain parts of the original quadratic's coefficients. Factoring reveals these relations, making it easier to solve for \(x\) in equations or simply to simplify the expression further. This process of simplifying and breaking down helps make calculations manageable, especially when dealing with higher degree polynomials.

Quadratic Trinomials

A quadratic trinomial is a polynomial with exactly three terms, typically written in the form \(ax^2 + bx + c\). In these expressions, the exponent of \(x\) decreases with each term, making them easy to identify. When factoring quadratic trinomials, the main goal is to find two binomials whose product equals the original trinomial. This approach simplifies solving or further manipulating the expression.

For instance, in our exercise with the expression \(6x^2 - x - 2\), the coefficients are \(a = 6\), \(b = -1\), and \(c = -2\). By looking for two numbers that multiply to \(a \times c\) (in this case, \(-12\)) and add to \(b\), we can split the middle term. The process involves finding such numbers, breaking the middle term into these numbers, and then grouping and factoring by grouping. Understanding the structure of quadratic trinomials makes factoring less daunting and more systematic.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. They form the basis of algebra and offer ways to represent various mathematical situations symbolically. The expression \(6x^2 - x - 2\) is an algebraic expression consisting of terms of varying powers of \(x\).

To manage these expressions, we use operations like addition, subtraction, multiplication, and factoring. Factoring is particularly crucial as it transforms expressions into a more manageable form. This manipulation is not just about simplifying expressions, but it also aids in solving algebraic equations, including finding roots and intercepts.

By successfully factoring an expression into simpler components, such as the result \((2x + 1)(3x - 2)\) from our problem, we reveal insights about the expression's behavior and its possible solutions. Engaging deeply with these expressions builds a strong foundation for tackling more advanced algebraic concepts and applications.