Problem 27 - Prealgebra and Introductory Algebra

Question

Find the LCM of the polynomials. $$\begin{aligned} &6+x-x^{2}\\\ &x+2\\\ &x-3 \end{aligned}$$

Step-by-Step Solution

Answer

The LCM of the polynomials is \(6 + x - x^2\).

Step 1: Analyzing the polynomials

First, we have to notice that the polynomial \(6 + x - x^2\) can be factorized to \((-1) \cdot (x^2 - x - 6)\), which further simplifies to \((-1) \cdot (x-3)(x+2)\). This implies that both \(x-3\) and \(x+2\) are factors of \(6 + x - x^2\). We also notice that \(x+2\) and \(x-3\) are already in their simplest form, so we can move on to the next step.

Step 2: Finding the LCM

The LCM of the given polynomials is the polynomial that each of the given polynomials divides evenly into. Because we already know from our analysis above that both \(x+2\) and \(x-3\) are factors of \(6 + x - x^2\), we can conclude that \(6 + x - x^2\) is the LCM of the given polynomials.

Key Concepts

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into a product of simpler polynomials. This process makes it easier to solve equations involving polynomials. Let's take a closer look at the polynomial given in the exercise:

\[6 + x - x^2\].
It's beneficial to rewrite the polynomial in a standard form, organized by decreasing powers of x, such as \(-x^2 + x + 6\). Recognizing it as a quadratic polynomial, we can attempt to factor it using suitable techniques.

• Factor by grouping: This involves rearranging terms and identifying common factors.
• Trial-and-error: Guessing reasonable factor pairs of the constant term and checking if they work.

For our polynomial, the trial-and-error method reveals that \((x-3)(x+2)\) are valid factors. Consequently, our polynomial can be expressed as:

\[(-1) \, (x-3)(x+2)\].
This makes solving polynomial equations much simpler as each factor represents a distinct linear function, helping to determine the roots of the equation.

Polynomial Divisibility

Understanding polynomial divisibility is key when working with expressions like polynomials. Simply put, one polynomial divides another if there exists a polynomial that multiplied by the divisor equals the dividend. Let’s illustrate this concept using our exercise:

We factorized the initial polynomial \(6 + x - x^2\) into \((-1) \, (x-3)(x+2)\). Notably:

• The polynomial \(x+2\) is a factor of \(6 + x - x^2\).
• The polynomial \(x-3\) is also a factor.
• Both of these indicate that they divide \(6 + x - x^2\) evenly without a remainder.

Because each of these polynomials is a factor of the larger polynomial, they're crucial in determining the Least Common Multiple (LCM). If a polynomial can be factorized into other smaller polynomials, then these smaller polynomials divide the original polynomial precisely.

Algebraic Expressions

Algebraic expressions, at their core, are mathematical phrases combining numbers, variables, and operations. In the context of the exercise, they manifest as polynomials like \(6 + x - x^2\), \(x+2\), and \(x-3\). Understanding how to manipulate and simplify these expressions is fundamental in algebra.

When trying to find the Least Common Multiple of these expressions, it's vital to comprehend how each component interacts within a polynomial. Consider the essential techniques:

• Simplifying expressions by combining like terms: This streamlines calculations and reveals hidden patterns.
• Knowing order of operations: Ensure all calculations follow a consistent rule set, such as PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).

In exercises like the given one, being adept with algebraic manipulation, such as factorization and divisibility, allows us to accurately find solutions like the LCM. This showcases the importance of algebraic expressions as tools for simplifying and addressing more complex mathematical tasks.