Problem 6 - Precalculus : Building Concepts and Connections

Question

Multiply. $$(2 x+1)(x-3)\left(\frac{x+7}{x-3}\right)$$

Step-by-Step Solution

Answer

The multiplication simplifies to \(2x^2 + 15x + 7\), for \(x \neq 3\)

Step 1: Identifying elements for simplification

Here notice that the factor \(x - 3\) is present both in the numerator and denominator of the fraction \(\frac{x+7}{x-3}\). So, they can be canceled out before multiplication. But remember, \(x \neq 3\) or the denominator would become zero. Now, let’s rewrite the expression a bit to show these similar factors clearly: \((2x + 1) * (x-3) * \frac{x+7}{x-3}\). Now cancel the factor \(x - 3\) from both the numerator and the denominator. The expression then changes to \((2x + 1) * (x+7)\).

Step 2: Multiplication of remaining expressions

Now we need to multiply the two remaining binomials: \((2x + 1) * (x+7)\). Utilize the distributive property, applying every term of the first binomial to each term in the second binomial. Distribute \(2x\) in \((2x + 1) * (x+7)\) to \(x\) and \(7\) to get \(2x^2 + 14x\). Then, distribute \(1\) in \((2x + 1) * (x+7)\) to \(x\) and \(7\) to get \(7 + x\). Combine these results to conclude that the final polynomial is \(2x^2 + 14x + x + 7\).

Step 3: Simplifying to the final answer

Simplify the polynomial from the previous step by combining like terms. We combine \(14x\) and \(x\) to obtain \(15x\). So the final answer simplifies to \(2x^2 + 15x + 7\).

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that helps us multiply expressions more efficiently. It states that a single term multiplied by a sum or difference can be distributed to multiply each addend separately. Here’s how it works: if you have an expression like

• \( a(b + c) \),
• the distributive property tells us that it equals \( ab + ac \).

In this exercise, we see the distributive property at play when multiplying the binomials

• \((2x + 1) \) and \((x + 7) \).

First, distribute \(2x\) across \(x\) and \(7\) to get:

• \(2x \cdot x = 2x^2\),
• \(2x \cdot 7 = 14x\).

Next, distribute \(1\) across \(x\) and \(7\):

• \(1 \cdot x = x\),
• \(1 \cdot 7 = 7\).

When you combine all these products, you get \(2x^2 + 14x + x + 7\), which simplifies to \(2x^2 + 15x + 7\). By mastering the distributive property, you can make quick work of multiplying complex polynomial expressions.

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. These expressions often seem intimidating because they include polynomials, but they have their rules and operations, similar to simple fractions. In this exercise, the fraction

• \(\frac{x+7}{x-3}\) represents a rational expression.

A crucial step is simplifying these expressions whenever possible. Simplification often involves canceling common factors in the numerator and denominator, provided the denominator doesn't become zero. Let’s look at the term

• \((x-3)\) present in both the numerator and denominator.

We can cancel

• \(x-3\) since it appears in both parts, but remember this only holds true when \(x eq 3\).

This simplifies the expression significantly, reducing it to

• \((2x+1)(x+7)\).

Understanding how to simplify rational expressions is key to working effectively with these types of mathematical expressions.

Binomial Expansion

Binomial expansion involves multiplying two binomial expressions to get a trinomial or polynomial. The process essentially means applying the distributive property systematically. In this exercise, we're multiplying the binomials

• \((2x + 1)\) and
• \((x + 7)\).

Here’s a breakdown:

• Multiply each term in the first binomial by each term in the second binomial. - First term: \(2x \cdot x = 2x^2\) - Second term: \(2x \cdot 7 = 14x\) - Third term: \(1 \cdot x = x\) - Fourth term: \(1 \cdot 7 = 7\).

By reorganizing, you get the expression: \(2x^2 + 14x + x + 7\). Then, combine like terms to simplify the expansion, which becomes \(2x^2 + 15x + 7\). Mastering binomial expansion through practice makes you comfortable with polynomial multiplications, paving the way to solve even more complex algebraic problems.