Problem 60 - Elementary and Intermediate Algebra

Question

Multiply. See Example 6. $$ (3 x+1)\left(2 x^{2}-3 x+1\right) $$

Step-by-Step Solution

Answer

The product is \(6x^3 - 7x^2 + 1\).

Step 1: Use the Distributive Property

To multiply the expression \((3x + 1)(2x^2 - 3x + 1)\), we'll use the distributive property, distributing each term from the first polynomial across the second. This means we'll separately distribute \(3x\) and \(1\) to each term in \(2x^2 - 3x + 1\).

Step 2: Multiply 3x by Each Term

First, distribute the \(3x\) across the terms of the second polynomial: - \(3x \cdot 2x^2 = 6x^3\)- \(3x \cdot (-3x) = -9x^2\)- \(3x \cdot 1 = 3x\)

Step 3: Multiply 1 by Each Term

Next, distribute the \(1\) across the terms of the second polynomial: - \(1 \cdot 2x^2 = 2x^2\)- \(1 \cdot (-3x) = -3x\)- \(1 \cdot 1 = 1\)

Step 4: Combine Like Terms

Now, combine all the distributed terms from the previous steps:- Group like terms together: - Combine \(6x^3\) - Combine \(-9x^2 + 2x^2 = -7x^2\) - Combine \(3x - 3x = 0x\) - Include the constant \(1\).- The resulting expression is \(6x^3 - 7x^2 + 1\).

Key Concepts

Distributive Property

The distributive property is a fundamental principle in algebra that allows you to multiply a single term by each term in a polynomial. It is a tool that simplifies expressions and solves equations. To understand how it works, imagine distributing or handing out a variable or number to everything inside parentheses. For example, if you have an expression like \((a + b)(c + d)\), the distributive property tells you to multiply \(a\) by every term in the second binomial and then \(b\) by every term as well.

Let's connect this to our example: we had \((3x + 1)(2x^2 - 3x + 1)\). Here, we distribute \(3x\) to each of the terms in the second polynomial, and then do the same with the 1.

This leads us through a more straightforward series of multiplications:

• \(3x \cdot 2x^2 = 6x^3\)
• \(3x \cdot (-3x) = -9x^2\)
• \(3x \cdot 1 = 3x\)
• \(1 \cdot 2x^2 = 2x^2\)
• \(1 \cdot (-3x) = -3x\)
• \(1 \cdot 1 = 1\)

In this way, the distributive property helps you to break complex multiplications into simpler steps, making it easier to handle.

Combining Like Terms

Once you have distributed all terms, it is time to combine like terms. But what exactly are 'like terms'? They are terms within an expression that have the same variables raised to the same power. For instance, \(2x^2\) and \(-9x^2\) are like terms because both have the variable \(x\) raised to the power of 2. However, \(3x\) and \(3x^2\) are not like terms.

Combining like terms means adding or subtracting their coefficients and leaving the variable part unchanged. This is crucial for simplifying equations. Let's apply this concept to our given example. After distributing, we got:

• \(6x^3\) (there are no other \(x^3\) terms to combine)
• \(-9x^2 + 2x^2 = -7x^2\) (both have \(x^2\) as part of them)
• \(3x - 3x = 0x\) (since these cancel each other out, it disappears)
• the constant term was \(1\)

After combining like terms, our simplified result is \(6x^3 - 7x^2 + 1\). The expression looks cleaner, and it's in its simplest form.

Algebraic Expressions

Algebraic expressions are compositions of numbers, variables, and operations. They form the building blocks of countless mathematical problems and solutions. Understanding how to work with them is essential in algebra and higher mathematics.

An expression like \((3x + 1)(2x^2 - 3x + 1)\) may look complex, but remember it’s just a set of operations applied on variables and constants! It doesn't have an equality sign (unlike equations), so it cannot be 'solved' in the typical sense, but it can be simplified or rearranged.

Key features include:

• **Variables**: placeholders for numbers; like \(x\) in our example.
• **Constants**: numbers without variables, such as \(1\) in our expression.
• **Operations**: arithmetic actions, including addition, subtraction, multiplication, and division.

Working with algebraic expressions involves using the distributive property and combining like terms to simplify or expand them. This knowledge helps you to manipulate and rearrange parts of mathematical statements efficiently. Whether you are multiplying terms or simplifying expressions, understanding the structure and handling of these expressions is core to success in algebra.