Problem 4 - Algebra for College Students

Question

Find each product. $$(2 x y)\left(-4 x^{2} y\right)$$

Step-by-Step Solution

Answer

The product is \(-8x^3y^2\).

Step 1: Recognize the Expression

We are presented with the expression \((2xy)(-4x^2y)\). This involves multiplying two binomial expressions, where the first term is the product of a number and variables \(2xy\) and the second term is also a product \(-4x^2y\).

Step 2: Distribute the Constants

Multiply the constants together. You have the constants 2 and -4. Thus, compute: \[2 imes (-4) = -8\]

Step 3: Multiply the Like Variables

Multiply the same variables together:- For the \(x\) terms: \(x imes x^2 = x^{1+2} = x^3\)- For the \(y\) terms: \(y imes y = y^{1+1} = y^2\)

Step 4: Combine the Products

Combine the results from Steps 2 and 3 to form the final product:The constant part is -8, the \(x\) part is \(x^3\), and the \(y\) part is \(y^2\). Thus, the product is:\[-8x^3y^2\]

Key Concepts

Multiplying Expressions

Multiplying algebraic expressions often requires combining numbers, known as constants, with variables that adhere to specific rules. When multiplying expressions like

• \((2xy)(-4x^2y)\),

we focus first on the numbers (constants) and then on the letters (variables).
The process kicks off by multiplying the constants straightforwardly. You deal with the numbers separately:

• multiply 2 and -4,
• which gives ---8.

With the constants combined, attention shifts to the variables. For each letter, multiply the bases by adding their exponents. The rules are handy:

• For example, \(x\) and \(x^2\) become \(x^{1+2}\),
• and \(y\) and \(y\) result in \(y^{1+1}\).

Finally, blend everything together in what is known as a product, bringing the constants and like terms alongside. Your finished answer is a new expression:

• \(-8x^3y^2\).

Polynomial Multiplication

Polynomial multiplication involves multiplying expressions that contain several terms. These terms might include constants, variables, or both, and are connected by addition or subtraction. The technique for multiplying polynomials requires applying the distributive property, often across each term.
In more complex polynomials, we carefully multiply each term in one polynomial by every term in the other. In our specific exercise, each ‘polynomial’ is a single term expression, such as \(2xy\) and \(-4x^2y\). When working through this, you continue the process of distributing like this:

• Multiply the constants and variables similarly,
• and if needed, repeat for more terms.

This principle also applies when there are more terms to manage, extending each multiplication step carefully.
After handling each part, combine like terms; in our example, the combination is more direct since you work with already minimized pieces. Simplifying sooner helps avoid unnecessary errors. Here, your completed polynomial expression is \(-8x^3y^2\), attained by attending to each term attentively and efficiently.

Variables and Constants

Understanding the role of variables and constants is crucial in simplifying expressions.
Variables represent unknown quantities and are usually denoted by letters like \(x\), \(y\), or \(z\). They might appear with exponents indicating repeated multiplication. Meanwhle, constants are numbers that remain fixed throughout the problem.
In calculations, we often first deal with constants, as they are similar to basic numeric multiplication. Then, variables are handled by combining them according to their rules. Each variable is multiplied with its like and summed exponents. For example:

• When multiplying \(x\) and \(x^2\),
• you add the exponents to get \(x^3\).

This operation helps define the term's degree and ensures the correct arrangement in solutions. By recognizing how variables and constants operate distinctly yet together, managing these elements in algebraic expressions becomes intuitive and more manageable.