Problem 68 - College Algebra with Modeling and Visualization

Question

Multiply the polynomials. $$-2 x\left(3-2 x+5 x^{2}\right)$$

Step-by-Step Solution

Answer

The result is \(-10x^3 + 4x^2 - 6x\).

Step 1: Distribute the First Term

To begin, distribute \(-2x\) to each term inside the parentheses, starting with the first term which is \(3\). Calculate \(-2x \times 3\).

Step 2: Calculate the First Distribution

Multiply the coefficients: \(-2 \times 3 = -6\). Attach the variable \(x\), so the result is \(-6x\).

Step 3: Distribute to the Second Term

Next, distribute \(-2x\) to the second term \(-2x\). Calculate \(-2x \times -2x\).

Step 4: Calculate the Second Distribution

Multiplying \(-2\) by \(-2\) gives \(4\), and \(x \times x\) is \(x^2\). So, the result is \(4x^2\).

Step 5: Distribute to the Third Term

Finally, distribute \(-2x\) to the third term \(5x^2\). Calculate \(-2x \times 5x^2\).

Step 6: Calculate the Third Distribution

Multiply the coefficients \(-2\) and \(5\) to get \(-10\). The variables are \(x \times x^2\), which is \(x^3\). Thus, the product is \(-10x^3\).

Step 7: Combine Like Terms

Add all the products together: \(-6x + 4x^2 - 10x^3\). Since these are unlike terms (each term has a different degree), write the expression in standard form, which is descending order of exponents.

Key Concepts

Distributive Property

The distributive property is a vital concept in algebra, helping simplify multiplication over addition or subtraction in expressions. When we use the distributive property, we multiply a single term by each term inside a set of parentheses. For the given exercise, the problem is to multiply \(-2x\) by each term inside the parentheses: \(3 - 2x + 5x^2\).
This property is particularly useful because it allows us to break down complex multiplication into a series of simpler steps. Remembering this key idea: distribute a term outside the parentheses to each term inside, will help solve many polynomial multiplication problems.
In simpler terms: \(a(b + c + d) = ab + ac + ad\). Therefore, applying \(-2x \) to each component separately ensures accurate results.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operators like addition and multiplication. They represent relationships or patterns using symbols and numbers. In this exercise, \(-2x(3 - 2x + 5x^2)\) is the algebraic expression we work with.
Dissecting it, we have:

• A variable term, \(-2x\), which includes a coefficient \(-2\) and a variable \(x\).
• The expression within the parentheses (also called the polynomial) \(3 - 2x + 5x^2\) contains constant, linear, and quadratic terms.

Understanding these components simplifies both manipulation and simplification of expressions, aiding in finding desired mathematical solutions.

Polynomial Expression

A polynomial is a specific type of algebraic expression which includes terms in the form of \(ax^n\), where \(a\) is a coefficient, and \(n\) is a non-negative integer called the degree. For instance, in the polynomial \(3 - 2x + 5x^2\), each term has a non-negative integer exponent, making it a polynomial expression.
During polynomial multiplication, each term of the first polynomial must be multiplied by every term of the second one, leveraging the distributive property. This results in a series of products which can then be combined into a single expression.
Recognizing and properly distributing terms ensures the polynomial maintains its structure, and helps achieve the correct standard form.

Standard Form

Writing an algebraic expression in standard form means arranging it so that terms are in descending order of their exponents. For polynomial expressions, this means beginning with the highest degree term and progressing to the lowest degree.
In the final step of the exercise, the products were combined to form \(-6x + 4x^2 - 10x^3\). This needs rearranging into the standard form, thus: \(-10x^3 + 4x^2 - 6x\).
Putting an expression in standard form makes it easier to interpret, compare and further manipulate algebraic expressions. This step is crucial, especially in preparing expressions for operations like addition, subtraction, and solving equations.