Problem 2 - Elementary Algebra

Question

The product is \(5 y^{2}+10 y^{3}-15 y^{2}\) and a factor is \(5 y^{2}\), Find the other factor.

Step-by-Step Solution

Answer

Answer: The other factor is \(2y-2\).

Step 1: Identify the given product and factor

The given product is \(5 y^{2}+10 y^{3}-15 y^{2}\) and the given factor is \(5 y^{2}\). We want to find the other factor.

Step 2: Divide the product by the given factor

Divide the product \(5 y^{2}+10 y^{3}-15 y^{2}\) by the given factor \(5 y^{2}\): \((5 y^2+10 y^3-15 y^2)/(5 y^2)\)

Step 3: Simplify the fraction

Now we can simplify the fraction by dividing each term by \(5 y^2\): \((5 y^2)/(5 y^2) + (10 y^3)/(5 y^2) - (15 y^2)/(5 y^2)\)

Step 4: Perform the division

Perform the division for each term to get: \(1 + 2y - 3\)

Step 5: Write the other factor

The other factor is the simplified expression after performing the division: \(1 + 2y - 3\) So, the other factor is \(2y-2\).

Key Concepts

Polynomials

Polynomials are algebraic expressions made up of variables and coefficients. They are combined using addition, subtraction, and multiplication but never division by a variable. In simple terms, a polynomial is like a multi-term expression that can involve powers of a variable. Consider the polynomial given in the exercise:

• The expression is: \(5y^2 + 10y^3 - 15y^2\)
• Here, the terms are: \(10y^3\), \(5y^2\), and \(-15y^2\).
• Each term consists of a coefficient (a number) and a variable raised to a power (like \(y^3\)).

Polynomials are important as they set the foundation for more complex algebraic operations and various aspects of mathematics, including polynomial equations and calculus. They also help model real-world situations smoothly due to their smooth curve graph representations.
When working with polynomials, always consider their degree, which is the highest power of the variable. For example, the degree of the polynomial \(10y^3 - 5y^2\) is 3 because the highest power is \(y^3\). Understanding polynomials is like learning the building blocks for further algebraic and mathematical studies.

Algebraic Expressions

Algebraic expressions are mathematical phrases combining numbers, variables, and arithmetic operations such as addition, subtraction, multiplication, and division. They represent values or quantities symbolically, which can be manipulated according to the rules of algebra. Let's break down a few elements:

• Variables are symbols, like \(y\), used to represent numbers whose exact values are not immediately known.
• Coefficients are the numbers multiplied by variables, seen in terms like \(5\) in \(5y^2\).
• Constants are fixed numbers with no attached variables, often appearing in expressions and equations.

Understanding how to manipulate and simplify algebraic expressions is crucial when solving mathematical problems. Simplification involves combining like terms, which are terms with the same variable to the same power. For instance, in the original problem, the expression \(5y^2 + 10y^3 - 15y^2\) can be simplified by recognizing that \(5y^2\) and \(-15y^2\) are like terms and can be combined.
Recognizing the structure of algebraic expressions aids in applying operations like factoring, distributing, and division effectively. Mastery of these concepts helps in progressing toward solving higher-level algebraic problems.

Factoring

Factoring is the process of breaking down an algebraic expression into smaller components, or "factors," that multiply to give the original expression. It is like unraveling a puzzle, revealing simpler expressions that combine to form more complex ones. In the exercise, one factor \(5y^2\) is already given, and we are tasked with finding its companion factor.
Here's how to approach factoring:

• Determine what needs to be factored out of each term if possible—often a common numerical factor or a common variable factor.
• Simplify by dividing each term by the identified factor, as seen in the division of \(5y^2 + 10y^3 - 15y^2\) by \(5y^2\).

The process involves recognizing that each term in the polynomial shares a common factor, pulling it out, and simplifying what's left. In the exercise, by dividing each part of \(5y^2 + 10y^3 - 15y^2\) by \(5y^2\), you obtain the simpler factor \(2y - 2\).
Factoring is a critical skill in algebra that simplifies solving equations and expressions. It also aids in graphing polynomial functions, finding zeros, and reducing complex fractions. Developing a keen eye for identifying factors quickly makes working through algebraic expressions more manageable and prepares you for advanced mathematical concepts.