Problem 103 - College Algebra

Question

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Knowing the difference between factors and terms is important: In \(\left(3 x^{2} y\right)^{2},\) I can distribute the exponent 2 on each factor, but in \(\left(3 x^{2}+y\right)^{2}, 1\) cannot do the same thing on each term.

Step-by-Step Solution

Answer

The first statement is true and the second statement is false.

Step 1: Understand Terms and Factors

In algebra, 'terms' are the elements that are added or subtracted in an equation. In the equation \(3x^{2} + y\), \(3x^{2}\) and \(y\) are the terms. 'Factors', on the other hand, are the quantities that multiply together to give the product. Hence, given the equation \((3x^{2}y)^{2}\) , the factors are \(3\), \(x^{2}\) , and \(y\).

Step 2: Distribute Exponent in First Statement

In the equation \((3x^{2}y)^{2}\), this can be distributed to every factor inside the parentheses. Using the rule \( (abc)^n = a^n b^n c^n \), the equation becomes \(3^{2} * (x^{2})^{2} * y^{2}\) which simplifies to \(9x^{4}y^{2}\).

Step 3: Distribute Exponent in Second Statement

In the equation \((3x^{2} + y)^{2}\), the stated theory suggests distributing the exponent over each term. However, this is incorrect. A correct application of the rule is \( (a + b)^2 = a^2 + 2ab + b^2 \), which will give us \((3x^{2})^{2} + 2*(3x^{2})*y + y^{2}\), i.e., \(9x^{4} + 6x^{2}*y + y^{2}\).

Step 4: Conclusion

The first statement is valid as exponents can be distributed over factors, resulting in \(9x^{4}y^{2}\). The second statement is incorrect since distributing the exponent over each term does not produce the valid expanded form of the equation.

Key Concepts

Terms and Factors

When we talk about algebraic expressions, understanding the difference between 'terms' and 'factors' is crucial. Confusing these can lead to errors in calculations.

• Terms: These are distinct elements separated by plus or minus signs in an expression. For example, in the expression \(3x^{2} + y\), there are two terms: \(3x^{2}\) and \(y\).
• Factors: These are numbers or variables that multiply together within a term. They form the parts of a term that create its product. In the expression \((3x^{2}y)\), the factors are \(3\), \(x^{2}\), and \(y\). Here, they multiply to form one term.

Recognizing these distinctions helps you correctly apply mathematical rules to algebraic expressions, especially when dealing with exponents.

Exponent Rules

Exponent rules are guidelines for simplifying expressions involving powers. Applying these correctly can prevent mistakes in calculations.

• Multiplying Factors in Parentheses: The rule \((abc)^n = a^n b^n c^n\) allows us to distribute the exponent over factors within one term. For example, in \((3x^{2}y)^{2}\), each factor is raised individually: \((3)^{2} = 9\), \((x^{2})^{2} = x^{4}\), and \(y^{2} = y^{2}\).
• Expanding Binomials: When dealing with an expression like \((a + b)^{2}\), one must apply the binomial expansion formula \((a + b)^{2} = a^2 + 2ab + b^2\). This principle helps avoid the common mistake of distributing the exponent over each term separately.

Understanding these rules ensures expressions are simplified correctly, reflecting the true mathematical relationships defined by exponents.

Expanding Expressions

Expanding expressions involves transforming expressions from a compact form to a more detailed one. It is crucial for understanding the underlying structure of algebraic equations.
To expand \((3x^{2} + y)^{2}\), follow the binomial expansion method:

• Square the first term: \((3x^{2})^{2} = 9x^{4}\).
• Twice the product of the terms: \(2 \times (3x^{2}) \times y = 6x^{2}y\).
• Square the second term: \(y^{2}\).

Combining these steps yields: \(9x^{4} + 6x^{2}y + y^{2}\).
This illustrates the importance of amplifying an expression correctly, as incorrect expansion can lead to drastically different results. Appropriately expanding expressions using these methods will safeguard against misinterpretation of algebraic structures.