Problem 55
Question
For the following problems, perform the multiplications and combine any like terms. $$ 6 a(a-5) $$
Step-by-Step Solution
Verified Answer
Answer: The simplified form of the algebraic expression is $$6a^2 - 30a$$.
1Step 1: Distributive Property
To perform the multiplication, we will use the distributive property. The distributive property states that a(b + c) = ab + ac. In our case, we have:
$$
6a(a-5) = 6a(a) - 6a(5)
$$
2Step 2: Multiply the terms
Now, we will multiply the terms in the expression we acquired in the previous step:
$$
6a(a) - 6a(5) = 6a^2 - 30a
$$
3Step 3: Combine like terms
Finally, we will combine like terms. In this case, we already have the terms in their simplest form, so there are no like terms to combine. Thus, our answer is:
$$
6a(a-5) = 6a^2 - 30a
$$
Key Concepts
Distributive PropertyCombining Like TermsMultiplying Variables
Distributive Property
Understanding the distributive property is essential for solving algebraic expressions involving multiplication. It allows you to multiply a single term by a group of terms added together within parentheses.
For example, take the expression \( 6a(a-5) \). Here, the distributive property is used to eliminate the parentheses by distributing the multiplication of \(6a\) to each term inside the parentheses, \(a\) and \( -5\). This results in \(6a \times a\) minus \(6a \times 5\), which simplifies to \(6a^2 - 30a\).
This property is a reliable method that can be applied to any similar problem: \(a(b + c) = ab + ac\), where \(a\), \(b\), and \(c\) are any algebraic terms. It's a fundamental concept in algebra that ensures terms are correctly expanded before further manipulation.
For example, take the expression \( 6a(a-5) \). Here, the distributive property is used to eliminate the parentheses by distributing the multiplication of \(6a\) to each term inside the parentheses, \(a\) and \( -5\). This results in \(6a \times a\) minus \(6a \times 5\), which simplifies to \(6a^2 - 30a\).
This property is a reliable method that can be applied to any similar problem: \(a(b + c) = ab + ac\), where \(a\), \(b\), and \(c\) are any algebraic terms. It's a fundamental concept in algebra that ensures terms are correctly expanded before further manipulation.
Combining Like Terms
After using the distributive property, you often need to simplify the expression further by combining like terms. Like terms are terms that have the same variable parts raised to the same powers, even if they have different coefficients.
In our example, \(6a^2\) and \( -30a\) are considered unlike terms because one is an \(a^2\) term and the other is an \(a\) term, so we cannot combine them further. If we had an expression like \(6a^2 - 5a^2 + 3a - 2a\), the process of combining like terms would look like this: combine the \(a^2\) terms and the \(a\) terms separately to get \(6a^2 - 5a^2 = 1a^2\), and \(3a - 2a = 1a\), resulting in \(1a^2 + 1a\).
Recognizing like terms and combining them correctly is vital for simplifying expressions and solving equations.
In our example, \(6a^2\) and \( -30a\) are considered unlike terms because one is an \(a^2\) term and the other is an \(a\) term, so we cannot combine them further. If we had an expression like \(6a^2 - 5a^2 + 3a - 2a\), the process of combining like terms would look like this: combine the \(a^2\) terms and the \(a\) terms separately to get \(6a^2 - 5a^2 = 1a^2\), and \(3a - 2a = 1a\), resulting in \(1a^2 + 1a\).
Recognizing like terms and combining them correctly is vital for simplifying expressions and solving equations.
Multiplying Variables
Multiplication involving variables follows the same rules as multiplication with numbers, with the consideration of variable powers. When we multiply variables, we apply the law of exponents, which states that when multiplying two powers that have the same base, we add the exponents.
Take the term \(6a \times a\) from our example. Since \(a\) is the same as \(a^1\), when we multiply \(6a\) (which is \(6a^1\)) by another \(a\), we add the exponents of \(a\): \(1 + 1 = 2\). Therefore, \(6a \times a\) simplifies to \(6a^2\).
It's crucial to remember this rule, especially when working with higher powers or multiple variables. Mistakes in multiplying variables can lead to incorrect solutions, so practice this skill to ensure accuracy in your algebraic operations.
Take the term \(6a \times a\) from our example. Since \(a\) is the same as \(a^1\), when we multiply \(6a\) (which is \(6a^1\)) by another \(a\), we add the exponents of \(a\): \(1 + 1 = 2\). Therefore, \(6a \times a\) simplifies to \(6a^2\).
It's crucial to remember this rule, especially when working with higher powers or multiple variables. Mistakes in multiplying variables can lead to incorrect solutions, so practice this skill to ensure accuracy in your algebraic operations.
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