Problem 42 - College Algebra

Question

For the following exercises, multiply the polynomials. $$ (6 k-5)\left(6 k^{2}+5 k-1\right) $$

Step-by-Step Solution

Answer

The product is \(36k^3 - 31k + 5\).

Step 1: Apply the Distributive Property

To multiply the polynomials, apply the distributive property. This means multiplying each term in the first polynomial \((6k - 5)\) by each term in the second polynomial \((6k^2 + 5k - 1)\).

Step 2: Multiply the First Term

Multiply the first term of the first polynomial, \(6k\), by each term of the second polynomial:1. \(6k \times 6k^2 = 36k^3\)2. \(6k \times 5k = 30k^2\)3. \(6k \times -1 = -6k\)

Step 3: Multiply the Second Term

Now multiply the second term of the first polynomial, \(-5\), by each term of the second polynomial:1. \(-5 \times 6k^2 = -30k^2\)2. \(-5 \times 5k = -25k\)3. \(-5 \times -1 = 5\)

Step 4: Combine Like Terms

Combine the results from Step 2 and Step 3 to combine like terms:1. \(36k^3\) remains as it is since there are no other cubic terms.2. \(30k^2 - 30k^2 = 0\), so these cancel each other out.3. \(-6k - 25k = -31k\).4. Add the constant term \(5\).Thus, the expression simplifies to \(36k^3 - 31k + 5\).

Key Concepts

The Distributive Property

The distributive property is a key concept in algebra that helps us handle situations like multiplying polynomials. It tells us that when we have a term outside a bracket, like \((a+b)(c+d)\), we need to multiply each term inside the bracket by the term outside. This is the same as saying \(a imes c + a imes d + b imes c + b imes d\).

For polynomial multiplication, we apply this property repeatedly. For our problem, the expression \( (6k - 5)(6k^2 + 5k - 1) \) uses the distributive property. Each term of \( (6k - 5) \) was multiplied by every term of \( (6k^2 + 5k - 1) \) to expand the expression fully.

• Multiply \( 6k \) with each term in \( 6k^2 + 5k - 1 \)
• Multiply \( -5 \) with each term in \( 6k^2 + 5k - 1 \)

The distributive property allows us to expand and break down expressions into manageable parts which are essential for simplifying complex algebraic expressions.

Combining Like Terms

Like terms in algebra are terms that have exactly the same variable raised to the same power. It allows us to simplify expressions by adding or subtracting these terms.

Once we distribute and multiply our polynomials, we'll often need to combine like terms to simplify the expression. In our exercise, results like \( 30k^2 \) and \( -30k^2 \) are like terms and can therefore be added or subtracted to zero.

While combining the like terms, we noticed the terms resulted from the multiplication were:

• \( 30k^2 - 30k^2 = 0 \)
• We were left withthe outcome \(-31k\) after combining \(-6k\) and \(-25k\)

It's important to carefully check and combine all like terms to make sure the final expression is as simplified as possible. That helps make the expression easier to use in other mathematical operations.

Understanding Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and operation symbols. They're fundamental in algebra for expressing mathematical ideas succinctly.

When we deal with polynomials like \((6k - 5)\) and \((6k^2 + 5k - 1)\), we're dealing with complex algebraic expressions. Each term in a polynomial is an algebraic expression in itself which might consist of:

• Numbers (constants)
• Variables (like \(k\) in our exercise)
• Operations (addition, subtraction, multiplication)

Understanding how these expressions work, and how to manipulate them through operations like multiplication and combining like terms, is crucial in algebra. It helps us model real-world scenarios and solve problems efficiently.

Polynomials and Their Characteristics

Polynomials are an important class of algebraic expressions consisting of variables and coefficients. They have several terms, each with a variable raised to a non-negative integer exponent. In our exercise, \(6k^2 + 5k - 1\) is a polynomial with three terms.

Polynomials have different characteristics:

• Each term is a combination of a constant and a variable with an exponent (e.g., \(6k^2\)).
• They can be classified by their degree which is determined by the highest exponent (in our case, the degree is 2).
• They are common in all levels of algebra, essential for higher math functions and calculus.

Understanding polynomials and how they form larger expressions is vital for solving algebra problems. Multiplying them, as shown in our exercise, helps improve comprehension of their structure and behavior under different operations. This foundational knowledge supports further studies in mathematics involving more complex algebraic methods.