Problem 58 Find each product. \(-5 r(r+1)... [FREE SOLUTION] (2024)

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Chapter 5: Problem 58

Find each product. \(-5 r(r+1)^{3}\)

Short Answer

Expert verified

-5r^4 - 15r^3 - 15r^2 - 5r

Step by step solution

01

Distribute the constant

Distribute the constant (-5r) to every term inside the parentheses (r+1)^{3}. To begin, leave (r+1)^{3} intact and multiply -5r by (r+1)^{3} as a whole.

03

Apply the distributive property

Multiply each pair of binomials. Start with (r+1)(r+1): (r)(r) + (r)(1) + (1)(r) + (1)(1) = r^2 + 2r + 1.

04

Continue expanding

Now, multiply the result from Step 3 (r^2 + 2r + 1) by (r+1): (r^2)(r) + (r^2)(1) + (2r)(r) + (2r)(1) + (1)(r) + (1)(1). This results in (r^3 + r^2 + 2r^2 + 2r + r + 1), which simplifies to (r^3 + 3r^2 + 3r + 1).

05

Final multiplication

Multiply the simplified expression by (-5r): (-5r)(r^3 + 3r^2 + 3r + 1). Apply the distributive property: (-5r)(r^3) + (-5r)(3r^2) + (-5r)(3r) + (-5r)(1).

06

Simplify the terms

Multiply each term separately: -5r^4, -15r^3, -15r^2, and -5r.

07

Combine final result

Combine all the terms from Step 6 to get the final product: -5r^4 - 15r^3 - 15r^2 - 5r.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

distribution

Let's tackle the concept of distribution. Distribution in algebra is the process of multiplying each term inside the parentheses by a factor outside the parentheses. In the exercise, we distribute the constant -5r, which means we multiply it with every term of the polynomial expression inside the parentheses.
This is the first step in breaking down complex expressions to make them more manageable. For example, if we have -5r(r+1)^3, we treat (r+1)^3 as a single entity first and multiply it by -5r, yielding -5r*(r+1)^3. Once we have distributed -5r to the binomial term, we can proceed to break down the binomial term further using other algebraic rules.
Ulitizing the distributive property ensures that every term within the parentheses is multiplied by the factor outside, which guarantees accurate results when simplifying algebraic expressions.

polynomial expansion

Polynomial expansion involves expanding expressions of polynomials, often through multiplication of binomials or higher-order terms. In our exercise, we need to expand (r+1)^3 to simplify the expression.
This involves breaking down the cubic binomial and multiplying it out step-by-step.
Here’s how you go from (r+1)^3:

  • First, rewrite (r+1)^3 as (r+1)(r+1)(r+1).
  • Next, multiply two of the binomials: (r+1)(r+1) to get r^2 + 2r + 1.
  • Finally, multiply this result by (r+1) once more to obtain r^3 + 3r^2 + 3r + 1.

This complete expansion process breaks down a higher-order polynomial into simpler terms, which can then be further manipulated in our calculations. Expansion follows systematic steps ensuring each term interrelates correctly as per algebraic multiplication principles.

algebraic multiplication

Algebraic multiplication is essentially multiplying algebraic expressions, terms, or polynomials systematically. This ensures accuracy and simplification of complex expressions.
After our polynomial expansion, we received an expression r^3 + 3r^2 + 3r + 1. Now we need to multiply this result by -5r.Apply distribution: (-5r) * (r^3 + 3r^2 + 3r + 1). Here’s the breakdown:

  • Multiply -5r by each term individually:
  • (-5r)(r^3) = -5r^4
  • (-5r)(3r^2) = -15r^3
  • (-5r)(3r) = -15r^2
  • (-5r)(1) = -5r

Combining these products gives us the final expression: -5r^4 - 15r^3 - 15r^2 - 5r.
This ensures each term is correctly multiplied and combined, providing the simplified result of our initial complex term.

binomials

A binomial is an algebraic expression containing two different terms connected by a plus or minus sign. Understanding binomials is essential because they are the building blocks of polynomial expressions and are frequently expanded or multiplied in algebra. In our exercise, we started with a binomial (r+1) which was raised to the third power, making (r+1)^3. To simplify, we broke it down and treated it as (r+1)(r+1)(r+1).

  • By knowing binomials, we were able to systematically expand this expression step-by-step:
  • (r+1)^2 became r^2 + 2r + 1, and
  • (r^2 + 2r + 1)(r+1) expanded into r^3 + 3r^2 + 3r + 1.

Grasping the properties and expansion methods of binomials allows us to handle more complex problems by breaking them down into these simpler components. This step ensures confidence with larger algebraic manipulations.

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Problem 58 Find each product. \(-5 r(r+1)... [FREE SOLUTION] (3)

Most popular questions from this chapter

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Problem 58 Find each product.  
\(-5 r(r+1)... [FREE SOLUTION] (2024)

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