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Chapter 4: Problem 75

Find each product. $$ -8 r^{3}\left(5 r^{2}+2\right)\left(5 r^{2}-2\right) $$

Short Answer

Expert verified
\(-200r^7 - 32r^3 \)

Step by step solution

01

- Use the Difference of Squares Formula

Notice that \(5r^2 + 2\) and \(5r^2 - 2\) form a difference of squares. The difference of squares formula is \(a^2 - b^2 = (a + b)(a - b)\). In this case, \(a = 5r^2\) and \(b = 2\). Thus: \( (5r^2+2)(5r^2-2) = (5r^2)^2 - 2^2 \)
02

- Simplify the Difference of Squares

Now, calculate each part of the expression: \( (5r^2)^2 = 25r^4 \) and \( 2^2 = 4 \). Thus, the expression becomes: \( 25r^4 - 4 \)
03

- Multiply by \(-8r^3\)

Distribute the \(-8r^3\) to the simplified difference of squares: \( -8r^3 \times (25r^4 - 4) = -8r^3 \times 25r^4 - 8r^3 \times 4 \)
04

- Simplify the Expression

Perform the multiplication: \(-8r^3 \times 25r^4 = -200r^7\) and \(-8r^3 \times 4 = -32r^3\). Thus, the final expression is: \(-200r^7 - 32r^3 \)

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Key Concepts

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

Algebraic Expression
An algebraic expression is a mathematical phrase that can include numbers, variables, and operation symbols. In our exercise, we deal with an expression involving the variable \( r \). Algebraic expressions are used to represent real-world scenarios and can simplify complex problems. For example, \( -8r^3(5r^2+2)(5r^2-2) \) is an algebraic expression. It contains coefficients (numbers), variables (\( r \)), and arithmetic operations (multiplication and addition/subtraction). To simplify such expressions, we follow a systematic approach, which involves applying mathematical rules and properties.
Polynomial Multiplication
Polynomial multiplication involves combining two or more polynomials to form a new, expanded polynomial. In our exercise, we deal with multiplying \( -8r^3 \) with a product of two binomials \( (5r^2+2) \) and \( (5r^2-2) \). Multiplying polynomials requires distributing each term in the first polynomial to every term in the second polynomial. Here are some key steps:
  • First, recognize the structure of the problem, such as factoring patterns.
  • Next, use algebraic identities. In this case, we use the difference of squares \( (a + b)(a - b) = a^2 - b^2 \).
  • Simplify the expression before multiplying.

This way, polynomial multiplication can be systematically worked through, arriving at the simplified form \( -200r^7 - 32r^3 \).
Distributive Property
The distributive property is a fundamental algebraic property that is used in polynomial multiplication. It states that for any numbers or expressions \( a \), \( b \), and \( c \): \( a(b + c) = ab + ac \). This property is essential when dealing with algebraic expressions and helps simplify them efficiently. In our exercise, we used the distributive property to handle the multiplication:
  • First, we multiplied \( -8r^3 \) by each term in the simplified polynomial.
  • Specifically, \( -8r^3 \times (25r^4 - 4) \) was broken down into two separate multiplications: \( -8r^3 \times 25r^4 \) and \( -8r^3 \times 4 \).
Applying the distributive property ensures that every term is multiplied correctly, leading to the final simplified expression \( -200r^7 - 32r^3 \). This property is a cornerstone of algebra and is fundamental in simplifying complex expressions.

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Most popular questions from this chapter

If it costs \(\left(4 x^{5}+3 x^{4}+2 x^{3}+9 x^{2}-29 x+2\right)\) dollars to fertilize a garden, and fertilizer costs \((x+2)\) dollars per square yard, write an expression, in square yards, for the area of the garden.

Graph each equation by completing the table of values. $$ \begin{aligned} &y=(x-4)^{2}\\\ &\begin{array}{|l|l|l|l|l|l} x & 2 & 3 & 4 & 5 & 6 \\ \hline y & & & & & \end{array} \end{aligned} $$

The special product \((x+y)(x-y)=x^{2}-y^{2}\) can be used to perform some multiplications.Example: $$\begin{array}{l|l}51 \times 49 & 102 \times 98 \\\=(50+1)(50-1) & =(100+2)(100-2) \\\=50^{2}-1^{2} & =100^{2}-2^{2} \\\=2500-1 & =10,000-4 \\\=2499 & =9996\end{array}$$ Use this method to calculate each product mentally. $$ 20 \frac{1}{2} \times 19 \frac{1}{2} $$

The special product \((x+y)(x-y)=x^{2}-y^{2}\) can be used to perform some multiplications.Example: $$\begin{array}{l|l}51 \times 49 & 102 \times 98 \\\=(50+1)(50-1) & =(100+2)(100-2) \\\=50^{2}-1^{2} & =100^{2}-2^{2} \\\=2500-1 & =10,000-4 \\\=2499 & =9996\end{array}$$ Use this method to calculate each product mentally. $$ 301 \times 299 $$

Find each product. Recall that \(a^{2}=a \cdot a\) and \(a^{3}=a^{2} \cdot a\). $$ -4 x^{3}\left(3 x^{4}+2 x^{2}-x\right)(-2 x+1) $$
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