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Chapter 5: Problem 39

Find each product. $$ q(5 q-1)(5 q+1) $$

Short Answer

Expert verified
25q^3 - q

Step by step solution

01

Identify the structure

The expression is given as \( q(5q - 1)(5q + 1) \). Notice that \((5q - 1)(5q + 1)\) is a product of two binomials in the form \((a - b)(a + b)\).
02

Apply the difference of squares formula

Apply the difference of squares formula: \((a - b)(a + b) = a^2 - b^2\). In this case, \(a = 5q\) and \(b = 1\). Substitute into the formula to get \((5q)^2 - 1^2\).
03

Simplify the binomials

Simplify the expression: \((5q)^2 - 1^2 \) becomes \( 25q^2 - 1 \).
04

Multiply by the remaining term

Multiply the simplified binomial product by \( q \), which was factored out initially: \( q(25q^2 - 1) \).
05

Distribute the term

Distribute \( q \) across \((25q^2 - 1)\): \( q \times 25q^2 - q \times 1 \). This results in: \( 25q^3 - q \).

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Key Concepts

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

Product of Binomials
To simplify the expression, we first need to understand how to handle the product of two binomials. In this case, we are dealing with (5q - 1)(5q + 1). This format is particularly neat because it follows the pattern of (a - b)(a + b).
When multiplying two binomials, each term in the first binomial multiplies with each term in the second binomial. But due to our special pattern, we can simplify it using the difference of squares formula. Knowing how to recognize these patterns makes solving these problems much easier and saves time.
Difference of Squares Formula
The difference of squares formula is handy when we have an expression like (a - b)(a + b). It tells us that we can simply write this as a^2 - b^2.
Let鈥檚 see how this works for our problem: for (5q - 1)(5q + 1), we identify a = 5q and b = 1. Using the difference of squares formula: (5q - 1)(5q + 1) becomes (5q)^2 - 1^2, which simplifies to 25q^2 - 1.
This step reduces our binomial product into a much simpler form that is easier to manage.
Distributive Property
The distributive property helps us finalize our solution. Initially, we had q(5q - 1)(5q + 1). After applying the difference of squares formula, we simplified it to q(25q^2 - 1).
Now we use the distributive property to multiply q across the terms in the simplified expression: q * 25q^2 and q * (-1).
So it becomes 25q^3 - q. Notice how we distribute the q to each term. This method ensures that each term is appropriately multiplied, simplifying our work effectively. Finally, this results in: 25q^3 - q.
The distributive property is powerful for breaking down complex multiplications into simpler, manageable steps.

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Most popular questions from this chapter

Use scientific notation to calculate the answer to each problem. Astronomers using the Spitzer Space Telescope discovered a twisted double- helix nebula, a conglomeration of dust and gas stretching across the center of the Milky Way galaxy. This nebula is \(25,000\) light-years from Earth. If one light-year is about \(6,000,000,000,000\) (that is, 6 trillion) miles, about how many miles is the twisted double-helix nebula from Earth?

Find each product. \((x+y)^{2}(x-y)^{2}\)

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Find each product. \((s+2)^{2}(s-2)^{2}\)

To understand how the special product \((a+b)^{2}=a^{2}+2 a b+b^{2}\) can be applied to a purely numerical problem. Evaluate \(35^{2},\) using either traditional paper-and-pencil methods or a calculator.
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