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Factor. Assume that variables used as exponents represent positive integers. $$ 36 x^{2 n}-49 $$

Short Answer

Expert verified
The expression is factored as \((6x^n - 7)(6x^n + 7)\).

Step by step solution

01

Recognize the Expression as a Difference of Squares

The expression \(36x^{2n} - 49\) is in the form \(a^2 - b^2\). We can identify the terms such that \(a^2 = 36x^{2n}\) and \(b^2 = 49\). This hints that the expression can be factored using the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b)\).
02

Find Each Square Root

To apply the formula, find the square roots of \(a^2\) and \(b^2\). For \(36x^{2n}\), the square root is \(a = 6x^n\) because \((6x^n)^2 = 36x^{2n}\). For \(49\), the square root is \(b = 7\) because \(7^2 = 49\).
03

Apply the Difference of Squares Formula

Use the difference of squares formula \((a - b)(a + b)\) with \(a = 6x^n\) and \(b = 7\). Substitute to get the factors: \((6x^n - 7)(6x^n + 7)\).
04

Verify the Factorization

To ensure the factorization is correct, multiply \((6x^n - 7)(6x^n + 7)\) back together. Using the distributive property or FOIL, we calculate:\[(6x^n - 7)(6x^n + 7) = (6x^n)^2 - (7)^2 = 36x^{2n} - 49\].This confirms the factorization is correct.

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Key Concepts

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

Factoring Polynomials
Factoring polynomials is an essential skill in algebra that involves breaking down a complex expression into simpler factors. These are simpler expressions that, when multiplied together, give the original polynomial. It's akin to reverse distributing, where you start with a product and work backwards to find the factors.

In our case, factoring revolves around recognizing the polynomial as a Difference of Squares. The Difference of Squares formula is written as \[ a^2 - b^2 = (a - b)(a + b) \].

This formula makes it simple to factor expressions like our exercise, where each term is a perfect square separated by subtraction.
  • The key is to identify your squares. In the example, \( 36x^{2n} \) is a perfect square, as is \( 49 \).
  • Recognize the structure: perfectly recognizing \( a^2 - b^2 \) allows you to use this formula.
  • Use square roots to identify \( a \) and \( b \) (in this case, \( 6x^n \) and \( 7 \)).
In summary, you manipulate polynomials into products of their squares, which allows for easier arithmetic or algebraic operations down the line.
Exponents in Algebra
In algebra, exponents are crucial for representing powers and simplifying complex expressions. They indicate how many times a number or variable is multiplied by itself.

For example, in the expression \(36x^{2n}\), the exponent \(2n\) tells us that the base \(x\) is squared and then possibly raised further depending on the value of \(n\).

Understanding exponents is critical because:
  • They help simplify multiplication—repeated multiplication can be rewritten as a single power.
  • They are used for expanding polynomials, which is essential for higher-level algebra.
  • When factoring, knowing how to deal with exponents allows you to find roots, as we did when identifying \(6x^n\) for our factorization.
Always remember for any exponentiation operation:1. The base number or variable is repeatedly multiplied.2. The exponent dictates the number of times the base is multiplied.

With practice, the manipulation of exponents becomes second nature, which is useful in factoring as well as other algebraic processes.
Distributive Property
The distributive property is a fundamental property in algebra that enables the multiplication of a sum or difference by a number. The general form is \(a(b + c) = ab + ac\). It assists in expanding expressions and supports verifying factorizations, like in our exercise.

When applying the difference of squares formula, you often need to "foil" or multiply the two binomials, verifying results using the distributive property:
  • First, multiply the first terms of each binomial (e.g., \((6x^n) \cdot (6x^n) = 36x^{2n}\)).
  • Then do the outer and inner terms (e.g., \((-7) \cdot (6x^n) + 6x^n \cdot 7\)), which cancel out in a difference of squares.
  • Finally, multiply the last terms (e.g., \((-7) \cdot 7 = -49\)).
Each of these products adds up to reshape the original expression.

The distributive property ensures that factoring is reversible, allowing an expression to be expanded and contracted seamlessly.

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Most popular questions from this chapter

Multiply. See Section 5.4. $$ (x-2)\left(x^{2}+2 x+4\right) $$

Find the value of \(c\) that makes each trinomial a perfect square trinomial. $$ x^{2}+c x+36 $$

Recall that a graphing calculator may be used to check addition, subtraction, and multiplication of polynomials. In the same manner, a graphing calculator may be used to check factoring of polynomials in one variable. For example, to see that $$ 2 x^{3}-9 x^{2}-5 x=x(2 x+1)(x-5) $$ graph \(\mathrm{Y}_{1}=2 x^{3}-9 x^{2}-5 x\) and \(\mathrm{Y}_{2}=x(2 x+1)(x-5) .\) Then trace along both graphs to see that they coincide. Factor the following and use this method to check your results. $$ x^{4}+6 x^{3}+5 x^{2} $$

The function \(f(x)=0.0007 x^{2}+0.24 x+7.98\) can be used to approximate the total cheese production in the United States from 2000 to \(2009,\) where \(x\) is the number of years after 2000 and \(y\) is pounds of cheese (in billions). Round answers to the nearest hundredth of a billion. (Source: National Agricultural Statistics Service, USDA) a. Approximate the number of pounds of cheese produced in the United States in 2000. b. Approximate the number of pounds of cheese produced in the United States in 2005. c. Use this function to estimate the pounds of cheese produced in the United States in 2015. d. From parts \((a),(b),\) and \((c),\) determine whether the number of pounds of cheese produced in the United States is increasing at a steady rate. Explain why or why not.

Factor each polynomial completely. See Examples 1 through 12. $$ 5 x^{2}-14 x-3 $$

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