/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 If \(n\) is an odd integer, show... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(n\) is an odd integer, show that \(n^{4}+4 n^{2}+11\) is of the form \(16 k\).

Short Answer

Expert verified
Therefore, \(n^4 + 4n^2 + 11\) is a multiple of 16 for any odd integer \(n\).

Step by step solution

01

Understand the Expression

We are given the expression \( n^4 + 4n^2 + 11 \) and need to show that it can be written as \( 16k \), where \( k \) is an integer. To achieve this, we need to demonstrate that the expression is a multiple of 16 for any odd integer \( n \).
02

Substitute Odd Integer Form

Note that any odd integer can be represented as \( n = 2m + 1 \), where \( m \) is an integer. Substitute \( n = 2m + 1 \) into the expression: \( (2m+1)^4 + 4(2m+1)^2 + 11 \).
03

Expand and Simplify the Terms

First, expand \((2m+1)^4\):\[(2m+1)^4 = (2m+1)(2m+1)(2m+1)(2m+1) = 16m^4 + 32m^3 + 24m^2 + 8m + 1.\]Next, expand \(4(2m+1)^2\):\[4(4m^2 + 4m + 1) = 16m^2 + 16m + 4.\]Combine these with 11:\[16m^4 + 32m^3 + 24m^2 + 8m + 1 + 16m^2 + 16m + 4 + 11.\]
04

Combine Like Terms

Add up all like terms from the expansion:\[16m^4 + 32m^3 + (24m^2 + 16m^2) + (8m + 16m) + (1 + 4 + 11).\]This simplifies to:\[16m^4 + 32m^3 + 40m^2 + 24m + 16.\]
05

Factor Out Common Factor

Notice that each term in \( 16m^4 + 32m^3 + 40m^2 + 24m + 16 \) is divisible by 16:\[ 16(m^4 + 2m^3 + 2.5m^2 + 1.5m + 1).\]Since the expression is factored by 16, the result is a multiple of 16.

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Key Concepts

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

Odd Integer Representation
Odd integers are numbers that are not divisible by 2. They appear as numbers like 1, 3, 5, and so on. Each odd integer can be expressed in a specific form: \( n = 2m + 1 \). In this representation, \( m \) is an integer. This formula arises from the fact that every odd integer is one more than an even integer.

By expressing odd integers in this way, we are better equipped to manipulate them mathematically. This representation helps simplify expressions and is often useful when applying algebraic techniques. Whenever you encounter a problem involving odd numbers, think \( n = 2m + 1 \)!

  • Key point: Odd Integer Representation: \( n = 2m + 1 \).
  • Why use it?: Simplifies mathematical expressions and calculations.
Polynomial Expansion
Polynomial expansion is the process of multiplying out the terms in a polynomial expression. For Example, consider the expression \((2m+1)^4\). When expanding this, we multiply \( (2m+1) \) by itself four times:

\((2m+1)(2m+1)(2m+1)(2m+1)\). This results in complex terms such as \(16m^4\), \(32m^3\), and so on. This is a systematic process that transforms a compact polynomial expression into an expanded form, which reveals the terms individually.

Through expansion, each term of the polynomial is exposed, making it easier to simplify and combine similar terms later. The expansion can seem cumbersome but is crucial for further simplification in algebraic problems.

  • Expanding: Breaking down polynomials to individual terms.
  • Example: \((2m+1)^4 = 16m^4 + 32m^3 + ... + 1\)
Factorization
Factorization is the process of breaking down an expression into a product of simpler terms, or factors. It's the opposite of expanding a polynomial. After expanding a polynomial like \((2m+1)^4 + 4(2m+1)^2 + 11\), you may have noticed terms like \(16m^4\), \(32m^3\), etc.

The objective here is to find a common factor for these terms. For the given expression, that common factor is 16. Each term in the expression \(16m^4 + 32m^3 + 40m^2 + 24m + 16\) is divisible by 16, showing that the entire expression can be rewritten as \(16k\), where \(k\) is an integer.

Why factorize? It simplifies the expression and more importantly, aids in identifying patterns or divisible elements, such as demonstrating divisibility by a specific number, like in this exercise with the number 16.

  • Process: Identify and extract common factors from terms.
  • Result: Expression can be expressed as a product \(16(m^4 + 2m^3 + 2.5m^2 + 1.5m + 1)\).
Divisibility by 16
Establishing an expression's divisibility by a specific number, like 16 in this problem, involves showing that the expression is a multiple of that number. In mathematical terms, if an expression can be written as \(16k\) where \(k\) is an integer, it is clear the expression is divisible by 16.

The full expanded expression \(16m^4 + 32m^3 + 40m^2 + 24m + 16\) was shown to be divisible by 16 by factorizing out the common factor. This factorization illustrates that, regardless of the value \(m\) takes, the entire original expression can be grouped into sets of 16, making it divisible by 16.

Determining divisibility is crucial for confirming that a formula holds for all cases, such as when all odd integers are considered in this exercise.

  • Key Insight: Factorize to show divisibility by a given number.
  • Conclusion: The expression is rigorously divisibly by 16.

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Most popular questions from this chapter

Use the Euclidean Algorithm to obtain integers \(x\) and \(y\) satisfying the following: (a) \(\operatorname{gcd}(56,72)=56 x+72 y\). (b) \(\operatorname{gcd}(24,138)=24 x+138 y\) (c) \(\operatorname{gcd}(119,272)=119 x+272 y\) (d) \(\operatorname{gcd}(1769,2378)=1769 x+2378 y\)

Prove that if \(a\) and \(b\) are both odd integers, then \(16 \mid a^{4}+b^{4}-2\).

For an arbitrary integer \(a\), verify the following: (a) \(2 \mid a(a+1)\), and \(3 \mid a(a+1)(a+2)\). (b) \(3 \mid a\left(2 a^{2}+7\right)\) (c) If \(a\) is odd, then \(32 \mid\left(a^{2}+3\right)\left(a^{2}+7\right)\).

(a) A man has \(\$ 4.55\) in change composed entirely of dimes and quarters. What are the maximum and minimum number of coins that he can have? Is it possible for the number of dimes to equal the number of quarters? (b) The neighborhood theater charges \(\$ 1.80\) for adult admissions and \(\$ .75\) for children. On a particular evening the total receipts were \(\$ 90\). Assuming that more adults than children were present, how many people attended? (c) A certain number of sixes and nines is added to give a sum of 126 ; if the number of sixes and nines is interchanged, the new sum is \(114 .\) How many of each were there originally?

Given integers \(a, b, c, d\), verify the following: (a) If \(a \mid b\), then \(a \mid b c\). (b) If \(a \mid b\) and \(a \mid c\), then \(a^{2} \mid b c\). (c) \(a \mid b\) if and only if \(a c \mid b c\), where \(c \neq 0\). (d) If \(a \mid b\) and \(c \mid d\), then \(a c \mid b d\).

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