/*! 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 98 Show that the equation $$ x^... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 98

Show that the equation $$ x^{5}-x^{4}-x^{3}-5 x^{2}-12 x-6=0 $$ has exactly one rational root, and then prove that it must have either two or four irrational roots.

Short Answer

Expert verified
The rational root is \(-1\). The equation must have two or four irrational roots.

Step by step solution

01

Apply Rational Root Theorem

The Rational Root Theorem states that any rational root of a polynomial equation \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0\) is a fraction \(\frac{p}{q}\) where \(p\) divides the constant term \(a_0\) and \(q\) divides the leading coefficient \(a_n\). For the polynomial \(x^5 - x^4 - x^3 - 5x^2 - 12x - 6\), the constant term is \(-6\) and the leading coefficient is \(1\). Therefore, any rational root must be a divisor of \(-6\) which are \(\pm 1, \pm 2, \pm 3, \pm 6\).
02

Test Rational Roots

Testing the potential rational roots, starting with \(x = 1\): \[1^5 - 1^4 - 1^3 - 5 \times 1^2 - 12 \times 1 - 6 = 1 - 1 - 1 - 5 - 12 - 6 = -24 \] This is not zero. Next, try \(x = -1\):\[(-1)^5 - (-1)^4 - (-1)^3 - 5(-1)^2 - 12(-1) - 6 = -1 - 1 + 1 - 5 + 12 - 6 = 0\]So, \(x = -1\) is a rational root.
03

Perform Polynomial Division

Since \(x = -1\) is a root, divide the polynomial by \(x + 1\) to factor it out. Using synthetic division, divide \(x^5 - x^4 - x^3 - 5x^2 - 12x - 6\) by \(x + 1\). The quotient is \(x^4 - 2x^3 - 3x^2 - 8x - 6\).
04

Determine Nature of Remaining Roots

To determine the nature of the roots of the quotient \(x^4 - 2x^3 - 3x^2 - 8x - 6 = 0\), note that it has degree 4. A degree 4 polynomial can have all real irrational roots or a mix of real and non-real complex roots. Since we've factored out the only rational root, the remaining roots cannot be rational if any were present they would have been found using possible rational roots. Thus, the remaining roots are either two pairs of irrational roots or a mix of irrational and possibly complex roots.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Polynomial Division
When we encounter a polynomial and find a root, particularly a rational one, we can use polynomial division to simplify the polynomial. Polynomial division works like long division for numbers.
It helps us factor out a found root to break down complex polynomials into smaller, more manageable parts.
  • Consider the polynomial: \[x^5 - x^4 - x^3 - 5x^2 - 12x - 6\]
  • Once a root, like \(x = -1\), is discovered, we can divide the polynomial by \(x + 1\), because \(x - (-1) = x + 1\).
  • The process involves synthetic or long division, which simplifies the polynomial to a lower degree.
Performing this division correctly results in something simpler, such as \(x^4 - 2x^3 - 3x^2 - 8x - 6\), allowing further analysis on the nature of remaining roots.
Rational Roots
The Rational Root Theorem is an essential tool when dealing with polynomial equations. It helps determine if a polynomial has rational roots and what those roots might be.
According to this theorem:
  • A rational root of a polynomial \(a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_0 = 0\) must be a fraction \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\), and \(q\) is a factor of the leading coefficient \(a_n\).
  • In our example polynomial: \(x^5 - x^4 - x^3 - 5x^2 - 12x - 6\), the constant term is \(-6\), and the leading coefficient is \(1\).
  • This means possible rational roots are simple because any rational root \(\frac{p}{q}\) must be a factor of \(-6\) alone. Thus, the possible rational roots are \(\pm 1, \pm 2, \pm 3, \pm 6\).
Testing these values helps identify if they satisfy the equation, pinpointing rational roots like \(x = -1\).
Irrational Roots
After uncovering a rational root and simplifying the polynomial, we often ask about the nature of any remaining roots. If our goal is to understand the character of these roots better, it's important to consider irrational roots.
Unlike rational roots, irrational roots cannot be expressed as fractions.
  • Once you've found and factored out a rational root through polynomial division, the degree of the resulting polynomial shows us how many roots remain.
  • In our particular example, we've reduced to a quartic polynomial (degree 4) \(x^4 - 2x^3 - 3x^2 - 8x - 6\).
  • This polynomial can have all real irrational roots or a mix of real and complex roots. But further rational roots won't exist as they've been exhausted by the theorem in the earlier step.
With a degree 4 polynomial, the most likely configurations are pairs of irrational roots. They often occur as conjugate pairs when derived from quadratic components within the polynomial's structure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Volume of a Box An open box with a volume of 1500 \(\mathrm{cm}^{3}\) is to be constructed by taking a piece of cardboard 20 \(\mathrm{cm}\) by \(40 \mathrm{cm},\) cutting squares of side length \(x \mathrm{cm}\) from each comer, and folding up the sides. Show that this can be done in two different ways, and find the exact dimensions of the box in each case.

Use Descartes鈥� Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros. $$ P(x)=x^{4}+x^{3}+x^{2}+x+12 $$

Use a graphing device to find all real solutions of the equation, rounded to two decimal places. $$ x^{4}-x-4=0 $$

Show that the polynomial does not have any rational zeros. $$ P(x)=2 x^{4}-x^{3}+x+2 $$

Transformations of \(y=1 / x^{2}\) In Example 2 we saw that some simple rational functions can be graphed by shifting, stretching, or reflecting the graph of \(y=1 / x\) . In this exercise we consider rational functions that can be graphed by transforming the graph of \(y=1 / x^{2},\) shown on the following page. (a) Graph the function $$ r(x)=\frac{1}{(x-2)^{2}} $$ by transforming the graph of \(y=1 / x^{2}\) . (b) Use long division and factoring to show that the function $$ s(x)=\frac{2 x^{2}+4 x+5}{x^{2}+2 x+1} $$ can be written as $$ s(x)=2+\frac{3}{(x+1)^{2}} $$ Then graph \(s\) by transforming the graph of \(y=1 / x^{2} .\) (c) One of the following functions can be graphed by trans- forming the graph of \(y=1 / x^{2} ;\) the other cannot. Use transformations to graph the one that can be, and explain why this method doesn't work for the other one. $$ p(x)=\frac{2-3 x^{2}}{x^{2}-4 x+4} \quad q(x)=\frac{12 x-3 x^{2}}{x^{2}-4 x+4} $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.