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Chapter 12: Problem 44

(a) Find a cubic polynomial \(f(x)\) with integer coefficients and leading coefficient 1 such that \(1+\sqrt[3]{2}\) is a root of the equation \(f(x)=0\) (b) Use a graphing utility to find out whether \(1-\sqrt[3]{2}\) is a root of the equation.

Short Answer

Expert verified
The cubic polynomial is \( x^3 - 3x^2 + 3x - 3 \); \( 1 - \sqrt[3]{2} \) is not a root.

Step by step solution

01

Express the Root as a Variable

Let \( x = 1 + \sqrt[3]{2} \). We need to eliminate the cube root by expressing it in terms of \(x\). So, \( \sqrt[3]{2} = x - 1 \).
02

Eliminate Cube Root

Cube both sides to eliminate the cube root: \( (\sqrt[3]{2})^3 = (x-1)^3 \). Simplifying gives us \( 2 = (x-1)^3 \).
03

Expand the Cubic Polynomial

Expand \((x-1)^3\) using the binomial theorem: \((x-1)^3 = x^3 - 3x^2 + 3x - 1\). Now, equate it to 2: \( x^3 - 3x^2 + 3x - 1 = 2 \).
04

Form the Polynomial

Bring 2 to the left side to form the polynomial equation: \( x^3 - 3x^2 + 3x - 3 = 0 \). Thus, the cubic polynomial \( f(x) \) is \( x^3 - 3x^2 + 3x - 3 \).
05

Verify the Second Root with Graphing Utility

Check if \( 1 - \sqrt[3]{2} \) is a root. Substitute \( x = 1 - \sqrt[3]{2} \) into the polynomial \( f(x) = x^3 - 3x^2 + 3x - 3 \) and evaluate. Using a graphing utility, find that the value is not 0, indicating it is not a root.

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Key Concepts

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

Integer Coefficients
Integer coefficients in polynomials are terms where each number representing the degree of a variable is a whole number. Whole numbers include all natural numbers, zero, and their negatives.

For example, in the cubic polynomial \(x^3 - 3x^2 + 3x - 3\), the coefficients are 1, -3, 3, and -3.
These values are integers, meaning the entire function consists of whole numbers. Polynomials with integer coefficients are commonly used because they are predictable and easier to manage computationally.
  • When constructing a polynomial with integer coefficients, ensure that every numerical component of the equation is a whole number.
  • Integer coefficients simplify the expression to facilitate solving polynomial equations and applying the Binomial Theorem.
Binomial Theorem
The Binomial Theorem is essential in expanding expressions raised to a power. It provides a systematic method of expanding a binomial raised to any given power. The theorem states that:
\[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]where \(n\) is a non-negative integer, and \(\binom{n}{k}\) is a binomial coefficient.

For the cubic polynomial \((x-1)^3\), the Binomial Theorem helps expand the expression:
  • First, expand \((x-1)^3\) to achieve \(x^3 - 3x^2 + 3x - 1\).
  • Each term represents a component based profoundly on the binomial expansion coefficients \(\binom{3}{k}\) for \(k = 0, 1, 2, 3\).
Understanding this concept ensures that you can handle polynomial expressions effectively and solve equations where transformations like cube expansion are necessary.
Polynomial Equations
Polynomial equations are algebraic expressions that equate a polynomial function to zero. The general form of a polynomial equation is:
\[a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0\]where \(a_n, a_{n-1}, \ldots, a_1, a_0\) are coefficients, and \(x\) represents the variable.
Cubic polynomials are a specific type with the highest degree of three, and they often result in equations like \(x^3 - 3x^2 + 3x - 3 = 0\).
These equations may contain real, complex, or irrational solutions.
  • Identifying roots of a polynomial is solving for values of \(x\) that make the equation true.
  • Graphing utilities can help in finding and verifying potential roots visually.
By exploring polynomial equations, you gain insight into solving and graphing these mathematical concepts used throughout calculus and advanced algebra.

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Most popular questions from this chapter

In a note that appeared in The College Mathematics Journal [vol. \(20(1989), \text { pp. } 139-141]\), Professor Don Redmond proved the following interesting result. Consider the polynomial equation \(f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}=0,\) and sup- pose that the degree of \(f(x)\) is at least 2 and that all of the coefficients are integers. If the three numbers \(a_{0}\) \(a_{n},\) and \(f(1)\) are all odd, then the given equation has no rational roots. Use this result to show that the following equations have no rational roots. (a) \(9 x^{5}-8 x^{4}+3 x^{2}-2 x+27=0\) (b) \(5 x^{5}+5 x^{4}-11 x^{2}-3 x-25=0\)

This exercise completes the discussion of improper rational expressions in this section. (a) Use long division to obtain the following result: \(\frac{2 x^{3}+4 x^{2}-15 x-36}{x^{2}-9}=(2 x+4)+\frac{3 x}{x^{2}-9}\) (b) Find constants \(A\) and \(B\) such that \(3 x /\left(x^{2}-9\right)=A /(x-3)+B /(x+3) .\) (According to the text, you should obtain \(A=B=3 / 2 .\) )

Let \(z=a+b i\) (a) Show that \((\bar{z})^{2}=\overline{z^{2}}\) (b) Show that \((\bar{z})^{3}=\overline{z^{3}}\)

Provides an example in which an error in a partial fraction decomposition is not easily detected with a graphical approach. Indeed, this may be an example of a case in which, to check your partial fractions work, it鈥檚 easier to repeat the algebra than to experiment with numerous viewing rectangles. Decide for yourself after completing the problem. There is an error in the following partial fraction decomposition: $$\frac{1}{(x+2)(x+5)(x-14)}=\frac{-1 / 48}{x+2}+\frac{1 / 57}{x+5}+\frac{1 / 305}{x-14}$$ (a) Let \(f\) and \(g\) denote the two functions defined by the expressions on the left side and the right side, respectively, in the above equation. Use a graphing utility to graph \(f\) and \(g,\) first in the standard viewing rectangle and then in the rectangle [-15,15,5] by \([-0.02,0.04,0.02] .\) In this latter rectangle, note that the graphs do appear to be identical. (People using a software graphing application and looking at the curves on a computer monitor may have a little advantage here over those drawing the graphs on a relatively small graphing calculator screen.) (b) Find a viewing rectangle clearly demonstrating that the graphs of \(f\) and \(g\) are not identical. (c) Find the correct partial fraction decomposition, given that the form is $$\frac{1}{(x+2)(x+5)(x-14)}=\frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x-14}$$

Determine the partial fraction decomposition for each of the given expressions. $$\frac{1}{(x-a)(x-b)} \quad(a \neq b)$$
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