/*! 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 20 Find the rational roots of each ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the rational roots of each equation, and then solve the equation. (Use the rational roots theorem and the upper and lower bound theorem, as in Example 2.) $$3 x^{3}-16 x^{2}+17 x-4=0$$

Short Answer

Expert verified
The rational roots are \(x = \frac{1}{3}, \frac{2}{3}, 6\).

Step by step solution

01

Identify Potential Rational Roots

Using the Rational Root Theorem, identify potential rational roots of the polynomial equation. The theorem states that any rational root of the polynomial equation \(ax^n + bx^{n-1} + ... + k = 0\) is of the form \(\pm \frac{p}{q}\), where \(p\) is a factor of the constant term \(k\) and \(q\) is a factor of the leading coefficient \(a\). For the polynomial \(3x^3 - 16x^2 + 17x - 4 = 0\), \(p\) are the factors of \(-4\) and \(q\) are the factors of \(3\). This gives potential roots: \(\pm 1, \pm 2, \pm 4, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}\).
02

Test Potential Roots Using Substitution

Substitute each potential rational root into the polynomial to see if it produces zero. When substituting \(x = 1\), the polynomial \(3(1)^3 - 16(1)^2 + 17(1) - 4 = 0\) does not satisfy. Testing \(x = 2\), we get \(3(2)^3 - 16(2)^2 + 17(2) - 4 = 0\), which is also not zero. Continue testing until a root is found.
03

Confirm Root and Factor Polynomial

After testing, you find that \(x = \frac{1}{3}\) is a root because substituting it gives zero. Use synthetic division or polynomial division to divide the original polynomial by \(x - \frac{1}{3}\) to simplify it into a quadratic, \(3x^2 - 16x + 12\).
04

Solve the Quadratic Equation

Solve the quadratic equation \(3x^2 - 16x + 12 = 0\) by either factoring, using the quadratic formula, or completing the square. Factoring gives \((3x - 2)(x - 6) = 0\). Thus the roots are \(x = \frac{2}{3}\) and \(x = 6\).
05

Compile the Roots

The rational roots of the original polynomial equation \(3x^3 - 16x^2 + 17x - 4 = 0\) are \(x = \frac{1}{3}, \frac{2}{3}, 6\).

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 91Ó°ÊÓ!

Key Concepts

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

Upper and Lower Bound Theorem
The Upper and Lower Bound Theorem is a useful tool in helping us understand the range within which all real roots of a polynomial can be found. Generally, this theorem gives us a sense of the interval we should be working with when hunting for roots. If we know a specific value does not yield trivial outcomes like zero when checked against a polynomial, it could serve as a boundary marker. Here’s how it works practically: - **Upper Bound**: If we divide the polynomial by \(x - c\) and all the resulting numbers in the synthetic division sequence are non-negative (all zeros included), \(c\) is considered an upper bound for the roots. - **Lower Bound**: Conversely, if each number in the synthetic division sequence alternates sign, then \(c\) is a lower bound.By understanding the bounds, we cut down the number of potential candidates when searching for rational roots using the Rational Root Theorem, greatly optimizing our process.
Synthetic Division
Synthetic division is a simpler and faster alternative to traditional long division for dividing polynomials, particularly handy when we're testing potential roots. It's especially useful with linear divisors like \(x - c\). This method helps us determine if a particular value is a root of the polynomial and can be used to simplify polynomials by dividing out roots. The process involves a few steps: - List the coefficients of the polynomial. - Use the potential root \(c\) in synthetic division: bring down the leading coefficient, multiply it by \(c\), and adjust each subsequent coefficient accordingly.- If the final result (remainder) is zero, then \(c\) is indeed a root.Synthetic division not only helps to confirm a rational root but provides a reduced polynomial, which can further be analyzed or decomposed into simpler factors.
Polynomial Equations
Polynomial equations take the form \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0\). Solving these equations involves finding the values of \(x\) (roots) that make the polynomial equal to zero. Here’s how you approach solving a polynomial equation: - **Identify possible rational roots** by using the Rational Root Theorem, which narrows down the possibilities to factors of the constant and leading coefficient. - **Test these potential roots** using substitution or synthetic division to verify which ones satisfy the equation.- **Simplify the polynomial** using techniques like synthetic division to break it down when a root is found.- **Solve the resulting equations**: After reducing the polynomial, solve the simpler quadratic or linear equations.As evidenced by the original example, polynomial equations sometimes require finding multiple feasible values, each of which corresponds to a different root.

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

Determine the partial fraction decomposition for each of the given rational expressions. Hint: In Exercises \(17,18,\) and \(26,\) use the rational roots theorem to factor the denominator. $$\frac{x}{x^{3}+8}$$

Determine the partial fraction decomposition for each of the given rational expressions. Hint: In Exercises \(17,18,\) and \(26,\) use the rational roots theorem to factor the denominator. $$\frac{x+1}{x^{4}-16}$$

(a) Let \(\theta=2 \pi / 7 .\) Use the reference angle concept to explain why \(\cos 3 \theta=\cos 4 \theta,\) then use your calculator to confirm the result. (b) For this portion of the exercise, assume as given the following two trigonometric identities: $$ \begin{array}{l} \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \\ \cos 4 \theta=8 \cos ^{4} \theta-8 \cos ^{2} \theta+1 \end{array} $$ Use these identities and the result in part (a) to show that \(\cos (2 \pi / 7)\) is a root of the equation $$ 8 x^{4}-4 x^{3}-8 x^{2}+3 x+1=0 $$ (c) List the prossibilities for the rational roots of equation (1). Then use synthetic division and the remainder theorem to show that there is only one rational root. Check that the reduced equation in this case is $$ 8 x^{3}+4 x^{2}-4 x-1=0 $$ (d) The work in parts (a) through (c) shows that the number \(\cos (2 \pi / 7)\) is a root of equation (2). By following the same technique, it can be shown that the numbers \(\cos (4 \pi / 7)\) and \(\cos (6 \pi / 7)\) also are roots of equation(2). Use this fact, along with Table 2 in Section \(12.4,\) to evaluate each of the following quantities. Then use a calculator to check your answers. (i) \(\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{6 \pi}{7}\) (ii) \(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}\)

Each polynomial equation has exactly one negative root. (a) Use a graphing utility to determine successive integer bounds for the root. (b) Use the method of successive approximations to locate the root between successive thousandths. (Make use of the graphing utility to generate the required tables. ) $$\frac{x^{5}}{10,000}-\frac{x^{3}}{50}+\frac{x}{1250}+\frac{1}{2000}=0$$

Find a quadratic equation with the given roots. Write your answers in the form \(A x^{2}+B x+C=0\) Suggestion: Make use of Table 2. $$r_{1}=1+i \sqrt{3}, r_{2}=1-i \sqrt{3}$$

See all solutions

Recommended explanations on Math Textbooks

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.