/*! 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 57 Consider the equation \(x^{3}+p ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 57

Consider the equation \(x^{3}+p x-q=0,\) where \(p\) and \(q\) are prime numbers. Observe that there are only four possible rational roots here: \(1,-1, q,\) and \(-q\) (a) Show that if \(x=1\) is a root, then we must have \(q=3\) and \(p=2 .\) What are the remaining roots in this case? (b) Show that none of the numbers \(-1, q,\) and \(-q\) can be a root of the equation. Hint: For each case, assume the contrary, and deduce a contradiction.

Short Answer

Expert verified
(a) Remaining roots are complex. (b) x = -1, q, -q are not roots.

Step by step solution

01

Verify the Root x=1

Assume \( x = 1 \) is a root of the equation \( x^3 + px - q = 0 \). Substitute \( x = 1 \) into the equation: \( 1^3 + p \cdot 1 - q = 0 \). Simplifying, we get \( 1 + p - q = 0 \).
02

Solve for p and q

From the equation \( 1 + p - q = 0 \), rearrange to get \( q = 1 + p \). Since \( p \) and \( q \) are prime, the smallest primes are \( p = 2 \) and \( q = 3 \). Verify: if \( p = 2 \), then \( q = 1 + 2 = 3 \), which are both primes.
03

Determine Remaining Roots for x=1

Given the verified roots \( p = 2 \) and \( q = 3 \), rewrite the equation as \( x^3 + 2x - 3 = 0 \). Use polynomial division or synthetic division with \( x = 1 \), confirming that \( (x-1) \) is a factor. The quotient will lead to other factors for remaining roots.
04

Calculate Remaining Roots

After division, the equation becomes \((x - 1) (x^2 + x + 3) = 0\), where \(x^2 + x + 3\) is the quadratic factor. Compute the roots using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = 3 \). This results in no real roots, indicating the remaining roots are complex.
05

Verify x = -1 as a Root

Assume \( x = -1 \) is a root. Substitute into the equation: \( (-1)^3 + p(-1) - q = 0 \) simplifies to \(-1 - p - q = 0\). Solving gives \( q = -1 - p \), implying a negative prime \( q \), which is impossible.
06

Verify x = q as a Root

Assume \( x = q \) is a root. Substitute \( x = q \) into the equation: \( q^3 + pq - q = 0 \), simplifying to \( q^3 = q(p + 1) \). Simplifying gives \( q^2 = p + 1 \), leading to non-prime results, contradicting the prime status of \( p \) and \( q \).
07

Verify x = -q as a Root

Assume \( x = -q \) is a root. Substitute \( x = -q \) into the equation: \( (-q)^3 - pq - q = 0 \) simplifies to \( -q^3 - pq - q = 0 \). Factor to get \( q^2 + p + 1 = 0 \), which is not possible for prime \( p \) and \( q \), again indicating contradiction.
08

Conclusion

For part (a), roots are one real root \( x = 1 \) and two complex roots from the factor \(x^2 + x + 3\). For part (b), negatives or positives of prime \(q\) do not satisfy as roots due to contradictions with integer factor forms and prime definitions.

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Key Concepts

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

Polynomial Division
Polynomial division is a mathematical technique used to divide one polynomial by another polynomial, similar to long division with numbers. When dealing with the polynomial provided, such as \[ x^3 + 2x - 3 \], the goal is to find its factors or roots. Begin with the assumption of known roots or verify them, for example, by using values like \( x = 1 \).
  • If \( x = 1 \) is a root, then \( (x - 1) \) becomes a factor of the polynomial.
  • Dividing the polynomial \( x^3 + 2x - 3 \) by \( (x - 1) \) simplifies it to \( (x - 1)(x^2 + x + 3) \).
  • Each quotient obtained in polynomial division, like \( x^2 + x + 3 \), can also be explored for further factors.
Utilizing polynomial division helps to break down complex polynomials into simpler, more manageable parts, which can assist in identifying all possible roots.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of quadratic equations of the form \( ax^2 + bx + c = 0 \). For the quadratic factor \( x^2 + x + 3 \) obtained from polynomial division, the formula is structured as follows:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this case:
  • \( a = 1 \)
  • \( b = 1 \)
  • \( c = 3 \)
Plug in these values into the formula. The calculation reveals:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 3}}{2 \times 1} = \frac{-1 \pm \sqrt{-11}}{2} \]
Since the discriminant is negative (\( b^2 - 4ac = -11 \)), the equation has no real roots, implying the presence of complex roots.
Complex Roots
Complex roots occur when the solutions to a polynomial equation involve imaginary numbers. This usually happens when the quadratic equation's discriminant (\( b^2 - 4ac \)) is negative. In our case:
  • The quadratic equation \( x^2 + x + 3 \) yields a discriminant of \(-11\), indicating that the roots are not real but complex.
The roots are represented in a general form using the imaginary unit \( i \), where \( i = \sqrt{-1} \):
\[ x = \frac{-1 \pm i\sqrt{11}}{2}\]Complex roots often appear in conjugate pairs, which helps ensure that all coefficients of a polynomial remain real. When dealing with complex roots:
  • It's important to understand their implications in the context of real-world applications.
  • They provide insight into systems that may have oscillatory or phase-shift components.
Prime Numbers in Polynomials
Prime numbers are integers greater than 1, having no divisors other than 1 and themselves. When dealing with polynomials and roots, particularly in equations like \( x^3 + px - q = 0 \), primes play a crucial role in defining possible roots.
Given \( p \) and \( q \) as primes, certain assumptions must be consistently checked for validity:
  • For instance, if \( x = 1 \) is a rational root, leading to \( q = 1 + p \), it must lead to legitimate, prime outcomes, like \( p = 2 \) and \( q = 3 \).
  • Proving that no other possible rational roots, like \( -1, q, -q \), work involves demonstrating contradictions in the prime constraints.
Prime numbers restrict the variety of choices when deducing roots, ensuring that only valid combinations are used when solving such polynomial equations.

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Most popular questions from this chapter

Scipio Ferro of Bologna well-nigh thirty years ago discovered this rule and handed it on to Antonio Maria Fior of Venice, whose contest with Niccol貌 Tartaglia of Brescia gave Niccol貌 occasion to discover it. He / Tartaglial gave it to me in response to my entreaties, though withholding the demonstration. Armed with this assistance, I sought out its demonstration in /various / forms. - Girolamo Cardano, Ars Magna (Nuremberg, 1545 ) The quotation is from the translation of Ars Magna by T. Richard Witmer (New York: Dover Publications, 1993 ). In his book Ars Magna (The Great Art) the Renaissance mathematician Girolamo Cardano \((1501-1576)\) gave the following formula for a root of the equation \(x^{3}+a x=b\). $$x=\sqrt[3]{\frac{b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}-\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}$$ (a) Use this formula and your calculator to compute a root of the cubic equation \(x^{3}+3 x=76\) (b) Use a graph to check the answer in part (a). That is, graph the function \(y=x^{3}+3 x-76,\) and note the \(x-\) intercept. Also check the answer simply by substituting it in the equation \(x^{3}+3 x=76\)

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. $$\frac{x^{3}+2}{x^{4}-8 x^{2}+16}$$

Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). Find a fourth-degree polynomial function that has zeros \(\sqrt{2},-\sqrt{2}, 1,\) and -1 and a graph that passes through (2,-20).

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 $$\frac{1}{(x+2)(x+5)(x-14)}=\frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x-14}$$

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. ) $$\frac{x^{5}}{10,000}-\frac{x^{3}}{50}+\frac{x}{1250}+\frac{1}{2000}=0$$
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