Problem 50
Question
Find all zeros of the polynomial. $$P(x)=x^{3}-x-6$$
Step-by-Step Solution
Verified Answer
The zeros are \(x = 2\), \(x = -1 + i\sqrt{2}\), and \(x = -1 - i\sqrt{2}\).
1Step 1: Understand the Problem
We are given a cubic polynomial \(P(x) = x^3 - x - 6\). The goal is to find all zeros of this polynomial, which are the values of \(x\) such that \(P(x) = 0\).
2Step 2: Use the Rational Root Theorem
The Rational Root Theorem suggests that any rational root of the polynomial, in the form \(\frac{p}{q}\), must have \(p\) as a factor of the constant term \(-6\) and \(q\) as a factor of the leading coefficient, which is \(1\). Thus, possible rational roots could be \(\pm 1, \pm 2, \pm 3, \pm 6\).
3Step 3: Test Possible Rational Roots
We will test each possible rational root by substituting them into \(P(x)\).\- Test \(x = 1\): \(P(1) = 1^3 - 1 - 6 = -6\).\- Test \(x = -1\): \(P(-1) = (-1)^3 + 1 - 6 = -6\).\- Test \(x = 2\): \(P(2) = 2^3 - 2 - 6 = 0\). Since \(x = 2\) makes the polynomial zero, it is a root.
4Step 4: Factor the Polynomial Using a Found Root
Since we found that \(x = 2\) is a root, we can factor the polynomial by \((x - 2)\). Perform synthetic division of \(P(x) = x^3 - x - 6\) by \(x - 2\). The quotient is \(x^2 + 2x + 3\), meaning \(P(x) = (x - 2)(x^2 + 2x + 3)\).
5Step 5: Solve the Quadratic Equation
To find other roots, solve the quadratic equation \(x^2 + 2x + 3 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 2\), \(c = 3\).\\[x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-2 \pm \sqrt{-8}}{2}\]
6Step 6: Calculate Complex Roots
The discriminant \(\Delta = -8\) is negative, indicating complex roots. Simplify \(\frac{-2 \pm \sqrt{-8}}{2}\):\\(x = \frac{-2 \pm 2i\sqrt{2}}{2} = -1 \pm i \sqrt{2}\). Hence, the polynomial has complex roots \(-1 + i\sqrt{2}\) and \(-1 - i\sqrt{2}\).
Key Concepts
Cubic PolynomialRational Root TheoremComplex Roots
Cubic Polynomial
A cubic polynomial is a polynomial of degree three, which means it is an expression that can be written in the form \( ax^3 + bx^2 + cx + d \).
This type of polynomial typically has more than one root, as it involves an equation with one or three real or complex solutions.
In simpler terms, we're dealing with an equation involving a variable raised to the third power. For the polynomial \( P(x) = x^3 - x - 6 \), the leading term is \( x^3 \). This term determines the general shape of the polynomial's graph.
Understanding the form of a cubic polynomial helps us analyze the number of possible solutions and their nature.
These polynomials often require concepts such as factoring, the use of the Rational Root Theorem, and even complex numbers when finding their zeros.
This type of polynomial typically has more than one root, as it involves an equation with one or three real or complex solutions.
In simpler terms, we're dealing with an equation involving a variable raised to the third power. For the polynomial \( P(x) = x^3 - x - 6 \), the leading term is \( x^3 \). This term determines the general shape of the polynomial's graph.
Understanding the form of a cubic polynomial helps us analyze the number of possible solutions and their nature.
These polynomials often require concepts such as factoring, the use of the Rational Root Theorem, and even complex numbers when finding their zeros.
Rational Root Theorem
The Rational Root Theorem is a handy tool when dealing with polynomials.
It allows us to identify possible rational roots (solutions in the form of fractions) of a polynomial equation.
According to the theorem, if a polynomial has a rational solution in the form \( \frac{p}{q} \), then \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient.
This theorem significantly narrows down the possible candidates we need to test when finding roots.
For example, in the polynomial \( P(x) = x^3 - x - 6 \), the Rational Root Theorem suggests possible rational roots like \( \pm 1, \pm 2, \pm 3, \pm 6 \) since they are factors of the constant term \(-6\).
By testing these values, we found that \( x = 2 \) is indeed a root. This root enables us to factor the polynomial and simplify the problem.
Utilizing the Rational Root Theorem can save a lot of time and effort when solving polynomial equations.
It allows us to identify possible rational roots (solutions in the form of fractions) of a polynomial equation.
According to the theorem, if a polynomial has a rational solution in the form \( \frac{p}{q} \), then \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient.
This theorem significantly narrows down the possible candidates we need to test when finding roots.
For example, in the polynomial \( P(x) = x^3 - x - 6 \), the Rational Root Theorem suggests possible rational roots like \( \pm 1, \pm 2, \pm 3, \pm 6 \) since they are factors of the constant term \(-6\).
By testing these values, we found that \( x = 2 \) is indeed a root. This root enables us to factor the polynomial and simplify the problem.
Utilizing the Rational Root Theorem can save a lot of time and effort when solving polynomial equations.
Complex Roots
Complex roots come into play when the polynomial equation doesn't have enough real solutions.
They occur when searching for solutions to polynomials results in taking the square root of a negative number, leading to imaginary numbers.
When dealing with quadratic equations like \( x^2 + 2x + 3 = 0 \) in the factorization of \( P(x) \), we use the quadratic formula.
To solve, we calculated: \[x = \frac{-2 \pm \sqrt{-8}}{2} \]\
The negative discriminant \(-8\) indicates the roots are not real. Instead, we find the solutions in terms of complex numbers: \(-1 \pm i \sqrt{2}\).
Therefore, complex roots explain scenarios where a cubic polynomial doesn't intersect the x-axis at any point in the real plane multiple times and hence, includes imaginary numbers.
Appreciating complex roots adds depth to our understanding of polynomials, allowing the complete analysis of all types of solutions, real or complex.
They occur when searching for solutions to polynomials results in taking the square root of a negative number, leading to imaginary numbers.
When dealing with quadratic equations like \( x^2 + 2x + 3 = 0 \) in the factorization of \( P(x) \), we use the quadratic formula.
To solve, we calculated: \[x = \frac{-2 \pm \sqrt{-8}}{2} \]\
The negative discriminant \(-8\) indicates the roots are not real. Instead, we find the solutions in terms of complex numbers: \(-1 \pm i \sqrt{2}\).
Therefore, complex roots explain scenarios where a cubic polynomial doesn't intersect the x-axis at any point in the real plane multiple times and hence, includes imaginary numbers.
Appreciating complex roots adds depth to our understanding of polynomials, allowing the complete analysis of all types of solutions, real or complex.
Other exercises in this chapter
Problem 50
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