Problem 13
Question
An equation is given, followed by one or more roots of the equation. In each case, determine the remaining roots. $$15 x^{3}-16 x^{2}+9 x-2=0 ; x=(1+\sqrt{2} i) / 3$$
Step-by-Step Solution
Verified Answer
The remaining root is \(x = \frac{2}{5}\).
1Step 1: Identify Complex Conjugate Roots
Since one of the roots is \(x = \frac{1+\sqrt{2}i}{3}\), its complex conjugate \(x = \frac{1-\sqrt{2}i}{3}\) is also a root of the polynomial because the coefficients are real.
2Step 2: Form a Quadratic From Complex Roots
The product of the conjugate roots forms a quadratic equation: \((x - \frac{1+\sqrt{2}i}{3})(x - \frac{1-\sqrt{2}i}{3})\). This can be simplified to \(x^2 - 2\left(\frac{1}{3}\right)x + \left(\left(\frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2\right) = x^2 - \frac{2}{3}x + \frac{1}{3}\).
3Step 3: Divide Original Polynomial by Quadratic
Now, divide the original polynomial \(15x^3 - 16x^2 + 9x - 2\) by the quadratic \(3x^2 - 2x + 1\). Use polynomial long division or synthetic division to find the quotient polynomial.
4Step 4: Solve for the Remaining Root
The quotient obtained from the division is \(5x - 2\). Set this equal to zero: \(5x - 2 = 0\). Solving for \(x\), we get \(x = \frac{2}{5}\).
5Step 5: Verify the Roots
Confirm the roots \(x = \frac{1+\sqrt{2}i}{3}\), \(x = \frac{1-\sqrt{2}i}{3}\), and \(x = \frac{2}{5}\) satisfy the original polynomial by substitution and verification.
Key Concepts
Complex ConjugatePolynomial DivisionRoots of Polynomial Equations
Complex Conjugate
When dealing with complex numbers, understanding the concept of complex conjugates is key. A complex conjugate is formed by changing the sign of the imaginary part of a complex number. For instance, if you have a complex number given by \(a + bi\), its complex conjugate is \(a - bi\).
Why do we care about complex conjugates in polynomials? Because if a polynomial has real coefficients, then the complex roots must appear in conjugate pairs. This happens since the imaginary parts cancel each other out when the polynomial is multiplied out, leaving a real result. In our exercise, the original equation has a root \(x = \frac{1+\sqrt{2}i}{3}\). Its complex conjugate is \(x = \frac{1-\sqrt{2}i}{3}\), which is also a root of the polynomial, ensuring the equation remains balanced with real coefficients.
Why do we care about complex conjugates in polynomials? Because if a polynomial has real coefficients, then the complex roots must appear in conjugate pairs. This happens since the imaginary parts cancel each other out when the polynomial is multiplied out, leaving a real result. In our exercise, the original equation has a root \(x = \frac{1+\sqrt{2}i}{3}\). Its complex conjugate is \(x = \frac{1-\sqrt{2}i}{3}\), which is also a root of the polynomial, ensuring the equation remains balanced with real coefficients.
Polynomial Division
Polynomial division is an essential tool when it comes to simplifying expressions and finding unknown roots. It works much like numerical long division but with variables.
In our example, we need to divide the original polynomial \(15x^3 - 16x^2 + 9x - 2\) by the quadratic polynomial formed from the complex conjugate pair: \(3x^2 - 2x + 1\). This process helps us to factor out part of the polynomial, revealing a simpler polynomial that can be solved more easily.
In our example, we need to divide the original polynomial \(15x^3 - 16x^2 + 9x - 2\) by the quadratic polynomial formed from the complex conjugate pair: \(3x^2 - 2x + 1\). This process helps us to factor out part of the polynomial, revealing a simpler polynomial that can be solved more easily.
- Start by arranging both the dividend and the divisor in descending powers of x.
- Determine how many times the leading term of the divisor can fit into the leading term of the dividend.
- Multiply the entire divisor by this term and subtract from the dividend.
- Repeat the process with the new polynomial until you reach a remainder that cannot be divided further by the divisor.
Roots of Polynomial Equations
The roots of a polynomial equation are the solutions that make the equation equal to zero. They are crucial in understanding the behavior of the polynomial.
In any polynomial equation, the Fundamental Theorem of Algebra assures us that a polynomial of degree \(n\) will have exactly \(n\) roots in the complex number system, counting multiplicity. For our cubic polynomial, this means there are three roots. Two of these roots are complex and conjugate, \(x = \frac{1+\sqrt{2}i}{3}\) and \(x = \frac{1-\sqrt{2}i}{3}\). The remaining root, uncovered through polynomial division, is real: \(x = \frac{2}{5}\).
Finding these roots is not just about solving for \(x\); it offers insights into the function’s shape and its intersection with the x-axis. For example, the existence of the two complex conjugates implies that this polynomial doesn't intersect the x-axis at those points, while the real root shows a point of intersection, marked by \(x = \frac{2}{5}\). Knowing roots helps us sketch graphs, analyze factors, and solve for unknowns in applied problems.
In any polynomial equation, the Fundamental Theorem of Algebra assures us that a polynomial of degree \(n\) will have exactly \(n\) roots in the complex number system, counting multiplicity. For our cubic polynomial, this means there are three roots. Two of these roots are complex and conjugate, \(x = \frac{1+\sqrt{2}i}{3}\) and \(x = \frac{1-\sqrt{2}i}{3}\). The remaining root, uncovered through polynomial division, is real: \(x = \frac{2}{5}\).
Finding these roots is not just about solving for \(x\); it offers insights into the function’s shape and its intersection with the x-axis. For example, the existence of the two complex conjugates implies that this polynomial doesn't intersect the x-axis at those points, while the real root shows a point of intersection, marked by \(x = \frac{2}{5}\). Knowing roots helps us sketch graphs, analyze factors, and solve for unknowns in applied problems.
Other exercises in this chapter
Problem 13
Express each polynomial in the form \(a_{n}\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{n}\right)\). $$4 x^{2}+23 x-6$$
View solution Problem 13
Show that each equation has no rational roots. $$x^{3}+x^{2}-x+1=0$$
View solution Problem 13
Evaluate each expression using the values \(z=2+3 i, w=9-4 i,\) and \(w_{1}=-7-i\). $$(z+w)+w_{1}$$
View solution Problem 13
Determine whether the given value is a zero of the function. \(f(x)=2 x^{3}-3 x+1\) (a) \(x=(\sqrt{3}-1) / 2\) (b) \(x=(\sqrt{3}+1) / 2\)
View solution