Problem 80
Question
Write a polynomial equation that has three solutions: \(0,2 i,\) and \(-2 i\)
Step-by-Step Solution
Verified Answer
The polynomial equation with the solutions 0, 2i, and -2i is \(x^3 + 4x = 0\).
1Step 1: Convert Solutions into Factors
The solutions of the polynomial equation are given as \(0, 2i,\) and \(-2i\). Each solution corresponds to a factor of the polynomial. If \(x = 0\) is a solution, the related factor is \(x\). If \(x = 2i\) or \(x = -2i\) is a solution, the related factors are \(x - 2i\) and \(x + 2i\), respectively. Therefore, the factors are \(x, (x - 2i),\) and \(x + 2i\).
2Step 2: Multiply the Imaginary Factors
The imaginary factors \((x - 2i)\) and \((x + 2i)\) can be multiplied using the difference of squares formula: \((a - b)(a + b) = a^2 - b^2\). Here, \(a = x\) and \(b = 2i\), so we have: \((x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 - 4(-1) = x^2 + 4\).
3Step 3: Form the Complete Polynomial
Now that we have multiplied the imaginary factors to get \(x^2 + 4\), we can form the complete polynomial by multiplying it by the remaining factor \(x\) from the first step. The complete polynomial is \(x \cdot (x^2 + 4) = x^3 + 4x\).
4Step 4: Check the Polynomial
Verify that the polynomial \(x^3 + 4x\) has the solutions \(0, 2i,\) and \(-2i\). The polynomial can be factored back into \(x(x - 2i)(x + 2i)\), confirming that it indeed has the given solutions.
Key Concepts
Complex NumbersImaginary NumbersFactoring Polynomials
Complex Numbers
Complex numbers are a fundamental concept in mathematics. They expand the idea of numbers beyond the real number line. A complex number is typically written in the form \(a + bi\), where \(a\) and \(b\) are real numbers. The unit \(i\) is the imaginary unit, defined by the property \(i^2 = -1\).
Complex numbers allow us to solve equations that don't have real solutions. For example, the equation \(x^2 + 1 = 0\) does not have any real number solutions since no real number squares to a negative value. However, it has complex solutions \(x = i\) and \(x = -i\).
Complex numbers allow us to solve equations that don't have real solutions. For example, the equation \(x^2 + 1 = 0\) does not have any real number solutions since no real number squares to a negative value. However, it has complex solutions \(x = i\) and \(x = -i\).
- The addition of complex numbers, \((a + bi) + (c + di)\), is straightforward: \((a+c) + (b+d)i\).
- The multiplication follows the distributive property: \((a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i\).
Imaginary Numbers
Imaginary numbers are a subset of complex numbers. They arise when we attempt to take the square root of a negative number. The basic unit of imaginary numbers is \(i\), which is defined such that \(i^2 = -1\).
Though some may find it abstract, imaginary numbers are important for many applications:
Though some may find it abstract, imaginary numbers are important for many applications:
- They are used in engineering, electromagnetism, and signal processing.
- Imaginary numbers appear naturally in contexts like the Fourier transform, simplifying many real-world problems.
Factoring Polynomials
Factoring polynomials involves expressing a polynomial as a product of its factors. This is a crucial skill in algebra that helps in solving polynomial equations.
Consider a polynomial equation with known solutions. Each solution \(x = r\) translates into a factor of the polynomial, \(x - r\). For example, if the solutions are \(x = 0\), \(x = 2i\), and \(x = -2i\), the factors of the polynomial are \(x\), \(x - 2i\), and \(x + 2i\).
Using the difference of squares formula, \((x - a)(x + a) = x^2 - a^2\), simplifies multiplying these factors:
Consider a polynomial equation with known solutions. Each solution \(x = r\) translates into a factor of the polynomial, \(x - r\). For example, if the solutions are \(x = 0\), \(x = 2i\), and \(x = -2i\), the factors of the polynomial are \(x\), \(x - 2i\), and \(x + 2i\).
Using the difference of squares formula, \((x - a)(x + a) = x^2 - a^2\), simplifies multiplying these factors:
- Multiply the imaginary factors: \((x - 2i)(x + 2i) = x^2 + 4\).
- Combine with the real factor to form the polynomial: \(x(x^2 + 4) = x^3 + 4x\).
Other exercises in this chapter
Problem 79
Simplify each expression. See Section 10.1. $$ \frac{6+4 \sqrt{5}}{2} $$
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Use the quadratic formula and a calculator to approximate each solution to the nearest tenth. $$ 3.6 x^{2}+1.8 x-4.3=0 $$
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Simplify each expression. $$ \frac{10-20 \sqrt{3}}{2} $$
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During the seventh stage of the 2010 Paris-Nice bicycle race, Thomas Voeckler posted the fastest average speed, but Alberto Contador won the race. The seventh s
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