Problem 40
Question
For the following exercises, find all complex solutions (real and non-real). \(x^{3}+x^{2}+x+1=0\)
Step-by-Step Solution
Verified Answer
The complex solutions are \(x = -1\), \(x = i\), and \(x = -i\).
1Step 1: Group Terms and Factor
To start solving the equation \( x^3 + x^2 + x + 1 = 0 \), first notice that you can group the terms in pairs as \((x^3 + x^2) + (x + 1) = 0\). Next, factor by grouping. For the first pair \(x^3 + x^2\), factor out \(x^2\), and for the second pair \(x + 1\), factor out \(1\). This gives us \(x^2(x + 1) + 1(x + 1) = 0\).
2Step 2: Factor Common Expressions Further
Notice that \( (x + 1) \) is a common factor. Thus, we can factor out \( (x + 1) \) from both terms to obtain \( (x + 1)(x^2 + 1) = 0\).
3Step 3: Solve Each Factor
Now, solve for each factor separately. For \( x + 1 = 0\), solve to find \( x = -1 \). For \( x^2 + 1 = 0 \), solve for \( x \) to find the complex solutions. Set up the equation \( x^2 = -1 \).
4Step 4: Find Complex Solutions
The equation \( x^2 = -1 \) can be solved by taking the square root of both sides. This yields \( x = i \) and \( x = -i \), where \( i \) is the imaginary unit.
Key Concepts
Polynomial EquationsFactoring by GroupingComplex NumbersImaginary Unit
Polynomial Equations
Polynomial equations are mathematical expressions equating a polynomial to zero, and we solve for values of the variable that satisfy this equality. In the case of the example equation, the polynomial is of degree three, denoted as \( x^3 + x^2 + x + 1 = 0 \). The degree, which is the highest power of the variable in the polynomial, indicates the number of solutions it potentially has. This is grounded in the fundamental theorem of algebra, which asserts that a polynomial of degree \( n \) has exactly \( n \) roots over the complex numbers. These roots can be real, non-real, or repeated.
To solve polynomial equations, one must identify the roots or solutions, including both real and complex. Identifying solutions may involve factoring, as shown in the original step-by-step solution, which is a common method to simplify polynomial equations into manageable parts.
To solve polynomial equations, one must identify the roots or solutions, including both real and complex. Identifying solutions may involve factoring, as shown in the original step-by-step solution, which is a common method to simplify polynomial equations into manageable parts.
Factoring by Grouping
Factoring by grouping is a technique used to simplify solving polynomial equations, especially when dealing with multi-term polynomials. This method involves rearranging terms and factoring in pairs or groups to find a common factor.
In the original problem, the polynomial \( x^3 + x^2 + x + 1 = 0 \) is grouped into two pairs, \((x^3 + x^2)\) and \((x + 1)\). In each group, common factors are identified and factored out. For \(x^3 + x^2\), \(x^2\) is factored out, resulting in \(x^2(x + 1)\). In the other group, \(1\) is factored resulting in \(1(x + 1)\).
After identifying \((x+1)\) as a common factor, the polynomial simplifies greatly to \((x+1)(x^2+1)=0\). This shows the effectiveness of factoring by grouping, transforming a complex polynomial into simpler expressions that are more straightforward to solve.
In the original problem, the polynomial \( x^3 + x^2 + x + 1 = 0 \) is grouped into two pairs, \((x^3 + x^2)\) and \((x + 1)\). In each group, common factors are identified and factored out. For \(x^3 + x^2\), \(x^2\) is factored out, resulting in \(x^2(x + 1)\). In the other group, \(1\) is factored resulting in \(1(x + 1)\).
After identifying \((x+1)\) as a common factor, the polynomial simplifies greatly to \((x+1)(x^2+1)=0\). This shows the effectiveness of factoring by grouping, transforming a complex polynomial into simpler expressions that are more straightforward to solve.
Complex Numbers
Complex numbers extend the idea of real numbers by incorporating an imaginary part. A complex number is typically represented as \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part, with \(i\) representing the imaginary unit.
In solving polynomial equations, we often encounter complex numbers when the roots involve taking square roots of negative numbers. The given equation \(x^3 + x^2 + x + 1 = 0\) simplifies to roots involving complex numbers when factoring results in expressions like \(x^2 + 1 = 0\).
Understanding complex numbers involves recognizing their geometric representation on the complex plane, where the real part \(a\) lies on the x-axis and the imaginary part \(bi\) on the y-axis. This visualization aids in comprehending operations such as addition, subtraction, and multiplication of complex numbers.
In solving polynomial equations, we often encounter complex numbers when the roots involve taking square roots of negative numbers. The given equation \(x^3 + x^2 + x + 1 = 0\) simplifies to roots involving complex numbers when factoring results in expressions like \(x^2 + 1 = 0\).
Understanding complex numbers involves recognizing their geometric representation on the complex plane, where the real part \(a\) lies on the x-axis and the imaginary part \(bi\) on the y-axis. This visualization aids in comprehending operations such as addition, subtraction, and multiplication of complex numbers.
Imaginary Unit
The imaginary unit, denoted as \(i\), is a mathematical concept representing the square root of \(-1\). It forms the foundation for complex numbers and is essential in solving equations that cannot be solved using only real numbers.
In the step-by-step solution, \(x^2 + 1 = 0\) results in \(x^2 = -1\), prompting the need to solve by finding the square root of both sides. Thus, we have \(x = i\) and \(x = -i\), showing that \(i\) allows us to express solutions involving square roots of negatives.
The imaginary unit follows simple arithmetic laws, such as \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\), creating a cyclical pattern. This cyclical nature of powers of \(i\) aids in simplifying expressions within the realm of complex numbers, making calculations manageable and intuitive.
In the step-by-step solution, \(x^2 + 1 = 0\) results in \(x^2 = -1\), prompting the need to solve by finding the square root of both sides. Thus, we have \(x = i\) and \(x = -i\), showing that \(i\) allows us to express solutions involving square roots of negatives.
The imaginary unit follows simple arithmetic laws, such as \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\), creating a cyclical pattern. This cyclical nature of powers of \(i\) aids in simplifying expressions within the realm of complex numbers, making calculations manageable and intuitive.
Other exercises in this chapter
Problem 39
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