Problem 414 - Precalculus

Question

For the following exercises, find all complex solutions (real and non-real). $$ x^{3}+x^{2}+x+1=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = -1, i, -i \).

Step 1: Factor by grouping

To factor the polynomial \( x^3 + x^2 + x + 1 \) by grouping, we first group terms: \( (x^3 + x^2) + (x + 1) \). Then, factor the greatest common factor from each group. The first group \( x^3 + x^2 \) can be factored as \( x^2(x + 1) \). The second group \( x + 1 \) is already factorable as \( 1(x + 1) \). Thus, the grouping yields \( x^2(x + 1) + 1(x + 1) \).

Step 2: Factor out the common binomial

The expression \( x^2(x + 1) + 1(x + 1) \) has a common factor \( (x + 1) \). Factor this out to obtain \( (x + 1)(x^2 + 1) \).

Step 3: Solve for real and non-real solutions

Now solve each factor separately: \( x + 1 = 0 \) gives the real solution \( x = -1 \). For \( x^2 + 1 = 0 \), rearrange to \( x^2 = -1 \). The solutions are \( x = i \) and \( x = -i \), which are non-real complex solutions, where \( i \) is the imaginary unit.

Key Concepts

Factor by Grouping

Factor by grouping is a common technique used to simplify or solve polynomial equations. This method involves grouping terms of the polynomial and factoring them separately, which can reveal common factors.

To use factor by grouping:

• Identify terms that can be grouped together.
• Factor out the greatest common factor from each group.
• If a common binomial factor appears in the grouped terms, factor it out.

In the given polynomial, \( x^3 + x^2 + x + 1 \), we grouped it as \((x^3 + x^2) + (x + 1)\).

The first group, \(x^3 + x^2\), was factored as \(x^2(x + 1)\), and the second was simply \(1(x + 1)\). The common factor \((x + 1)\) was then factored out, yielding \((x + 1)(x^2 + 1)\). This technique is powerful as it turns complex polynomials into simpler factors.

Polynomial Equations

Polynomial equations are expressions composed of variables raised to whole number powers and their coefficients. Solving these equations involves finding the values of the variable that make the equation true.

Polynomial equations can range from the simple linear, like \(2x + 3 = 0\), to more complicated ones like the cubic equation \(x^3 + x^2 + x + 1 = 0\).

These equations can often be solved by factoring, as shown in the given exercise. Factoring breaks the equation into simpler pieces, making it easier to find the solution.

• If the polynomial is factored into a product of simpler polynomials, each factor can be set to zero to find possible solutions.
• A solution can be real or complex, leading to a set of values called the roots of the equation.

Thus, understanding polynomial equations is a key step in many problem-solving approaches in algebra.

Complex Numbers

Complex numbers extend our number system beyond real numbers to include numbers that involve the square root of negative numbers. A complex number is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.

• When solving polynomial equations, complex numbers are essential, especially when real solutions are insufficient to satisfy the equation.
• The complex part, \(bi\), allows us to work with solutions not possible with just real numbers.

In the polynomial \(x^3 + x^2 + x + 1 = 0\), besides the real solution \(x = -1\), we identified complex solutions \(x = i\) and \(x = -i\). Both include \(i\), the imaginary unit, showcasing how complex numbers solve equations that real numbers cannot alone.

Imaginary Unit

The imaginary unit, denoted by \(i\), is defined as the square root of \(-1\), i.e., \(i^2 = -1\). This concept is fundamental to understanding and working with complex numbers.

Here’s what makes the imaginary unit special:

• It enables the extension of real numbers to complex numbers, introducing an expanded scope for solving equations.
• The imaginary unit solves problems involving negative square roots, which real numbers cannot handle directly.

In the polynomial equation \(x^2 + 1 = 0\), solving for \(x\) leads to \(x^2 = -1\). This requires taking the square root of \(-1\), and utilizing the imaginary unit \(i\), we derive \(x = i\) and \(x = -i\). Thus, the imaginary unit plays a critical role in complex solutions, bridging gaps left by real numbers.