Problem 34 - College Algebra with Modeling and Visualization

Question

Solve the polynomial equation. $$ x^{2}+x+2=x^{3} $$

Step-by-Step Solution

Answer

The solutions are \(x = 2\), \(x = \frac{-1 + i\sqrt{3}}{2}\), and \(x = \frac{-1 - i\sqrt{3}}{2}\).

Step 1: Bring All Terms to One Side

Start by rewriting the equation so that all terms are on one side. Move the term on the right-hand side to the left. This results in the equation: \[ x^3 - x^2 - x - 2 = 0 \] Now, we have a polynomial in standard form that we can solve.

Step 2: Use Factorization to Simplify

To solve this cubic equation, we first try to factor it. Look for a possible root using the Rational Root Theorem or testing small integer values. Let's test \(x = 1\):\(1^3 - 1^2 - 1 - 2 = 1 - 1 - 1 - 2 = -3 \)It’s not a root. Now test with \(x = -1\): \((-1)^3 - (-1)^2 - (-1) - 2 = -1 - 1 + 1 - 2 = -3 \)Also not a root. Finally, test with \(x = 2\): \(2^3 - 2^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0 \)\(x = 2\) is a root.

Step 3: Use Synthetic Division

Now that we know \(x = 2\) is a root, perform synthetic division of the polynomial \(x^3 - x^2 - x - 2\) by \(x - 2\):\[ 2 | 1 \, -1 \, -1 \, -2 \]\[ | \, 2 \, 2 \, 2 \]\[ ------------------ \]\[ 1 \, 1 \, 1 \, 0 \]The quotient is \(x^2 + x + 1\) and the remainder is \(0\). Thus, the polynomial factors are \( (x - 2)(x^2 + x + 1) = 0 \).

Step 4: Solve the Quadratic Equation

Next, solve the quadratic equation \(x^2 + x + 1 = 0\). Use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 1\), \(b = 1\), and \(c = 1\). Plug these into the formula: \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} \]The solutions are complex numbers: \[ x = \frac{-1 \pm i\sqrt{3}}{2} \].

Step 5: Determine Final Solutions

Combine the real solution found through factorization and the complex solutions derived from the quadratic formula. The complete solutions to the polynomial equation \( x^3 - x^2 - x - 2 = 0 \) are:\( x = 2, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \).

Key Concepts

Cubic Equations

Cubic equations are polynomial equations of degree three. This means the highest power of the variable in the equation is three. Solving cubic equations can be a bit more challenging than solving linear or quadratic equations, but the process is often similar: rearrange, factor, and solve. In the provided exercise, we start with a cubic equation of the form:

• Move all terms to one side to form a standard polynomial equation: \[ x^3 - x^2 - x - 2 = 0 \]
• Attempt to factor the cubic equation. This often begins with finding at least one real solution by testing small values using the Rational Root Theorem.
• Once a real root is found, use it to simplify the polynomial by performing synthetic division.

This reduces the cubic equation to a quadratic equation that can be solved using methods familiar from quadratic equations. For more complex cubic equations, additional methods such as Cardano's formula may be necessary, but those go beyond basic algebra.

Quadratic Formula

The quadratic formula is a powerful tool for finding solutions to any quadratic equation of the form \[ ax^2 + bx + c = 0 \]. It gives us the roots directly by applying a specific formula:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula can yield real or complex solutions depending on the value of the discriminant \( b^2 - 4ac \). In our case, the quadratic equation resulting from the cubic equation's factorization was \[ x^2 + x + 1 = 0 \]. Substituting the coefficients \( a = 1 \), \( b = 1 \), and \( c = 1 \) into the formula gives us:

• \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} \]

Since the discriminant is negative, the solutions are complex numbers, which leads us into the next concept.

Complex Numbers

Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using an "imaginary" unit \( i \) where \( i^2 = -1 \). They are expressed in the form \( a + bi \). The solutions we get using the quadratic formula in the exercise are \[ x = \frac{-1 \pm i\sqrt{3}}{2} \]. These are complex conjugates, which is typical when dealing with real coefficients in polynomial equations. Complex numbers are essential for solving polynomial equations that do not have all real roots. They allow for the roots to reflect in the complex plane, making it possible to solve every polynomial equation. Understanding complex numbers is crucial, especially because they often arise in real-world applications, such as electrical engineering and control systems.