Problem 10 - Algebra Form and Function

Question

Give all the solutions of the equations. $$ x^{3}-2 x^{2}+2^{2} x-2^{3}=0 $$

Step-by-Step Solution

Answer

Answer: The solutions are \(x = 2, 2i, -2i\).

Step 1: Rewrite the equation

To make it easier to solve the equation, rewrite the given equation as: $$ x^3 - 2x^2 + 4x - 8 = 0 $$

Step 2: Try to Factor the Equation by Grouping

Group the terms in pairs and try to factor out the common terms: $$ (x^3 - 2x^2) + (4x - 8) = 0 $$ Factor out the common terms: $$ x^2(x - 2) + 4(x - 2) = 0 $$

Step 3: Factor Out the Common Binomial

Notice that both terms have a common binomial factor \((x - 2)\). Factor this out to get: $$ (x^2 + 4)(x - 2) = 0 $$

Step 4: Solve for x using the Zero-Product Property

According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. So, we have two cases: Case 1: \((x^2 + 4) = 0\) Solve for x in this case: $$ x^2 = -4 $$ This equation has complex solutions, as the square root of a negative number is imaginary. Therefore, we have two complex solutions: $$ x_1 = 2i, x_2 = -2i $$ Case 2: \((x - 2) = 0\) Solve for x in this case: $$ x = 2 $$ This is the real solution to the equation. Putting all the solutions together, we have: $$ x = 2, 2i, -2i $$

Key Concepts

Polynomial Equations

Polynomial equations can sometimes seem daunting, but they are simply equations that involve sums of powers of an unknown variable. Let's break down our example: the equation \( x^3 - 2x^2 + 4x - 8 = 0 \) is a polynomial equation of the third degree, also known as a cubic equation. The highest power of the variable, \( x \), tells us it's a cubic equation. Here are some key points about polynomial equations:

• They can have real and/or complex solutions, depending on the degree and the coefficients.
• The fundamental theorem of algebra states that a polynomial equation of degree \( n \) will have exactly \( n \) solutions in the complex number system, though some may be repeated.
• These solutions can be found using several methods, including factoring, the quadratic formula if applicable, or numerical methods for higher-degree polynomials.

Understanding polynomial equations allows us to predict behaviors such as where graphs will cross the x-axis (the solutions) and can provide insights into the symmetry and turning points of the graph.

Factoring by Grouping

Factoring by grouping is an extremely helpful technique when faced with polynomial equations, especially when direct factoring seems unclear. In essence, we rearrange and group terms to find common factors, making it easier to simplify the equation. Here's how it works in our context:

• Initially, we worked with the polynomial \( x^3 - 2x^2 + 4x - 8 \).
• This was grouped as \((x^3 - 2x^2) + (4x - 8) = 0\), allowing us to factor each group separately.
• From \( (x^3 - 2x^2) \), the common factor is \( x^2 \), leading to \( x^2(x - 2) \).
• Similarly, in \((4x - 8)\), the common factor is \( 4 \), giving us \( 4(x - 2) \).

With both groups showing a common binomial factor \((x - 2)\), we combined it: \((x^2 + 4)(x - 2) = 0\). Factoring by grouping reduces the complexity, making it easier to find solutions using further techniques like the zero-product property.

Complex Solutions

Complex solutions arise when the solutions to an equation include imaginary numbers. This happens when we are tasked with finding the square root of a negative number. In our example, solving \((x^2 + 4) = 0\) gives us \(x^2 = -4\), leading to:

• Imaginary unit \( i \), where \( i = \sqrt{-1} \).
• The solutions \( x_1 = 2i \) and \( x_2 = -2i \) because \( x = \pm \sqrt{-4} = \pm 2i \).

These complex solutions might seem abstract at first, but they have real-world applications in fields like engineering and physics. Here are some simplified pointers about complex solutions:

• Any polynomial with real coefficients will have complex solutions appearing in conjugate pairs, if they are not real.
• Complex numbers are expressed as \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part.
• Understanding complex solutions can help in graphing and analyzing polynomial functions, predicting oscillations, and in other calculations that require the full set of solutions.