Problem 38 - College Algebra with Modeling and Visualization

Question

Solve the polynomial equation. $$ x^{5}+9 x^{3}=x^{4}+9 x^{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = 1\), \(x = 3i\), and \(x = -3i\).

Step 1: Rearrange the Equation

Move all terms to one side of the equation to set it equal to zero. The equation becomes:\[ x^5 + 9x^3 - x^4 - 9x^2 = 0 \]

Step 2: Factor by Grouping

Group the terms in pairs and factor each pair. First, factor the common terms in the first two terms and the last two terms. This gives:\[ x^3(x^2 + 9) - x^2(x^2 + 9) = 0 \]

Step 3: Extract the Common Factor

Notice that \(x^2 + 9\) is a common factor. Factor \(x^2 + 9\) out:\[ (x^3 - x^2)(x^2 + 9) = 0 \]

Step 4: Factor the First Polynomial Further

Factor \(x^3 - x^2\) by extracting the common factor of \(x^2\), resulting in:\[ x^2(x - 1) = 0 \]

Step 5: Set Each Factor to Zero

Set each factor equal to zero and solve:1. \(x^2 = 0\) implies \(x = 0\).2. \(x - 1 = 0\) implies \(x = 1\).3. \(x^2 + 9 = 0\) implies \(x^2 = -9\), which results in complex solutions \(x = \pm 3i\).

Step 6: List All Solutions

The solutions to the equation are the real solutions \(x = 0\), \(x = 1\) and the complex solutions \(x = 3i\), \(x = -3i\).

Key Concepts

Factoring Polynomials

When faced with a polynomial equation, one powerful technique to solve it is factoring. Factoring takes a complex expression and breaks it down into simpler parts (factors) that when multiplied together give the original polynomial. This step is crucial as it helps to simplify the problem, making it easier to identify each possible solution.

Consider the polynomial equation provided: \[x^5 + 9x^3 = x^4 + 9x^2\]By rearranging all terms to one side, we get:\[x^5 + 9x^3 - x^4 - 9x^2 = 0\]The goal is to write this equation in the form of a product, where each factor can be set to zero. The process starts with grouping terms. Grouping similar terms like this helps in identifying common factors.

• First Pair: \(x^3(x^2 + 9)\)
• Second Pair: \(-x^2(x^2 + 9)\)

This gives us: \[x^3(x^2 + 9) - x^2(x^2 + 9) = 0\] Notice how we can extract \(x^2 + 9\) as a common factor to simplify the polynomial further. Thus leading to:\[(x^3 - x^2)(x^2 + 9) = 0\]Each of these factors needs further simplification, making the factoring process a continuous simplification cycle until no more common factors are identifiable.

Complex Solutions

When solving polynomial equations, sometimes you might encounter solutions that are not real numbers. These are known as complex solutions. Complex numbers arise when dealing with polynomials that include negative numbers beneath the square root in the process of solving them.

In the step-by-step solution, after factoring down to:\[x^2(x - 1)(x^2 + 9) = 0\]you get a factor:\[x^2 + 9 = 0\]To solve for \(x\), you subtract 9 from both sides:\[x^2 = -9\]This equation doesn't have real solutions because square roots of negative numbers are not real. Instead, we use the imaginary unit \(i\), where \(i^2 = -1\).

Thus, solving the above, we find:\[x = \pm 3i\]These are complex solutions because they involve \(i\). Complex solutions often come in conjugate pairs like \(3i\) and \(-3i\), helping complete the set of solutions, especially in polynomial equations with complex roots.

Solving Equations

Solving polynomial equations is about finding values of \(x\) that make the equation true. Once factored, each part of the equation can be addressed separately by applying the zero product property. This property states if a product of multiple factors equals zero, at least one of the factors must be zero.

Returning to:\[x^2(x - 1)(x^2 + 9) = 0\]apply the zero product property. Here’s how:

• Set \(x^2 = 0\) leading to \(x = 0\).
• Set \(x - 1 = 0\) leading to \(x = 1\).
• Set \(x^2 + 9 = 0\) leading to complex solutions from a previous section.

By solving each equation independently, you're finding the complete set of solutions for the original polynomial equation. This approach ensures all potential solutions are considered, whether they are real or complex.

Real Solutions

Real solutions of a polynomial equation are those that do not involve the imaginary unit \(i\). They are essentially values of \(x\) that satisfy the equation using only real numbers.

In our provided equation, the real solutions are derived from the factors that are easily solvable without delving into complex numbers:

• From \(x^2 = 0\), we find \(x = 0\).
• From \(x - 1 = 0\), we find \(x = 1\).

Both these values (0 and 1) are termed as real solutions as they do not require the concept of imaginary numbers to solve.

Using real solutions is beneficial for understanding the behavior of the polynomial on a real number line, such as identifying where a polynomial crosses an axis or changes direction. Real solutions are particularly useful in graphing contexts.