Solving 4x^5 = 64x^3 Comprehensive Guide To Solutions

In this article, we will delve into the process of solving the equation $4x^5 = 64x^3$. This equation is a polynomial equation, and our goal is to find all values of $x$ that satisfy it. Polynomial equations like this one appear frequently in various fields of mathematics, physics, and engineering. Understanding how to solve them is a fundamental skill. To tackle this problem effectively, we'll employ algebraic manipulation and factoring techniques. We'll start by rearranging the equation to bring all terms to one side, setting the equation equal to zero. This is a crucial first step in solving polynomial equations. Then, we'll look for common factors that can be factored out, simplifying the equation. Factoring is a powerful tool that allows us to break down complex expressions into simpler ones, making them easier to solve. Next, we'll apply the zero-product property, which states that if the product of several factors is zero, then at least one of the factors must be zero. This property is the cornerstone of solving factored polynomial equations. By setting each factor equal to zero, we can find the individual solutions. We'll carefully consider all possible solutions, including real and complex roots. Finally, we'll verify our solutions by substituting them back into the original equation to ensure they satisfy it. This is a critical step in the problem-solving process, as it helps us catch any errors we might have made along the way. Through this step-by-step approach, we'll gain a clear understanding of how to solve this specific equation and similar polynomial equations.

Step-by-Step Solution

1. Rearrange the Equation

Our initial equation is $4x^5 = 64x^3$. To begin, we need to rearrange the equation so that all terms are on one side, setting the equation equal to zero. This is a standard practice when solving polynomial equations, as it allows us to use factoring techniques more effectively. To do this, we subtract $64x^3$ from both sides of the equation. This gives us:

$4x^5 - 64x^3 = 0$

This form of the equation is now suitable for factoring, which is the next step in our solution process. By setting the equation to zero, we've created a scenario where the zero-product property can be applied later on. This property is essential for finding the solutions to polynomial equations, as it allows us to break down the equation into simpler parts. Rearranging the equation is a crucial first step that sets the stage for the rest of the solution.

2. Factor out Common Factors

Now that we have the equation $4x^5 - 64x^3 = 0$, we can identify common factors in both terms. Factoring is a fundamental technique in algebra that simplifies equations and makes them easier to solve. In this case, we observe that both terms have a common factor of $4x^3$. Factoring out $4x^3$ from the equation, we get:

$4x^3(x^2 - 16) = 0$

This step significantly simplifies the equation. We've reduced a fifth-degree polynomial to a product of a cubic term and a quadratic term. The term $(x^2 - 16)$ is a difference of squares, which can be further factored. Recognizing and factoring out common factors is a crucial skill in algebra. It allows us to break down complex expressions into simpler ones, making them more manageable. In this case, factoring out $4x^3$ has made the equation much easier to solve.

3. Factor the Difference of Squares

We've arrived at the equation $4x^3(x^2 - 16) = 0$. Now, we notice that the term $(x^2 - 16)$ is a difference of squares. Recognizing and factoring patterns like the difference of squares is a valuable skill in algebra. The difference of squares pattern is $a^2 - b^2 = (a - b)(a + b)$. Applying this pattern to $(x^2 - 16)$, where $a = x$ and $b = 4$, we get:

$x^2 - 16 = (x - 4)(x + 4)$

Substituting this back into our equation, we have:

$4x^3(x - 4)(x + 4) = 0$

Now, the equation is fully factored. This is a crucial step because it allows us to use the zero-product property to find the solutions. By factoring the difference of squares, we've broken down the quadratic term into two linear factors. This makes it much easier to identify the values of $x$ that will make the equation equal to zero. Factoring is a powerful tool for solving polynomial equations, and recognizing patterns like the difference of squares is key to successful factoring.

4. Apply the Zero-Product Property

We've successfully factored the equation to get $4x^3(x - 4)(x + 4) = 0$. Now, we apply the zero-product property, which states that if the product of several factors is zero, then at least one of the factors must be zero. This property is the cornerstone of solving factored polynomial equations. It allows us to break down the equation into a set of simpler equations, each of which can be solved individually. Applying the zero-product property to our equation, we set each factor equal to zero:

• $4x^3 = 0$
• $x - 4 = 0$
• $x + 4 = 0$

This gives us three simpler equations to solve. Each equation represents a possible solution to the original equation. By setting each factor to zero, we're essentially saying that if any of these factors is zero, the entire product will be zero, which satisfies the original equation. The zero-product property is a fundamental concept in algebra, and it's essential for solving factored polynomial equations.

5. Solve for x

Now we have three equations to solve:

1. $4x^3 = 0$
2. $x - 4 = 0$
3. $x + 4 = 0$

Let's solve each equation separately:

1. For $4x^3 = 0$, we divide both sides by 4 to get $x^3 = 0$. Taking the cube root of both sides gives us $x = 0$. This is one solution to the original equation. It's important to note that this solution has a multiplicity of 3, meaning it appears three times as a root of the equation. The multiplicity of a root can have implications for the behavior of the polynomial function near that root.
2. For $x - 4 = 0$, we add 4 to both sides to get $x = 4$. This is another solution to the original equation. It's a simple linear equation, and solving it is straightforward. This solution represents a point where the graph of the polynomial function crosses the x-axis.
3. For $x + 4 = 0$, we subtract 4 from both sides to get $x = -4$. This is the final solution to the original equation. Similar to the previous solution, this is a simple linear equation, and solving it is straightforward. This solution also represents a point where the graph of the polynomial function crosses the x-axis.

6. List the Solutions

We have found the following solutions for the equation $4x^5 = 64x^3$:

• $x = 0$
• $x = 4$
• $x = -4$

These are the values of $x$ that satisfy the equation. It's important to list all the solutions, as each one is a valid answer. In this case, we have three distinct solutions. It's also worth noting that the degree of the polynomial (5 in this case) indicates the maximum number of solutions the equation can have. However, some solutions may have multiplicities greater than 1, as we saw with the solution $x = 0$. Listing the solutions clearly and accurately is the final step in solving the equation.

Final Answer

The solutions to the equation $4x^5 = 64x^3$ are: -4, 0, 4

In summary, we solved the equation $4x^5 = 64x^3$ by rearranging it, factoring out common factors, factoring the difference of squares, applying the zero-product property, and solving for $x$. We found three solutions: $x = -4$, $x = 0$, and $x = 4$. This process demonstrates the power of algebraic manipulation and factoring techniques in solving polynomial equations. By breaking down a complex equation into simpler parts, we can systematically find all the solutions. Understanding these techniques is crucial for anyone studying mathematics, physics, or engineering. Polynomial equations appear in many different contexts, and being able to solve them is a valuable skill.