Solve Polynomial Equation: $\frac{9 X^3-x}{1-3 X}=0$

Hey guys! Today, we're diving deep into the fascinating world of mathematics to tackle a rather juicy equation: $\frac{9 x^3-x}{1-3 x}=0$. If you're a fan of algebra and love the thrill of cracking numerical puzzles, then this one's for you. We'll break it down step-by-step, making sure everyone can follow along, whether you're a math whiz or just starting to explore the algebraic universe. So, grab your calculators, sharpen your pencils, and let's get this equation solved!

Understanding the Equation

Before we jump into solving, let's take a moment to appreciate the equation we're dealing with: $\frac{9 x^3-x}{1-3 x}=0$. At its heart, this is a rational equation, meaning it's a fraction where the numerator and denominator are polynomials. Our main goal here is to find the value(s) of 'x' that make this entire expression equal to zero. It might look a little intimidating with that $x^3$ term, but trust me, it's all about understanding the properties of equations and how to manipulate them. The key principle we'll be using is that for a fraction to be equal to zero, the numerator must be zero, provided that the denominator is not zero. This is a crucial distinction, as division by zero is undefined in mathematics, and we need to keep that in mind.

So, the first major step involves setting the numerator equal to zero: $9x^3 - x = 0$. This is where the real algebraic work begins. We need to find the roots of this cubic polynomial. Remember, a cubic equation can have up to three real roots. We'll use factoring techniques, which are super handy for simplifying and solving polynomial equations. Factoring helps us break down complex expressions into simpler ones, making it easier to isolate 'x'. The process of factoring often involves looking for common factors, applying difference of squares or cubes formulas, or grouping terms. For this specific numerator, $9x^3 - x$, we can immediately spot a common factor. This is usually the best place to start when factoring any polynomial. Identifying and pulling out common factors simplifies the equation significantly and often reveals further opportunities for factoring.

Furthermore, we also need to consider the denominator, $1 - 3x$. It's absolutely vital that this part of the equation never equals zero. If it does, our original fraction would be undefined, and any 'x' value that makes the denominator zero would be an extraneous solution – meaning it's a solution we found mathematically but doesn't actually work in the original equation. So, we'll find the value of 'x' that makes the denominator zero and make sure to exclude it from our final set of answers. This check is a non-negotiable step in solving rational equations. It ensures the validity of our solutions. Think of it as a gatekeeper, preventing invalid answers from passing through. This careful consideration of the denominator is what separates a correct solution from an incorrect one in rational equations.

Factoring the Numerator

Alright, mathletes, let's get down to business with the numerator: $9x^3 - x = 0$. The first thing that jumps out at us is a common factor of 'x' in both terms. So, let's pull that 'x' out: $x(9x^2 - 1) = 0$. This step is super powerful because it immediately tells us that one possible solution is $x=0$. If $x$ is zero, the entire numerator becomes zero, satisfying our condition. But we're not done yet! We still have the second part, $9x^2 - 1$, which we need to factor further. This expression, $9x^2 - 1$, is a classic example of the difference of squares pattern. Remember that pattern? It's in the form $a^2 - b^2 = (a - b)(a + b)$. In our case, $9x^2$ is $(3x)^2$ and $1$ is $1^2$. So, we can rewrite $9x^2 - 1$ as $(3x)^2 - 1^2$.

Applying the difference of squares formula, we get $(3x - 1)(3x + 1)$. So, our factored numerator now looks like: $x(3x - 1)(3x + 1) = 0$. This fully factored form is gold, guys! It breaks down the cubic polynomial into three simple linear factors. Now, for the product of these three factors to be zero, at least one of the factors must be zero. This is the zero product property in action. It's one of the most fundamental and useful properties in algebra. It states that if the product of two or more factors is zero, then at least one of the factors must be zero. We'll apply this property to each of our factors.

This means we have three potential scenarios to consider:

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

Each of these will give us a possible value for 'x'. We've already found one solution: $x=0$. Now, let's solve the other two linear equations. These are straightforward to solve for 'x'. For $3x - 1 = 0$, we add 1 to both sides to get $3x = 1$, and then divide by 3 to find $x = \frac{1}{3}$. For $3x + 1 = 0$, we subtract 1 from both sides to get $3x = -1$, and then divide by 3 to find $x = -\frac{1}{3}$.

So, from factoring the numerator, we've found three potential solutions: $x = 0$, $x = \frac{1}{3}$, and $x = -\frac{1}{3}$. Pretty neat, huh? This is where the understanding of factoring and basic algebraic manipulation really pays off. The difference of squares pattern is a common one, and recognizing it can save you a lot of time and effort when solving equations. Keep an eye out for it!

Checking the Denominator

Now, here's where we bring in the critical second part of solving rational equations: checking the denominator. Remember, we established that the denominator, $1 - 3x$, cannot be equal to zero. So, let's find out which value(s) of 'x' would make it zero. We set the denominator equal to zero: $1 - 3x = 0$. This is a simple linear equation. To solve for 'x', we can add $3x$ to both sides, giving us $1 = 3x$. Then, we divide both sides by 3 to get $x = \frac{1}{3}$.

This means that $x = \frac{1}{3}$ is the value that makes our denominator zero. In the context of our original equation $\frac{9 x^3-x}{1-3 x}=0$, if $x = \frac{1}{3}$, the denominator becomes $1 - 3(\frac{1}{3}) = 1 - 1 = 0$. Division by zero is undefined! Therefore, $x = \frac{1}{3}$ cannot be a valid solution to our equation. It's an extraneous solution.

When we are solving rational equations, finding solutions that make the denominator zero is a really important step. These values are not actual solutions to the original equation because they lead to an undefined expression. It's like trying to divide by zero on a calculator – it just throws an error! So, we must discard any potential solution that causes the denominator to be zero. This is why we perform this check after finding all the potential solutions from setting the numerator to zero.

Let's recap our potential solutions from the numerator: $x = 0$, $x = \frac{1}{3}$, and $x = -\frac{1}{3}$. Now, we compare these with the value that makes the denominator zero, which is $x = \frac{1}{3}$. We see that $x = \frac{1}{3}$ is present in both lists. So, we must eliminate it from our set of solutions.

The remaining potential solutions are $x = 0$ and $x = -\frac{1}{3}$. Let's quickly check these in the denominator to make sure they are valid:

• For $x = 0$: The denominator is $1 - 3(0) = 1 - 0 = 1$. Since $1 \neq 0$, this solution is valid.
• For $x = -\frac{1}{3}$: The denominator is $1 - 3(-\frac{1}{3}) = 1 - (-1) = 1 + 1 = 2$. Since $2 \neq 0$, this solution is also valid.

So, after carefully checking the denominator and eliminating any extraneous solutions, we are left with our true solutions. This process of checking the denominator is a fundamental part of solving any equation involving fractions, especially rational equations. It ensures that the answers we provide are mathematically sound and valid within the original equation's constraints. Always remember this step, guys!

The Final Solution

After navigating the twists and turns of algebraic manipulation and diligently checking for extraneous solutions, we've arrived at the final answer! We started with the equation $\frac{9 x^3-x}{1-3 x}=0$. Our primary strategy was to set the numerator equal to zero, since any fraction equaling zero must have a zero numerator. This led us to the cubic equation $9x^3 - x = 0$. Through careful factoring, we pulled out the common factor 'x', resulting in $x(9x^2 - 1) = 0$. Then, recognizing $9x^2 - 1$ as a difference of squares, we further factored it into $(3x - 1)(3x + 1)$.

This gave us the fully factored numerator: $x(3x - 1)(3x + 1) = 0$. Applying the zero product property, we found three potential solutions: $x=0$, $3x-1=0$ (which yields $x=\frac{1}{3}$), and $3x+1=0$ (which yields $x=-\frac{1}{3}$). These were our candidates for the solution.

However, the crucial next step was to check the denominator, $1 - 3x$. We found that if $x = \frac{1}{3}$, the denominator becomes zero, making the original expression undefined. Therefore, $x = \frac{1}{3}$ is an extraneous solution and must be discarded. This is a really important step in solving rational equations, so never forget to check your denominator!

By eliminating the extraneous solution, we are left with the valid solutions. The values of 'x' that satisfy the equation $\frac{9 x^3-x}{1-3 x}=0$ are $x = 0$ and $x = -\frac{1}{3}$.

So, the solution set for this equation is { $0, -\frac{1}{3}$ }. We can write this formally as $x \in \{0, -\frac{1}{3}\}$. These are the only values of 'x' that make the given rational expression equal to zero without causing division by zero. It's always satisfying when you can fully solve an equation and be confident in your answers. Remember, math is all about patience, precision, and understanding the underlying principles. Keep practicing, and you'll master these types of problems in no time!

Keep exploring the incredible world of mathematics, and don't hesitate to tackle more challenging equations. Every problem solved is a step forward in your mathematical journey. Happy solving, everyone!