Solving The Equation: $3x^3(9x^2 + 4)(8x^3 + 27)(3x^2 + 10x + 8) = 0$

Hey math enthusiasts! Today, we're diving deep into the equation: $3x^3(9x^2 + 4)(8x^3 + 27)(3x^2 + 10x + 8) = 0$. Don't worry, it looks a bit intimidating, but we'll break it down step-by-step to find all the solutions. This is a great opportunity to flex our algebra muscles and understand how to solve polynomial equations. Let's get started, shall we?

Understanding the Basics of Polynomial Equations

Alright guys, before we jump into the nitty-gritty, let's refresh our memory on some fundamental concepts. A polynomial equation is, simply put, an equation that involves variables raised to non-negative integer powers, multiplied by coefficients, and summed together. The degree of a polynomial is determined by the highest power of the variable in the equation. In our case, the equation $3x^3(9x^2 + 4)(8x^3 + 27)(3x^2 + 10x + 8) = 0$ is a product of several factors, each contributing to the overall degree. Understanding this helps us predict the number of potential solutions. For example, a quadratic equation (degree 2) usually has two solutions, and a cubic equation (degree 3) typically has three. Keep in mind that these solutions can be real numbers or complex numbers.

Zero Product Property: The Zero Product Property is our best friend in these scenarios. It states that if the product of several factors equals zero, then at least one of the factors must be zero. This lets us break down a complex equation into smaller, more manageable ones. This is exactly what we're going to do. By setting each factor equal to zero, we can find the values of x that satisfy the original equation. We'll be using this property extensively as we solve the given equation. It's the key to unlocking the solutions.

Complex Numbers: Remember, not all solutions will be real numbers. We might encounter complex numbers, which are numbers of the form a + bi, where a and b are real numbers, and i is the imaginary unit (√-1). Complex solutions often come in conjugate pairs (a + bi and a - bi). When we deal with quadratic equations, especially those with negative discriminants, we're sure to find complex solutions.

Alright, with these basics in mind, let’s move on to the actual solving process. We’ll take each factor one by one and find its roots. Let the fun begin!

Step-by-Step Solution Breakdown

Factor 1: $3x^3 = 0$

First up, we have $3x^3 = 0$. This one's pretty straightforward, right? To solve for x, we divide both sides by 3, which gives us $x^3 = 0$. Taking the cube root of both sides gives us $x = 0$. This is a real solution, and since the exponent is 3, this root has a multiplicity of 3, meaning it appears three times. So, we've got one set of solutions already!

Factor 2: $9x^2 + 4 = 0$

Next, let’s tackle $9x^2 + 4 = 0$. This is a quadratic equation. To solve for x, we start by isolating the term with x. Subtracting 4 from both sides gives us $9x^2 = -4$. Then, we divide by 9: $x^2 = -4/9$. Now, we take the square root of both sides: $x = ±√(-4/9)$. Since we have a negative number under the square root, we know we'll be dealing with complex solutions. Simplifying, we get $x = ± (2/3)i$, where i is the imaginary unit. These are two complex conjugate solutions.

Factor 3: $8x^3 + 27 = 0$

Now, let's look at $8x^3 + 27 = 0$. This is a cubic equation, but it's a special type – it's a sum of cubes. We can rewrite it as $(2x)^3 + 3^3 = 0$. This allows us to use the sum of cubes factorization formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. Applying this formula, we get $(2x + 3)((2x)^2 - (2x)(3) + 3^2) = 0$, or $(2x + 3)(4x^2 - 6x + 9) = 0$. Now we have two factors to solve.

• Solving 2x + 3 = 0: Subtracting 3 from both sides gives us $2x = -3$, and dividing by 2 gives us $x = -3/2$. This is another real solution.
• Solving 4x² - 6x + 9 = 0: This is a quadratic equation. We can solve it using the quadratic formula: $x = (-b ± √(b² - 4ac)) / 2a$. In our case, a = 4, b = -6, and c = 9. Plugging these values into the formula, we get $x = (6 ± √((-6)² - 4 * 4 * 9)) / (2 * 4)$. This simplifies to $x = (6 ± √(36 - 144)) / 8$, or $x = (6 ± √(-108)) / 8$. Further simplification gives us $x = (6 ± 6√3 * i) / 8$, which reduces to $x = (3 ± 3√3 * i) / 4$. These are two complex conjugate solutions.

Factor 4: $3x^2 + 10x + 8 = 0$

Finally, let’s solve the quadratic equation $3x^2 + 10x + 8 = 0$. We can try factoring this one. We need two numbers that multiply to give us (3 * 8) = 24 and add up to 10. These numbers are 6 and 4. So, we can rewrite the equation as $3x^2 + 6x + 4x + 8 = 0$. Factoring by grouping, we get $3x(x + 2) + 4(x + 2) = 0$, which gives us $(3x + 4)(x + 2) = 0$. Now we have two factors:

• Solving 3x + 4 = 0: Subtracting 4 from both sides gives us $3x = -4$, and dividing by 3 gives us $x = -4/3$.
• Solving x + 2 = 0: Subtracting 2 from both sides gives us $x = -2$.

Summary of Solutions

Alright, guys, let’s put all our solutions together! We’ve worked through each factor of the original equation and found the following solutions:

• From $3x^3 = 0$: $x = 0$ (multiplicity of 3)
• From $9x^2 + 4 = 0$: $x = (2/3)i$ and $x = -(2/3)i$
• From $8x^3 + 27 = 0$: $x = -3/2$, $x = (3 + 3√3 * i) / 4$, and $x = (3 - 3√3 * i) / 4$
• From $3x^2 + 10x + 8 = 0$: $x = -4/3$ and $x = -2$

So, in total, we have three real solutions and four complex solutions, giving us a grand total of seven solutions when considering the multiplicity of roots. Isn’t it cool how many different types of solutions we can get from just one equation? This comprehensive approach is typical for advanced math problems, where understanding the nature of each factor is key.

Conclusion: Mastering Polynomial Equations

Congratulations! You've successfully navigated the complex world of polynomial equations. We've explored real and complex solutions, used the Zero Product Property, and applied the sum of cubes factorization. This example shows that even complicated-looking equations can be solved with a methodical approach and a good understanding of fundamental concepts. The process involves identifying each factor, using appropriate methods such as factoring, the quadratic formula, or the sum/difference of cubes factorization, and then compiling the solutions. Remember, practice is key. The more you work through these problems, the more confident and proficient you’ll become. Feel free to try some similar problems and challenge yourselves with more complex equations. Keep up the great work, and happy solving! If you enjoyed this explanation, let me know, and I’ll explain another one soon! See ya!