Solving: 6(x-3)(x²+4)(x+1) = 0 - Find Real Solutions

The challenge before us is to determine the number of solutions for the equation $6(x-3)(x^2+4)(x+1)=0$ and, more specifically, to identify its real solutions. This involves understanding how each factor contributes to the overall solution set. Let's break down the equation step by step to reveal its secrets.

Understanding the Equation

At its heart, the equation is a product of several factors set equal to zero. The beauty of this setup lies in the zero-product property, which states that if any of these factors equals zero, the entire product becomes zero. Thus, we can find the solutions by setting each factor individually to zero and solving for $x$.

Our equation consists of the following factors:

1. A constant: $6$
2. A linear term: $(x-3)$
3. A quadratic term: $(x^2+4)$
4. Another linear term: $(x+1)$

The constant $6$ doesn't contribute to the solutions, as it can never equal zero. So, we focus on the remaining factors. We need to analyze each of these factors separately to determine the values of $x$ that make the entire equation true. By carefully examining each component, we can systematically uncover the solutions that satisfy the given equation. Understanding each term’s behavior is crucial for correctly identifying all possible solutions, especially distinguishing between real and complex roots.

Finding Real Solutions from Linear Factors

Let's start with the linear factors, as they typically yield straightforward real solutions.

Solving $(x-3) = 0$

To find the value of $x$ that makes this factor zero, we simply add 3 to both sides of the equation:

$x - 3 = 0$
$x = 3$

So, $x = 3$ is one real solution.

Solving $(x+1) = 0$

Similarly, we solve for $x$:

$x + 1 = 0$
$x = -1$

Thus, $x = -1$ is another real solution. These linear factors provide us with two immediate real solutions, which are critical components of the overall solution set. Identifying these simple roots first helps in narrowing down the possibilities and simplifying the analysis of the more complex factors.

Analyzing the Quadratic Factor

Now, let's consider the quadratic factor $(x^2+4)$. To find its solutions, we set it equal to zero:

$x^2 + 4 = 0$
$x^2 = -4$

To solve for $x$, we take the square root of both sides:

$x = \pm \sqrt{-4}$

Since we're taking the square root of a negative number, the solutions will be complex numbers. Specifically:

$x = \pm 2i$

Where $i$ is the imaginary unit, defined as $i = \sqrt{-1}$. Thus, the solutions from this factor are $x = 2i$ and $x = -2i$. Understanding the nature of these solutions is important; they do not lie on the real number line and therefore, do not contribute to the real solutions of the original equation. The quadratic factor introduces complex roots, which expand the solution set beyond real numbers. This distinction is crucial in the context of the problem, as we are specifically interested in identifying only the real solutions.

Counting the Solutions

Now that we've analyzed each factor, we can determine the total number of solutions and identify the real solutions.

• From $(x-3)$, we found one real solution: $x = 3$.
• From $(x+1)$, we found another real solution: $x = -1$.
• From $(x^2+4)$, we found two complex solutions: $x = 2i$ and $x = -2i$.

Therefore, the equation has a total of four solutions (two real and two complex). However, the question specifically asks for the real solutions. The distinction between real and complex solutions is fundamental in this context. While the equation as a whole has multiple solutions, only those that are real numbers are relevant to the specific question being asked. This highlights the importance of carefully considering the domain and nature of the solutions when solving equations.

Final Answer

The equation $6(x-3)(x^2+4)(x+1)=0$ has four solutions. Its real solutions are $x = 3$ and $x = -1$.

In summary, solving equations involves breaking them down into manageable factors, applying appropriate algebraic techniques, and carefully distinguishing between different types of solutions. In this case, understanding the zero-product property and the nature of real versus complex numbers was crucial in arriving at the correct answer.

Additional Insights and Tips

When dealing with polynomial equations, remember to consider the following:

1. Factorization: Always try to factor the equation first. Factoring simplifies the equation and allows you to use the zero-product property.
2. Real vs. Complex Solutions: Be mindful of whether you're looking for real solutions only or all possible solutions (including complex ones).
3. Quadratic Formula: For quadratic factors that don't easily factor, the quadratic formula can be a lifesaver.
4. Graphical Interpretation: Visualizing the equation as a graph can provide insights into the number and nature of the solutions (real roots are where the graph intersects the x-axis).

By applying these techniques and keeping these considerations in mind, you'll be well-equipped to tackle a wide range of equations and find their solutions effectively.

Practice Problems

To reinforce your understanding, try solving these similar equations:

1. $(x+2)(x-5)(x^2+9) = 0$
2. $4(x-1)(x^2+1)(x-4) = 0$
3. $(x+3)(x-7)(x^2+25) = 0$

Identify the real solutions for each equation. Working through these practice problems will help solidify your understanding of how to identify and solve for real solutions in polynomial equations, enhancing your problem-solving skills and building confidence in your abilities.

Conclusion

In conclusion, solving the equation $6(x-3)(x^2+4)(x+1)=0$ involves a multi-faceted approach that requires understanding factorization, the zero-product property, and the distinction between real and complex solutions. By systematically analyzing each factor, we successfully identified the real solutions $x = 3$ and $x = -1$. This process not only answers the specific question but also reinforces broader problem-solving skills applicable to various mathematical contexts.

For further learning and exploration of related topics, consider visiting Khan Academy's Algebra Section.