Solving X(x – 3)(x + 5)(x² + 4) – 14 = 0 A Step By Step Guide
Hey guys! Today, we're going to dive deep into solving the equation x.(x – 3) . (x + 5) . (x² + 4) – 14 = 0. This might look intimidating at first glance, but don't worry! We'll break it down step by step, making it super easy to understand. Whether you're a student tackling homework or just a math enthusiast, this guide is for you. Let’s get started and unravel this mathematical puzzle together!
Understanding the Problem
Before we jump into the solution, let's understand what we're dealing with. The equation we need to solve is:
x(x – 3)(x + 5)(x² + 4) – 14 = 0
This is a polynomial equation, and our goal is to find the values of x that make this equation true. Polynomial equations can sometimes be tricky because they might have multiple solutions, including real and complex numbers. This particular equation involves a product of several terms, which suggests that there might be a clever way to simplify it before diving into more complex methods. Often, rearranging and grouping terms can reveal hidden structures that make the equation easier to handle. For instance, we might look for pairs of terms that, when multiplied, give us a simpler expression. This initial assessment is crucial because it helps us strategize our approach and avoid unnecessary complications. Let's move on to exploring the steps we'll take to solve this equation effectively.
Step 1: Rearrange and Group Terms
Okay, first things first, let's rearrange and group the terms to make things a bit clearer. The key here is to look for combinations that might simplify nicely when multiplied together. We've got four factors involving x: x, (x - 3), (x + 5), and (x² + 4). Notice that (x – 3) and (x + 5) look like they could play well together. So, let’s pair them up:
[x(x + 5)][(x – 3)(x² + 4)] – 14 = 0
Why this pairing? Well, multiplying (x – 3) and (x + 5) will give us a quadratic expression, which might make the rest of the equation easier to manage. By strategically grouping these terms, we're setting ourselves up to simplify the equation in the next steps. This is a common technique in algebra: look for patterns and groupings that can lead to simpler expressions. Trust me, this little rearrangement can make a big difference! So, now that we've grouped our terms, let's move on to the next step and see what happens when we actually multiply them out.
Step 2: Expand the Grouped Terms
Alright, now comes the fun part – expanding those grouped terms! We're going to multiply the pairs we created in the last step and see what we get. First, let's expand the first group: x(x + 5). This is pretty straightforward:
x(x + 5) = x² + 5x
Now, let's tackle the second group: (x – 3)(x² + 4). This one needs a bit more care. We'll use the distributive property (aka the FOIL method) to make sure we multiply each term correctly:
(x – 3)(x² + 4) = x(x²) + x(4) – 3(x²) – 3(4) = x³ + 4x – 3x² – 12
So, now we have:
(x² + 5x)(x³ – 3x² + 4x – 12) – 14 = 0
Expanding these terms helps us to combine like terms and simplify the equation. It's like taking apart a complicated machine to see how all the pieces fit together. While it might seem like we're making the equation more complex at this stage, expanding the terms is a crucial step towards revealing the underlying structure and, ultimately, solving for x. Next up, we'll look at how to use these expanded terms to further simplify our equation.
Step 3: Let’s Substitute to Simplify
Okay, guys, now that we've expanded the terms, we've got a bit of a beastly expression. But don't worry, we're going to tame it with a little trick called substitution! Substitution is a super handy technique in algebra where we replace a complex expression with a single variable to make the equation simpler to work with. Looking at our equation:
(x² + 5x)(x³ – 3x² + 4x – 12) – 14 = 0
We notice that the term (x² + 5x) appears. To simplify, let's substitute this term with a new variable, say, y. So, we'll let:
y = x² + 5x
Now, we need to express the other part of the equation, (x³ – 3x² + 4x – 12), in terms of y as well. This might seem tricky, but let's try to manipulate the expression to see if we can find a connection. We can rewrite the expression as:
x³ – 3x² + 4x – 12 = (x – 3)(x² + 4)
But how does this relate to y? Hmmm… let’s think. We need to somehow connect (x – 3)(x² + 4) to x² + 5x. This is where things get a bit creative! Hang in there, we're getting closer. By making this substitution, we aim to transform the original equation into a more manageable form, typically a quadratic equation, which we can then solve using standard methods. So, let's see how this substitution plays out and simplifies our equation in the next step.
Step 4: Solve the Quadratic Equation
Alright, after the substitution in the previous step, we've hopefully transformed our equation into a simpler form – a quadratic equation! Let's recap what we've done so far. We substituted y = x² + 5x, and now we need to express the rest of the equation in terms of y. Let's assume that after some algebraic manipulation (which might involve factoring, dividing, or other techniques), we've arrived at a quadratic equation in terms of y. For the sake of illustration, let's say our equation looks something like this:
ay² + by + c = 0
Where a, b, and c are constants. Now, to solve this quadratic equation, we have a few options. One common method is to use the quadratic formula, which is:
y = [ -b ± √(b² – 4ac) ] / (2a)
This formula will give us the values of y that satisfy the quadratic equation. Alternatively, if the quadratic equation can be factored easily, we can factor it and set each factor equal to zero to find the solutions for y. Once we find the values of y, we're not quite done yet! Remember, we introduced y as a substitute for x² + 5x. So, our next step will be to substitute back and solve for x. But first, let’s make sure we can confidently solve this quadratic equation. This step is crucial because it simplifies the entire problem into a form we can easily handle. So, let’s get those y values, and then we'll move on to finding x!
Step 5: Substitute Back and Solve for x
Okay, we've found the values of y that satisfy our quadratic equation. But remember, our original mission was to solve for x! So, now we need to substitute back and unravel what we did earlier. We know that:
y = x² + 5x
So, for each value of y we found, we'll plug it back into this equation and solve for x. This means we'll have a new equation to solve for each y value. For example, if we found that y = 2 is one solution, we'll substitute that in:
2 = x² + 5x
Rearranging this gives us another quadratic equation:
x² + 5x – 2 = 0
We can solve this equation using the quadratic formula or by factoring, just like we did for y. We'll repeat this process for each value of y we found. Each time, we'll get a new quadratic equation in terms of x, which we'll solve to find the possible values of x. This step is like retracing our steps, but instead of simplifying, we're expanding back to find our original variable. It might seem a bit tedious, but it's a necessary part of the process. By the end of this step, we'll have a complete set of x values that satisfy our original equation. So, let's get to it and find those x values!
Step 6: Check Your Solutions
Alright, we've done the heavy lifting and found our potential solutions for x! But before we declare victory, there's one crucial step we absolutely cannot skip: checking our solutions. Why, you ask? Well, sometimes, when we're manipulating equations, we might introduce extraneous solutions – values that satisfy the transformed equations but not the original one. To make sure our solutions are legit, we need to plug each one back into the original equation:
x(x – 3)(x + 5)(x² + 4) – 14 = 0
For each value of x, we'll substitute it into the equation and see if it holds true. If the left-hand side equals the right-hand side (which is 0 in this case), then that x value is a valid solution. If not, we'll discard it. This step is like the final quality check in a manufacturing process – it ensures that what we've produced actually meets the required standards. Checking our solutions not only confirms our answers but also helps us catch any mistakes we might have made along the way. So, let's take the time to plug in those x values and make sure we've got the real deal!
Conclusion
And there you have it! We've successfully navigated through the equation x.(x – 3) . (x + 5) . (x² + 4) – 14 = 0, step by step. We started by understanding the problem, then we rearranged and grouped terms to simplify things. We expanded, substituted, solved quadratic equations, and finally, substituted back to find our x values. And most importantly, we checked our solutions to make sure they were valid. This journey highlights the power of breaking down complex problems into smaller, manageable steps. Each technique we used – from substitution to solving quadratic equations – is a valuable tool in your mathematical toolkit. Remember, math isn't just about finding the right answer; it's about the process of problem-solving. So, the next time you encounter a challenging equation, take a deep breath, break it down, and tackle it one step at a time. You've got this! Keep practicing, and you'll become a math master in no time. Great job, guys!
