Finding Roots Solve X(x – 3)(x + 5)(x² + 4) – 38 = 0

by Omar Yusuf 53 views

Hey guys, let's dive into this intriguing math problem together! We're tasked with finding the values of 'x' that make the expression x(x – 3)(x + 5)(x² + 4) – 3 – 4 – 26 – 5 = 0 equal to zero. This looks like a fun challenge, so let's break it down step by step.

Understanding the Equation

Before we jump into solving, let's simplify and understand the equation better. The expression can be rewritten as x(x – 3)(x + 5)(x² + 4) – 38 = 0. We need to find the roots of this polynomial, which are the values of 'x' that satisfy the equation. The roots are the values where the polynomial intersects the x-axis on a graph, making the entire expression equal to zero. To successfully find the roots of the expression x(x – 3)(x + 5)(x² + 4) – 38 = 0, we need to understand the nature of polynomial equations and how their factors contribute to the roots. This equation is a quintic polynomial (degree 5), which is formed by multiplying several factors together. The factors include linear terms like (x), (x - 3), and (x + 5), as well as a quadratic term (x² + 4). Each of these factors plays a crucial role in determining the roots of the equation. When we talk about roots, we mean the values of x that make the entire expression equal to zero. For a product of factors to be zero, at least one of the factors must be zero. Therefore, by setting each factor to zero, we can find potential roots of the polynomial. The linear factors are straightforward; for example, x = 0, x - 3 = 0, and x + 5 = 0 give us roots directly. However, the quadratic factor x² + 4 is a bit different because it does not have real roots. This is because x² is always non-negative for real x, and adding 4 makes it strictly positive, so it can never be zero in the real number system. To find the actual roots, we need to substitute the given options into the equation and see which one makes the expression zero. This method is effective because it directly tests potential solutions against the original equation. By substituting each option, we can quickly determine which values of x satisfy the equation. This approach is especially useful when dealing with complex polynomial equations where finding roots analytically might be challenging. For instance, if we substitute x = 0 into the equation, we can evaluate whether the entire expression equals zero. Similarly, we can test x = 3, x = -5, and x = 5 to see if any of these values satisfy the equation. The goal here is to identify which substitution results in the entire expression equaling zero, thereby confirming the root. This direct substitution method is a practical way to solve polynomial equations, especially in multiple-choice scenarios where the options provide specific values to test.

Testing the Options

Now, let's plug in the given options one by one and see what happens:

  • Option a) x = -5

    Substituting x = -5 into the equation, we get: -5((-5) – 3)((-5) + 5)((-5)² + 4) – 38 -5(-8)(0)(29) – 38 0 – 38 = -38 So, x = -5 does not make the expression zero. When we substitute x = -5 into the equation x(x – 3)(x + 5)(x² + 4) – 38 = 0, we are essentially testing whether this value is a root of the polynomial. A root is a value that, when substituted for x, makes the entire expression equal to zero. This is because the roots correspond to the points where the polynomial function intersects the x-axis on a graph. By substituting x = -5, we replace each instance of x in the equation with -5. The equation then becomes -5((-5) – 3)((-5) + 5)((-5)² + 4) – 38. We proceed by simplifying each term within the parentheses and brackets. The term (-5) – 3 simplifies to -8, and the term (-5) + 5 simplifies to 0. The term (-5)² is equal to 25, so (-5)² + 4 becomes 25 + 4, which equals 29. Now the equation looks like this: -5(-8)(0)(29) – 38. Since one of the factors is 0, the entire product -5(-8)(0)(29) becomes zero. Therefore, the equation simplifies to 0 – 38. Finally, 0 – 38 equals -38, which means that when x = -5, the expression does not equal zero. This result is crucial because it tells us that x = -5 is not a root of the polynomial equation. The expression evaluates to -38, which is a non-zero value, indicating that -5 does not satisfy the condition for a root. In the context of the problem, this step is part of a process of elimination. We are given several options, and by testing each one, we can identify which values of x make the equation true. Since substituting x = -5 did not result in zero, we move on to testing the next option. This method of direct substitution is a common and effective strategy for solving polynomial equations, particularly in scenarios where multiple-choice answers are provided. It allows us to quickly check potential roots and narrow down the possibilities until we find the correct solution.

  • Option b) x = 0

    Let's see what happens when we substitute x = 0: 0((0) – 3)((0) + 5)((0)² + 4) – 38 0 – 38 = -38 Nope, x = 0 doesn't work either. Substituting x = 0 into the equation x(x – 3)(x + 5)(x² + 4) – 38 = 0 is a straightforward way to test if zero is a root of the polynomial. When we replace each x in the equation with 0, we get 0((0) – 3)((0) + 5)((0)² + 4) – 38. This simplifies to 0(-3)(5)(4) – 38. The term 0(-3)(5)(4) is a product that includes zero, which means the entire product becomes zero. So the equation reduces to 0 – 38. Calculating 0 – 38 gives us -38, which is not equal to zero. This result tells us that x = 0 does not satisfy the equation, and therefore, zero is not a root of the given polynomial. The reason zero is a significant value to test is that it often simplifies expressions due to the property that any number multiplied by zero is zero. In this case, while zero made the product of the factors zero, the constant term -38 prevented the entire expression from being zero. This constant term is a crucial element of the equation because it shifts the polynomial function vertically on a graph. If the constant term were zero, then x = 0 would indeed be a root. The process of substituting and simplifying is a fundamental technique in algebra for solving equations. It allows us to directly evaluate the expression for specific values and determine if those values are solutions. In this instance, the clear and simple substitution of x = 0 quickly demonstrated that it is not a solution to the equation. This method is particularly useful in situations where we have multiple potential solutions to test, such as in a multiple-choice question. By systematically substituting each option, we can efficiently narrow down the possibilities and find the correct answer.

  • Option c) x = 3

    Plugging in x = 3: 3((3) – 3)((3) + 5)((3)² + 4) – 38 3(0)(8)(13) – 38 0 – 38 = -38 Still not zero. It seems x = 3 is not a root. Substituting x = 3 into the equation x(x – 3)(x + 5)(x² + 4) – 38 = 0 is another step in our process of identifying the roots of the polynomial. When we replace every instance of x with 3, the equation becomes 3((3) – 3)((3) + 5)((3)² + 4) – 38. The first step in simplifying is to evaluate the expressions within the parentheses. The term (3) – 3 equals 0, and the term (3) + 5 equals 8. The term (3)² is 9, so (3)² + 4 becomes 9 + 4, which equals 13. Now, the equation looks like this: 3(0)(8)(13) – 38. Notice that one of the factors in the product is 0, which means the entire product 3(0)(8)(13) becomes zero. This simplifies the equation to 0 – 38. The result of 0 – 38 is -38, which is not equal to zero. This outcome indicates that x = 3 does not make the expression equal to zero, and therefore, 3 is not a root of the polynomial equation. The significance of this step lies in the systematic approach we are taking to test each potential solution. By substituting each given option and evaluating the expression, we can definitively determine whether the value is a root or not. The fact that x = 3 resulted in -38 rather than zero confirms that it is not a solution to the equation. This process of elimination is a crucial strategy in solving polynomial equations, especially when faced with multiple-choice questions. Each substitution provides valuable information, allowing us to narrow down the possibilities and focus on the remaining options. In this case, having tested x = -5, x = 0, and x = 3, we have ruled out these values as roots and can proceed to test the final option, if necessary, with the assurance that we are methodically approaching the solution.

  • Option d) x = 5

    Finally, let's try x = 5: 5((5) – 3)((5) + 5)((5)² + 4) – 38 5(2)(10)(29) – 38 2900 – 38 = 2862 Definitely not zero. So, x = 5 is not a root. Substituting x = 5 into the equation x(x – 3)(x + 5)(x² + 4) – 38 = 0 is the final step in our systematic approach to finding the roots of the polynomial. When we replace each x in the equation with 5, we get 5((5) – 3)((5) + 5)((5)² + 4) – 38. As before, we begin by simplifying the expressions within the parentheses. The term (5) – 3 simplifies to 2, and the term (5) + 5 simplifies to 10. For the term (5)² + 4, we first calculate (5)² which is 25, and then add 4 to get 29. Now, the equation looks like this: 5(2)(10)(29) – 38. Next, we multiply the factors together: 5 * 2 * 10 * 29. This product equals 2900. So the equation becomes 2900 – 38. Subtracting 38 from 2900 gives us 2862, which is clearly not equal to zero. This result confirms that x = 5 does not satisfy the equation and is therefore not a root of the polynomial. The significance of this final substitution is that we have now tested all the given options and none of them resulted in the expression equaling zero. This comprehensive approach is crucial in problem-solving, as it ensures that we have thoroughly explored all potential solutions provided. In the context of a multiple-choice question, if none of the given options are roots, it might indicate a mistake in the question itself or the need to re-evaluate our calculations. However, based on our calculations, none of the values x = -5, x = 0, x = 3, or x = 5 make the equation x(x – 3)(x + 5)(x² + 4) – 38 = 0 true. This methodical testing approach not only helps in finding the correct answer but also reinforces the understanding of how roots function within polynomial equations. By systematically substituting and evaluating, we gain a clearer insight into the behavior of the polynomial and the conditions necessary for a value to be considered a root.

Conclusion

After testing all the options, none of them make the expression equal to zero. This could mean there might be a mistake in the question, or the roots are not among the given options. In such cases, it's always a good idea to double-check the equation and the calculations. Keep practicing, guys, and you'll become polynomial-solving pros in no time!