Finding The X-Intercepts Of F(x) = X^3 + 6x^2 - 25x - 150

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In the realm of mathematics, particularly in algebra and calculus, finding the x-intercepts of a function is a fundamental task. These intercepts, also known as roots or zeros, represent the points where the function's graph intersects the x-axis. In simpler terms, they are the values of x for which the function's output, f(x), equals zero. This exploration delves into the process of determining the x-intercepts of the cubic function f(x)=x3+6x2−25x−150f(x) = x^3 + 6x^2 - 25x - 150. We will navigate through various techniques, including factoring, the rational root theorem, and synthetic division, to systematically uncover these crucial points. Understanding the x-intercepts not only provides insights into the function's behavior but also serves as a cornerstone for solving equations, sketching graphs, and tackling more complex mathematical problems.

The Significance of X-Intercepts

Before we dive into the specifics of solving for the x-intercepts of our given function, it's crucial to understand their significance. X-intercepts are the points where the graph of a function crosses the x-axis. At these points, the y-value (or f(x) value) is zero. They are the solutions to the equation f(x) = 0. These intercepts provide critical information about the function's behavior: they tell us where the function's output is zero, which can represent real-world scenarios such as the equilibrium points in a system or the times when a projectile hits the ground. Identifying x-intercepts is also essential for sketching the graph of a function, as they serve as key anchor points. Furthermore, finding x-intercepts is a fundamental step in solving polynomial equations and inequalities, making it a cornerstone skill in mathematics.

Exploring Factoring Techniques

Our first approach to finding the x-intercepts of f(x)=x3+6x2−25x−150f(x) = x^3 + 6x^2 - 25x - 150 involves exploring factoring techniques. Factoring is the process of breaking down a polynomial expression into a product of simpler expressions. If we can factor the cubic function into linear factors, we can easily find the x-intercepts by setting each factor equal to zero and solving for x. Let's try factoring by grouping, a technique that involves grouping terms together and factoring out common factors. We can group the first two terms and the last two terms: (x3+6x2)+(−25x−150)(x^3 + 6x^2) + (-25x - 150). Now, we factor out the greatest common factor (GCF) from each group. From the first group, we can factor out x2x^2, and from the second group, we can factor out -25: x2(x+6)−25(x+6)x^2(x + 6) - 25(x + 6). Notice that we now have a common factor of (x+6)(x + 6) in both terms. We can factor this out, resulting in (x+6)(x2−25)(x + 6)(x^2 - 25). The expression (x2−25)(x^2 - 25) is a difference of squares, which can be further factored into (x+5)(x−5)(x + 5)(x - 5). Thus, the fully factored form of the function is f(x)=(x+6)(x+5)(x−5)f(x) = (x + 6)(x + 5)(x - 5).

Applying the Rational Root Theorem

When factoring by grouping or other elementary techniques doesn't readily reveal the x-intercepts, we can turn to the Rational Root Theorem. This theorem provides a systematic way to identify potential rational roots of a polynomial equation. A rational root is a root that can be expressed as a fraction p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. In our case, the constant term is -150, and the leading coefficient is 1. Therefore, the possible rational roots are the factors of -150 divided by the factors of 1. The factors of -150 are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±25, ±30, ±50, ±75, and ±150. The factors of 1 are ±1. Thus, the possible rational roots are simply the factors of -150. To test these potential roots, we can use synthetic division or direct substitution. We'll demonstrate how to use synthetic division in the next section.

Utilizing Synthetic Division

Synthetic division is a streamlined method for dividing a polynomial by a linear factor of the form (x - c). It's particularly useful for testing potential rational roots identified by the Rational Root Theorem. Let's demonstrate how to use synthetic division to test if x = 5 is a root of our function f(x)=x3+6x2−25x−150f(x) = x^3 + 6x^2 - 25x - 150. We set up the synthetic division table with the coefficients of the polynomial (1, 6, -25, -150) and the potential root (5). We bring down the first coefficient (1), multiply it by the potential root (5), and add the result to the next coefficient (6). This gives us 11. We then multiply 11 by 5, add the result to -25, and get 30. Finally, we multiply 30 by 5, add the result to -150, and get 0. The last number in the result is the remainder. A remainder of 0 indicates that x = 5 is indeed a root. The other numbers in the result (1, 11, 30) are the coefficients of the quotient, which is a quadratic polynomial. In this case, the quotient is x2+11x+30x^2 + 11x + 30. We can now try to factor this quadratic or use the quadratic formula to find its roots. If we factor the quadratic, we get (x+5)(x+6)(x + 5)(x + 6). This confirms our earlier factoring by grouping result.

Determining the X-Intercepts

Having factored the function as f(x)=(x+6)(x+5)(x−5)f(x) = (x + 6)(x + 5)(x - 5), we can now easily determine the x-intercepts. The x-intercepts occur when f(x) = 0. This happens when any of the factors are equal to zero. So, we set each factor equal to zero and solve for x:

  • x + 6 = 0 => x = -6
  • x + 5 = 0 => x = -5
  • x - 5 = 0 => x = 5

Therefore, the x-intercepts of the function f(x)=x3+6x2−25x−150f(x) = x^3 + 6x^2 - 25x - 150 are x = -6, x = -5, and x = 5. We can express these intercepts as ordered pairs: (-6, 0), (-5, 0), and (5, 0).

Verification and Graphical Representation

To verify our solution, we can substitute these x-intercept values back into the original function and confirm that the result is zero. For x = -6: f(−6)=(−6)3+6(−6)2−25(−6)−150=−216+216+150−150=0f(-6) = (-6)^3 + 6(-6)^2 - 25(-6) - 150 = -216 + 216 + 150 - 150 = 0. For x = -5: f(−5)=(−5)3+6(−5)2−25(−5)−150=−125+150+125−150=0f(-5) = (-5)^3 + 6(-5)^2 - 25(-5) - 150 = -125 + 150 + 125 - 150 = 0. For x = 5: f(5)=(5)3+6(5)2−25(5)−150=125+150−125−150=0f(5) = (5)^3 + 6(5)^2 - 25(5) - 150 = 125 + 150 - 125 - 150 = 0. This confirms that our x-intercepts are correct. Graphically, these intercepts represent the points where the curve of the function crosses the x-axis. A sketch of the graph would show the curve intersecting the x-axis at x = -6, x = -5, and x = 5.

Conclusion

In conclusion, we have successfully determined the x-intercepts of the cubic function f(x)=x3+6x2−25x−150f(x) = x^3 + 6x^2 - 25x - 150. We employed a combination of factoring techniques, the Rational Root Theorem, and synthetic division to systematically identify the roots of the function. The x-intercepts are x = -6, x = -5, and x = 5, which can be expressed as the ordered pairs (-6, 0), (-5, 0), and (5, 0). This exploration highlights the importance of understanding various algebraic techniques and their applications in solving mathematical problems. The ability to find x-intercepts is a fundamental skill in mathematics with far-reaching applications in various fields, including engineering, physics, and economics.