- Set the cubic equation to zero: \(x^3 + 6x^2 - 40x - 192 = 0\).
- Test the provided values: \(x = -4\) and \(x = -8\) satisfy the equation, while \(x = 192\) does not.
- Factor the polynomial to find all roots, revealing the solutions are \(x \in \{-8, -4, 6\}\).
- The identified solutions from the list are \(\boxed{-4, -8}\).
Explanation
- Problem Analysis and Setup
We are given the system of equations \(y = x^3 + 6x^2 - 40x\) and \(y = 192\). To find the intersection points (which represent the solutions for \(x\)), we set the two expressions for \(y\) equal to each other:
\[x^3 + 6x^2 - 40x = 192\]
Rearranging this equation into the standard polynomial form \(P(x) = 0\), we get:
\[x^3 + 6x^2 - 40x - 192 = 0\]
We are specifically asked to check if the values \(x = -4\), \(x = 192\), and \(x = -8\) are solutions to this equation.
- Checking Potential Solutions
To determine which of the values \(-4, 192, -8\) are solutions, we substitute each into the polynomial \(P(x) = x^3 + 6x^2 - 40x - 192\):
- For \(x = -4\): \((-4)^3 + 6(-4)^2 - 40(-4) - 192 = -64 + 6(16) + 160 - 192 = -64 + 96 + 160 - 192 = 0\). Thus, \(x = -4\) is a solution.
- For \(x = -8\): \((-8)^3 + 6(-8)^2 - 40(-8) - 192 = -512 + 6(64) + 320 - 192 = -512 + 384 + 320 - 192 = 0\). Thus, \(x = -8\) is a solution.
- For \(x = 192\): \((192)^3 + 6(192)^2 - 40(192) - 192 = 7077888 + 221184 - 7680 - 192 = 7291200
eq 0\). Thus, \(x = 192\) is not a solution.
- Finding the Complete Set of Solutions
Since we found two roots \(x = -4\) and \(x = -8\), the polynomial \(P(x)\) must be divisible by \((x + 4)\) and \((x + 8)\). The product is \((x + 4)(x + 8) = x^2 + 12x + 32\).
Dividing \(x^3 + 6x^2 - 40x - 192\) by \(x^2 + 12x + 32\), we obtain the quotient:
\[\frac{x^3 + 6x^2 - 40x - 192}{x^2 + 12x + 32} = x - 6\]
Setting the quotient equal to zero gives the third root, \(x - 6 = 0\), which results in \(x = 6\). Thus, the complete set of solutions to the equation \(x^3 + 6x^2 - 40x = 192\) is \(x \in \{-8, -4, 6\}\). Among the provided values, only \(-4\) and \(-8\) are solutions.
Examples
This problem demonstrates how to solve for intersection points in a system involving a cubic function and a constant. In real-world engineering or physics, such problems arise when determining points of equilibrium or critical thresholds where a process governed by a cubic relationship (e.g., fluid dynamics, stress analysis) reaches a specific constant value. Recognizing how to identify solutions by testing candidates and factoring ensures accuracy in modeling these systems.