To find the solutions to the equation [tex]\(x^3 + 6x^2 - 40x = 192\)[/tex], we need to understand where the graph of the function [tex]\( y = x^3 + 6x^2 - 40x \)[/tex] intersects the horizontal line [tex]\( y = 192 \)[/tex].
Let's rephrase the problem into one equation by setting the two equations equal to each other:
[tex]\[ x^3 + 6x^2 - 40x = 192 \][/tex]
But for clarity of understanding, we'll convert it to:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
Now, let's solve this polynomial equation step-by-step to find the values of [tex]\( x \)[/tex].
### Step-by-Step Solution
1. Equation Setup:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
2. Finding Rational Roots:
By the Rational Root Theorem, potential rational roots of the polynomial are the factors of the constant term (-192) divided by the factors of the leading coefficient (1). Hence, the possible rational roots are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 32, \pm 48, \pm 64, \pm 96, \pm 192 \][/tex]
3. Testing Potential Roots:
Start by checking some of the potential roots using substitution.
- Testing [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 + 6(-8)^2 - 40(-8) - 192 = -512 + 384 + 320 - 192 \][/tex]
[tex]\[ = 0 \][/tex]
Since this yields 0, [tex]\( x = -8 \)[/tex] is a root.
4. Polynomial Division:
Use polynomial long division or synthetic division to divide the original polynomial by [tex]\( (x + 8) \)[/tex], since [tex]\( x + 8 \)[/tex] is a factor.
Dividing [tex]\( x^3 + 6x^2 - 40x - 192 \)[/tex] by [tex]\( x + 8 \)[/tex] gives:
[tex]\[ x^3 + 6x^2 - 40x - 192 = (x + 8)(x^2 - 2x - 24) \][/tex]
5. Factoring the Quadratic Expression:
We now need to solve the quadratic equation [tex]\( x^2 - 2x - 24 = 0 \)[/tex].
To factor this quadratic, find two numbers that multiply to -24 and add to -2.
[tex]\[ x^2 - 2x - 24 = (x - 6)(x + 4) \][/tex]
6. Finding Additional Roots:
Set the factors equal to zero:
[tex]\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
### Summary and Verification
We have found the roots of the polynomial: [tex]\(x = -8\)[/tex], [tex]\(x = -4\)[/tex], and [tex]\(x = 6\)[/tex]. Verify these solutions by substituting back into the original equation:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
- For [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 + 6(-8)^2 - 40(-8) - 192 = -512 + 384 + 320 - 192 = 0 \][/tex]
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ (-4)^3 + 6(-4)^2 - 40(-4) - 192 = -64 + 96 + 160 - 192 = 0 \][/tex]
- For [tex]\( x = 6 \)[/tex]:
[tex]\[ 6^3 + 6(6^2) - 40(6) - 192 = 216 + 216 - 240 - 192 = 0 \][/tex]
Thus, the solutions to the equation [tex]\( x^3 + 6x^2 - 40x = 192 \)[/tex] are:
[tex]\[ \boxed{x = -8, x = -4, x = 6} \][/tex]
