To find the number of units that will yield a profit of $80,000, we first need to convert the profit value into the form consistent with the given profit function \( P(x) \). Since \( P(x) \) is given in hundreds of dollars, we can express \( 80,000 \) dollars as:
\[ P(x) = \frac{80,000}{100} = 800 \]
Now, we set the profit function equal to 800:
\[ P(x) = -x^3 + 5x^2 + 400x - 1200 = 800 \]
We can rearrange this equation to:
\[ -x^3 + 5x^2 + 400x - 1200 - 800 = 0 \]
This simplifies to:
\[ -x^3 + 5x^2 + 400x - 2000 = 0 \]
Multiplying through by -1 to simplify the calculation:
\[ x^3 - 5x^2 - 400x + 2000 = 0 \]
Next, we can use factoring by grouping to solve the equation. To factor it, we first look for possible rational roots. We can test some integer values to find one that satisfies the equation.
Testing \( x = 10 \):
\[ 10^3 - 5(10^2) - 400(10) + 2000 = 1000 - 500 - 4000 + 2000 = 1000 - 500 - 4000 + 2000 = -1500 \ \text{(not a root)} \]
Testing \( x = 20 \):
\[ 20^3 - 5(20^2) - 400(20) + 2000 = 8000 - 2000 - 8000 + 2000 = 0 \ \text{(is a root)} \]
Now, since \( x = 20 \) is a root, we can factor \( (x - 20) \) out of the cubic polynomial using synthetic division or polynomial long division.
Using synthetic division:
20 | 1 -5 -400 2000
| 20 300 -2000
-------------------------
1 15 -100 0
After division, we have:
\[ x^3 - 5x^2 - 400x + 2000 = (x - 20)(x^2 + 15x - 100) \]
Next, we can factor \( x^2 + 15x - 100 \) by finding two numbers that multiply to \(-100\) and add to \(15\):
These numbers are \(20\) and \(-5\), hence we can factor it:
\[ x^2 + 15x - 100 = (x + 20)(x - 5) \]
Therefore, we can express the original cubic polynomial as:
\[ (x - 20)(x + 20)(x - 5) = 0 \]
This gives us three potential solutions:
- \( x - 20 = 0 \) → \( x = 20 \)
- \( x + 20 = 0 \) → \( x = -20 \) (not feasible as production can't be negative)
- \( x - 5 = 0 \) → \( x = 5 \)
So the feasible numbers of units that will give a profit of $80,000 are:
\[ \boxed{20, 5} \]