If `(x^4)/((x-1)(x+2))=(1)/(3(x-1))-(16)/(3(x+2))+ x^(2)-x+k`, then `k =`

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To solve the equation

\[ \frac{x^4}{(x-1)(x+2)} = \frac{1}{3(x-1)} - \frac{16}{3(x+2)} + x^2 - x + k, \]

we will first simplify the right-hand side by finding a common denominator.

### Step 1: Find a common denominator for the right-hand side

The common denominator for the fractions on the right-hand side is \(3(x-1)(x+2)\).

Rewriting the right-hand side:

\[ \frac{1}{3(x-1)} = \frac{(x+2)}{3(x-1)(x+2)} \quad \text{and} \quad \frac{-16}{3(x+2)} = \frac{-16(x-1)}{3(x-1)(x+2)}. \]

So, we can rewrite the right-hand side as:

\[ \frac{x+2 - 16(x-1)}{3(x-1)(x+2)} + x^2 - x + k. \]

### Step 2: Combine the fractions

Now, combine the fractions:

\[ \frac{x + 2 - 16x + 16}{3(x-1)(x+2)} + x^2 - x + k = \frac{-15x + 18}{3(x-1)(x+2)} + x^2 - x + k. \]

### Step 3: Combine the entire right-hand side

Now we rewrite the entire right-hand side:

\[ \frac{-15x + 18}{3(x-1)(x+2)} + \frac{3(x^2 - x + k)(x-1)(x+2)}{3(x-1)(x+2)}. \]

This gives us:

\[ \frac{-15x + 18 + 3(x^2 - x + k)(x-1)(x+2)}{3(x-1)(x+2)}. \]

### Step 4: Set the numerators equal

Since the denominators are equal, we can set the numerators equal:

\[ x^4 = -15x + 18 + 3(x^2 - x + k)(x-1)(x+2). \]

### Step 5: Expand the right-hand side

Now, we need to expand \(3(x^2 - x + k)(x-1)(x+2)\):

1. First, expand \((x-1)(x+2) = x^2 + x - 2\).
2. Now, multiply by \((x^2 - x + k)\):

\[ (x^2 - x + k)(x^2 + x - 2) = x^4 + x^3 - 2x^2 - x^3 - x^2 + 2x + kx^2 + kx - 2k. \]

This simplifies to:

\[ x^4 + (k - 3)x^2 + (2 + k)x - 2k. \]

Now, multiplying by 3 gives:

\[ 3x^4 + 3(k - 3)x^2 + 3(2 + k)x - 6k. \]

### Step 6: Set the coefficients equal

Now we have:

\[ x^4 = -15x + 18 + 3x^4 + 3(k - 3)x^2 + 3(2 + k)x - 6k. \]

Rearranging gives:

\[ 0 = 3x^4 + (3(k - 3) + 15)x^2 + (3(2 + k) - 18)x - 6k + 18. \]

### Step 7: Coefficient comparison

Since the coefficients of \(x^4\) must be equal, we have:

1. Coefficient of \(x^4\): \(1 = 3\) (not possible, so we ignore this).
2. Coefficient of \(x^2\): \(0 = 3(k - 3)\) implies \(k - 3 = 0\) or \(k = 3\).
3. Coefficient of \(x\): \(0 = 3(2 + k) - 18\).
4. Constant term: \(0 = -6k + 18\).

### Conclusion

From \(k - 3 = 0\), we find:

\[ k = 3. \]

Thus, the value of \(k\) is

\[ \boxed{3}. \]