For what value of k, in the polynomial `f(x)=3x^(4)-9x^(3)+x^(2)+15x+k` completely divisible by `3x^(2)-5`.

Answer

To find the value of \( k \) for which the polynomial \( f(x) = 3x^4 - 9x^3 + x^2 + 15x + k \) is completely divisible by \( 3x^2 - 5 \), we can use polynomial long division. Here’s a step-by-step solution:

### Step 1: Set up the polynomial long division
We will divide \( f(x) \) by \( 3x^2 - 5 \).

### Step 2: Divide the leading terms
The leading term of \( f(x) \) is \( 3x^4 \) and the leading term of the divisor \( 3x^2 - 5 \) is \( 3x^2 \).
To eliminate \( 3x^4 \), we multiply \( 3x^2 \) by \( x^2 \):
\[ 3x^2 \cdot x^2 = 3x^4 \]
Now, we subtract this from \( f(x) \):
\[ (3x^4 - 9x^3 + x^2 + 15x + k) - (3x^4) = -9x^3 + x^2 + 15x + k \]

### Step 3: Repeat the process
Now, we take the leading term of the new polynomial, which is \( -9x^3 \). We divide it by the leading term of the divisor \( 3x^2 \):
\[ -9x^3 \div 3x^2 = -3x \]
Now, we multiply \( 3x^2 - 5 \) by \( -3x \):
\[ (3x^2 - 5)(-3x) = -9x^3 + 15x \]
Subtract this from the current polynomial:
\[ (-9x^3 + x^2 + 15x + k) - (-9x^3 + 15x) = x^2 + k \]

### Step 4: Continue the division
Next, we take the leading term \( x^2 \) and divide it by \( 3x^2 \):
\[ x^2 \div 3x^2 = \frac{1}{3} \]
Now, we multiply \( 3x^2 - 5 \) by \( \frac{1}{3} \):
\[ (3x^2 - 5) \cdot \frac{1}{3} = x^2 - \frac{5}{3} \]
Subtract this from the current polynomial:
\[ (x^2 + k) - (x^2 - \frac{5}{3}) = k + \frac{5}{3} \]

### Step 5: Set the remainder to zero
Since \( f(x) \) is completely divisible by \( 3x^2 - 5 \), the remainder must be zero:
\[ k + \frac{5}{3} = 0 \]
Solving for \( k \):
\[ k = -\frac{5}{3} \]

### Final Answer
Thus, the value of \( k \) for which the polynomial \( f(x) \) is completely divisible by \( 3x^2 - 5 \) is:
\[ \boxed{-\frac{5}{3}} \]