What real number should be subtracted from the polynomial `(3x^(3)+10x^(2)-14x+9)` so that (3x-2) divides it exactly?

Explanation

To determine what real number should be subtracted from the polynomial \(3x^3 + 10x^2 - 14x + 9\) so that \(3x - 2\) divides it exactly, we can follow these steps:

### Step 1: Perform Polynomial Long Division
We need to divide the polynomial \(3x^3 + 10x^2 - 14x + 9\) by \(3x - 2\).

1. **Divide the leading term**: Divide the leading term of the dividend \(3x^3\) by the leading term of the divisor \(3x\):
\[ \frac{3x^3}{3x} = x^2 \]
2. **Multiply and subtract**: Multiply \(x^2\) by the entire divisor \(3x - 2\):
\[ x^2(3x - 2) = 3x^3 - 2x^2 \]
Now subtract this from the original polynomial:
\[ (3x^3 + 10x^2 - 14x + 9) - (3x^3 - 2x^2) = (10x^2 + 2x^2) - 14x + 9 = 12x^2 - 14x + 9 \]

### Step 2: Repeat the Process
Now, we repeat the process with the new polynomial \(12x^2 - 14x + 9\).

1. **Divide the leading term**: Divide \(12x^2\) by \(3x\):
\[ \frac{12x^2}{3x} = 4x \]
2. **Multiply and subtract**: Multiply \(4x\) by \(3x - 2\):
\[ 4x(3x - 2) = 12x^2 - 8x \]
Now subtract:
\[ (12x^2 - 14x + 9) - (12x^2 - 8x) = (-14x + 8x) + 9 = -6x + 9 \]

### Step 3: Final Division
Now, we divide \(-6x + 9\) by \(3x - 2\).

1. **Divide the leading term**: Divide \(-6x\) by \(3x\):
\[ \frac{-6x}{3x} = -2 \]
2. **Multiply and subtract**: Multiply \(-2\) by \(3x - 2\):
\[ -2(3x - 2) = -6x + 4 \]
Now subtract:
\[ (-6x + 9) - (-6x + 4) = 9 - 4 = 5 \]

### Step 4: Conclusion
The remainder of the division is \(5\). To make the polynomial \(3x^3 + 10x^2 - 14x + 9\) exactly divisible by \(3x - 2\), we need to subtract this remainder from the polynomial.

Thus, the real number that should be subtracted is:
\[ \boxed{5} \]