If `x=(4)/(3)` is a zero of the polynomial `p(x)=6x^(3)-11x^(2)+kx-20` , then find the value of k .

Answer

To find the value of \( k \) such that \( x = \frac{4}{3} \) is a zero of the polynomial \( p(x) = 6x^3 - 11x^2 + kx - 20 \), we can follow these steps:

### Step 1: Substitute \( x = \frac{4}{3} \) into the polynomial
Since \( x = \frac{4}{3} \) is a zero of the polynomial, we can substitute this value into the polynomial and set it equal to zero.

\[ p\left(\frac{4}{3}\right) = 6\left(\frac{4}{3}\right)^3 - 11\left(\frac{4}{3}\right)^2 + k\left(\frac{4}{3}\right) - 20 = 0 \]

### Step 2: Calculate \( \left(\frac{4}{3}\right)^2 \) and \( \left(\frac{4}{3}\right)^3 \)
First, we calculate \( \left(\frac{4}{3}\right)^2 \) and \( \left(\frac{4}{3}\right)^3 \):

\[ \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
\[ \left(\frac{4}{3}\right)^3 = \frac{64}{27} \]

### Step 3: Substitute these values back into the polynomial
Now we substitute these values back into the polynomial equation:

\[ p\left(\frac{4}{3}\right) = 6\left(\frac{64}{27}\right) - 11\left(\frac{16}{9}\right) + k\left(\frac{4}{3}\right) - 20 = 0 \]

### Step 4: Simplify the equation
Now, simplify each term:

1. \( 6 \cdot \frac{64}{27} = \frac{384}{27} \)
2. \( 11 \cdot \frac{16}{9} = \frac{176}{9} = \frac{528}{27} \) (converting to a common denominator)
3. \( k \cdot \frac{4}{3} = \frac{4k}{3} = \frac{36k}{27} \) (converting to a common denominator)
4. \( -20 = -20 = -\frac{540}{27} \) (converting to a common denominator)

Now, substituting these into the equation gives:

\[ \frac{384}{27} - \frac{528}{27} + \frac{36k}{27} - \frac{540}{27} = 0 \]

### Step 5: Combine the fractions
Combine the fractions:

\[ \frac{384 - 528 + 36k - 540}{27} = 0 \]

This simplifies to:

\[ \frac{36k - 684}{27} = 0 \]

### Step 6: Solve for \( k \)
Setting the numerator equal to zero:

\[ 36k - 684 = 0 \]
\[ 36k = 684 \]
\[ k = \frac{684}{36} = 19 \]

### Final Answer
Thus, the value of \( k \) is:

\[ \boxed{19} \]