If `x=4/3` is a root of the polynomial `f(x)=6x^3-11 x^2+k x-20 ,` find the value of `k`

Options

A. `k = 19`
B. `k = 29`
C. `k = 9`
D. `k = 49`

Explanation

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

### Step 1: Substitute the root into the polynomial
Since \( x = \frac{4}{3} \) is a root, we substitute this value into the polynomial and set it equal to zero:
\[ f\left(\frac{4}{3}\right) = 6\left(\frac{4}{3}\right)^3 - 11\left(\frac{4}{3}\right)^2 + k\left(\frac{4}{3}\right) - 20 = 0 \]

### Step 2: Calculate \( \left(\frac{4}{3}\right)^3 \) and \( \left(\frac{4}{3}\right)^2 \)
Calculating these powers:
\[ \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
\[ \left(\frac{4}{3}\right)^3 = \frac{64}{27} \]

### Step 3: Substitute these values back into the equation
Now substituting these values into the polynomial:
\[ 6 \cdot \frac{64}{27} - 11 \cdot \frac{16}{9} + k \cdot \frac{4}{3} - 20 = 0 \]

### Step 4: Simplify each term
Calculating each term:
- The first term:
\[ 6 \cdot \frac{64}{27} = \frac{384}{27} \]
- The second term:
\[ -11 \cdot \frac{16}{9} = -\frac{176}{27} \]
- The third term remains as \( \frac{4k}{3} \).
- The fourth term:
\[ -20 = -\frac{540}{27} \]

### Step 5: Combine the terms
Now combine all the terms:
\[ \frac{384}{27} - \frac{176}{27} + \frac{4k}{3} - \frac{540}{27} = 0 \]
Combining the fractions:
\[ \frac{384 - 176 - 540}{27} + \frac{4k}{3} = 0 \]
Calculating the numerator:
\[ 384 - 176 - 540 = -332 \]
So, we have:
\[ \frac{-332}{27} + \frac{4k}{3} = 0 \]

### Step 6: Isolate \( k \)
Rearranging gives:
\[ \frac{4k}{3} = \frac{332}{27} \]
Multiplying both sides by 3:
\[ 4k = \frac{332 \cdot 3}{27} \]
Calculating the right side:
\[ 4k = \frac{996}{27} \]
Now, divide both sides by 4:
\[ k = \frac{996}{27 \cdot 4} = \frac{996}{108} = \frac{83}{9} \]

### Step 7: Final value of \( k \)
Thus, the value of \( k \) is:
\[ k = \frac{83}{9} \]