MathematicsIf x=43 is a root of the polynomial f(x)=6×3-11×2+kx-20, then find the value of k.

If $x = \frac{4}{3}$ is a root of the polynomial $f (x) = 6 x^{3} - 11 x^{2} + kx - 20$ , then find the value of $k$ .

Given

x = \frac{4}{3}

is a root of the polynomial

f (x) = 6 x^{3} - 11 x^{2} + kx - 20

.
As,

f (x) = 6 x^{3} - 11 x^{2} + kx - 20

Then,

f (\frac{4}{3}) = 6 {(\frac{4}{3})}^{3} - 11 {(\frac{4}{3})}^{2} + k (\frac{4}{3}) - 20 = 0

\Rightarrow 6 (\frac{64}{27}) - 11 (\frac{16}{9}) + \frac{4 k}{3} - 20 = 0

\Rightarrow 128 - 176 + 12 k - 180 = 0

\Rightarrow 12 k + 128 - 356 = 0

\Rightarrow 12 k = 228

\Rightarrow k = 19 .

Therefore, the value of k is 19.
Hence, the correct option is 2.

JEE