Question

Let ( f(x)=6 x^{3}-11 x^{2}+k x-20 )
( fleft(frac{4}{3}right)=6left(frac{4}{3}right)^{3}-11left(frac{4}{3}right)^{2}+kleft(frac{4}{3}right)-20=0 )
( Rightarrow quad 6 . frac{64}{27}-11 . frac{16}{9}+frac{4 k}{3}-20=0 )
( Rightarrow quad 128-176+12 k-180=0 )
( Rightarrow quad 12 k+128-356=0 Rightarrow 12 k=228 Rightarrow k=19 )

# If x= root of the polynomial f(x)=6r? - 11.x² + kx-20, then find the value of k.

Solution