Question

( f(x)=2 x^{3}+x^{2}+k x-6 ) if aveded thy ( x-3 )
there bemender ix 36
i.e ( f(x=3)=36 )
( S_{0} Rightarrow 2(3)^{3}+3^{2}+k(3)-6=36 )
( 2(27)+9+3 k-6=36 )
[
begin{aligned}
63-6+3 k &=36
3 k &=36-57
end{aligned}
]
valueds ( k quad(x-7) ) ( 3 x=-21 )
( k= )

# cuhen 223+*+ Kx-6 ů divided by 2.6 the demainder is 36. find the value of

Solution