Question

find the value of k in order that one zero of 3x³+(1+4k)x+ k²+5 may be one third of other.

Answer 1

Should it be 6x^3-kx^2-5x+6?divide (x-3) into this polynomial and you get 6x^2+(18-k)x+[3(18-k)-5] with the remainder of3[3(18-k)-5]+6 which must equal 03[3(18-k)-5]+6=03[54-3k-5]+6=03[49-3k]+6=03[49-3k]=-649-3k=-249+2=3k51=3kk=51/3=17or more simply, use Mark M.'s method: f(3) must equal 0 so replace each x with 3 to get the value of k

hope it helped, regards