(x -1) is a factor, so we substitute x = 1 in each case and solve for k by making p(1) equal to 0.
(i) p(x)=x2+x+kp(1)=1+1+k=0⇒k=−2
(ii) p(x)=2x2+kx+√2p(1)=2×12+k×1+√2=0⇒2+k+√2=0⇒k=−2−√2=−(2+√2)
(iii) p(x)=kx2−√2x+1p(1)=k−√2+1=0⇒k=√2−1
(iv) p(x)=kx2−3x+kp(1)=k−3+k=0⇒2k−3=0⇒k=32