Answer above questions fast please Share with your friends Share 0 Atul Shaw answered this Dear student,12. Assuming the eqn. to be:p(x)=ax4+bx3+cx2+dx+ex2-1=(x-1)(x+1) [a2-b2=(a+b)(a-b)]so, x=1, -1 are zeroes of the p(x)By factor theorem,p(1)=0 ⇒a(1)4+b(1)3+c(1)2+d(1)+e=0⇒a+b+c+d+e=0⇒a+b+c+d+e=0......(1)and also,p(-1)=0⇒a(-1)4+b(-1)3+c(-1)2+d(-1)+e=0⇒a-b+c-d+e=0.......(2)⇒a+c+e=b+d......(3)eq. (1)+eq. (2)a+b+c+d+e+a-b+c-d+e=02(a+c+e)=0a+c+e=0......(4)From eq. 3 and eq. 4a+c+e=0+b+d⇒a+c+e=b+d=0For remaining queries we request you to post them in separate threads to have rapid assistance from our experts.Regards 0 View Full Answer
