Question

Show that –1, 3, 6 are the zeroes of the polynomial f(x)= x^3-8x^2+9x+18

Answer 1

f(x)=x³-8x²+9x+18

Put x=-1,3,6 and check

First put x=-1

f(-1)=(-1)³-8(-1)²+9(-1)+18
=-1-8(1)-9+18
=-1-8-9+18
=-18+18
=0

Put x=3,
f(3)=(3)³-8(3)²+9(3)+18
=27-8(9)+27+18
=72-72
=0

Put x=6,
f(6)=(6)³-8(6)²+9(6)+18
=216-8(36)+54+18
=216-288+54+18
=-72+72
=0

Hence proved.

Hope it helps.....