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Question
If the zeros of the polynomial f(x) = ax3 + 3bx2 + 3cx + d are in A.P., prove that 2b3 − 3abc + a2d = 0.
Solution
Let a - d, a and a + d be the zeros of the polynomial f(x). Then,
Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`
`a-d+a+a+d=(-3b)/a`
`3a=(-3b)/a`
`a=(-3b)/axx1/3`
`a=(-b)/a`
Since a is a zero of the polynomial f(x).
Therefore,
f(x) = ax3 + 3bx2 + 3cx + d
f(a) = 0
f(a) = aa3 + 3ba2 + 3ca + d
aa3 + 3ba2 + 3ca + d = 0
`a((-b)/a)^3+3bxx((-b)/a)^2+3cxx((-b)/a)+d=0`
`axx(-b)/axx(-b)/axx(-b)/a+3xxbxx(-b)/axx(-b)/a+3xxcxx(-b)/a+d=0`
`(-b^3)/a^2+(3b^3)/a^2-(3cb)/a+d=0`
`(-b^3+3b^3-3abc+a^2d)/a^2=0`
2b3 − 3abc + a2d = 0 xa2
2b3 − 3abc + a2d = 0
Hence, it is proved that 2b3 − 3abc + a2d = 0
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