66@arshad, no its not c
@;) the factorization should be x(x4 + ax3 + ...)
Let P(x)=x^5+ax^4+bx^3+cx^2+dx+e.
If the graph of y = P(x) cuts the x-axis at five distinct points and P(0)=0, then which of the coefficients cannot be zero?
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6 Answers
1P(0)=0 \\ means \ e =0 \\ x(ax^4+....... ) \\ since\ there \ are \4 \ roots \ of \ the \ residual \ polynomial \ equation \ it \ must \ be \ 4th \ degree \ \\ hence \ a \ cannot \ be \ 0
661oh sorry sir
P(x)=x.(x-α)(x-β)(x-γ)(x-δ)
NOW AS ALL ROOTS ARE DISTINCT
α,β,γ,δ NONE OF THEM CAN BE 0
HENCE α.β.γ.δ ≠0
HENCE d≠0
341you are reversing the claim: he's proving p→q and you are asking about q→p