f(x)=x^3-81x/(2x=1)^2

find the zeros of the function by factoring or algebraically
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1 Answer

Do you mean f(x)=(x3-81x)/(2x-1)2? If so:

then we can write f(x)=x(x2-81)/(2x-1)2=x(x-9)(x+9)/(2x-1)2, which can only be zero if the numerator is zero.

so x(x-9)(x+9)=0, making the zeroes x=0, 9 or -9.

by Top Rated User (1.2m points)

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