Find the domain and the zeros f(x)= 3x^3-x^2-2x/x(x+1)^2

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Guest Feb 4, 2015

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Best Answer

$$\\f(x)= 3x^3-x^2-\frac{2x}{x}(x+1)^2\\
x\ne0\\
f(x)= 3x^3-x^2-2(x+1)^2\\$$

Zero is approximately x=1.891

domain: x in the set ot reals where x is not equal to 0

Melody Feb 5, 2015

Melody · Answer 1 · 2015-02-05T10:02:19

$$\\f(x)= 3x^3-x^2-\frac{2x}{x}(x+1)^2\\
x\ne0\\
f(x)= 3x^3-x^2-2(x+1)^2\\$$

Zero is approximately x=1.891

domain: x in the set ot reals where x is not equal to 0