Answer :

the zeroes of f(x) = 12x^3−32x^2−145x+25

x = 06.70

x = -22.70

To find the zeros of a function, we must first find the derivative of the function. This one is not very difficult; we will just apply the power rule to each individual term

f(x)=12x3−32x2-145x+25

f'(x)=[tex]36x^{2}[/tex]−64x-120

Then we set the derivative equal to zero and solve for x

[tex]36x^{2}[/tex]−64x-120=0

We get two values

x=06.70

x= -22.70

Using the formula, we find the zeroes by:

x = −b±√b2−4ac2

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