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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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