How do you find the zeroes of the function #g(x)= 2x^3 - 5x^2 + 4#?

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Answer 1 · 2016-06-28T14:43:48

#x=(1+-sqrt(17))/4 and x=2#

let's consider the known term 4: it is divided by

#+-1, +-2, +-4#

By the remainder theorem, let's substitute each of the dividers in x in the function g(x), so

#g(1)=2-5+4!=0#

#g(-1)=-2-5+4!=0#

#g(2)=16-20+4=0#

So, one of the zeroes is 2

Then we can divide the polynomial g(x) by the bynomial x-2

and obtain:

#2x^2-x-2=0#

by which you can find two zeroes

#x=(1+-sqrt(17))/4#