What are all the possible rational zeros for #f(x)=2x^3-3x^2+1# and how do you find all zeros?

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Answer 1 · 2016-09-14T15:01:12

#x=1#

#2x^3-3x^2+1=0#

#x=1# is zero. Divide # 2x^3-3x^2+1# at x-1. It will turn: #2x^2+x+1#

#2x^2+x+1=0# have not rational zeros (discriminant<0)

Answer 2 · 2016-09-14T15:09:31

The Zeroes of #f(x)# are #1,1,-1/2#.

#f(x)=2x^3-3x^2+1#

Here, the sum of the co-effs.#=2-3+1=0.#

Hence, #(x-1)# is a factor of #f(x)#. Rewriting #f(x)# as,

#f(x)=ul(2x^3-2x^2)-ul(x^2+x)-ul(x+1)#

#=2x^2(x-1)-x(x-1)-1(x-1)#

#=(x-1)(2x^2-x-1)#

In the quadr., we have the same story #[2-1-1=0]# !

#:. f(x)=(x-1){ul(2x^2-2x)+ul(x-1)}#

#=(x-1){2x(x-1)+1(x-1)}#

#=(x-1)(x-1)(2x+1)#

#=(x-1)^2(2x+1)#

Clearly, the Zeroes of #f(x)# are #1,1,-1/2#.

Enjoy Maths.!