What are all the possible rational zeros for #f(x)=2x^3-5x^2+4x-1# and how do you find all zeros?

Question

4537 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T19:30:04

So, all the zeroes of #f# are #1,1,1/2#.

We observe that, the sum of the co-effs. of

#f" "is=2-5+4-1=0.#

We conclude that #(x-1)# is a factor of #f(x)#. Now,

#f(x)=2x^3-5x^2+4x-1#

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

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

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

#=(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)#

So, all the zeroes of #f# are #1,1,1/2#.