As #1/2# is a zero of function #f(x)=2x^3-x^2-10x+5#, #(x-1/2)# is a factor of #f(x)#. Note that coefficient of #x^3# is #2#, hence one can say #2(x-1/2)=2x-1# is a factor.
We can easily divide #f(x)=2x^3-x^2-10x+5# by #2x-1# as follows:
#f(x)=2x^3-x^2-10x+5#
= #x^2(2x-1)-5(2x-1)#
= #(2x-1)(x^2-5)#
It is easy to factorize #x^2-5# using #a^2-b^2=(a+b)(a-b)# and
#x^2-5=x^2-(sqrt5)^2=(x+sqrt5)(x-sqrt5)#
And hence #(x+sqrt5)# and #(x-sqrt5)# are also factors of #f(x)# and therefore
#sqrt5# and #-sqrt5# are other two zeros of #f(x)=2x^3-x^2-10x+5#.
