Question 834474
{{{ f(x)=red(4)x^4-8x^3-19x^2+23x-green(6)}}}
The possible rational zeros can be written as {{{n/d}}} or {{{-n/d}}} where {{{n}}} is a factor of {{{green(6)}}} and {{{d}}} is a factor of {{{red(4)}}}.
They are:
-6,-3,-2,-1,1,2,3,6,
-3/2,-3/4,-1/2,-1/4,1/4,1/2,3/4,3/2.
Trying the easiest ones first, we find two zeros:
{{{f(1)=4-8-19+23-6=-6}}}
{{{f(-1)=4+8-19-23-6=-36}}}
{{{f(2)=64-64-76+46-6=-36}}}
{{{f(-2)=64+64-76-46-6=0}}}
{{{f(3)=324-216-171+69-6=0}}}
Since {{{f(-2)=0}}} and {{{f(3)=0}}} ,
{{{(x-(-2))=(x+2)}}} and {{{(x-3)}}} are factors of {{{f(x)}}} .
So, we can divide {{{f(x)}}} by {{{(x+2)}}} and by {{{(x-3)}}} one after the other,
or we can divide {{{f(x)}}} by {{{(x+2)(x-3)=x^2-x-6}}} .
Either way, the final quotient is {{{4x^2-4x+1=(2x-1)^2}}} ,
which has {{{x=1/2}}} as a double zero.
So {{{f(x)=(x+2)(x-3)(2x-1)^2}}} , and the zeros of {{{f(x)}}} are
{{{x=-2}}}, {{{x=3}}}, and {{{x=1/2}}} .