Question 835252
{{{f(x)= x^5+x^3+2x^2-12x+8=(x+2)(x-1)^2(x^2+2)}}}
{{{x^2+2}}} is not a linear factor.
It is a quadratic polynomial, and can only be factored using imaginary numbers.
 
HOW DO WE GET THAT?
Noticing that {{{f(1)=1^5+1^3+2*1^2-12*1+8=1+1+2-12+8=0}}} ,
we realize that {{{x-1}}} is a factor of {{{f(x)}}} .
Dividing, we find that
{{{f(x)/(x-1)=(x^5+x^3+2x^2-12x+8)/(x-1)=x^4+x^3+2x^2+4x-8=g(x)}}}
Since {{{g(1)=1^4+1^3+2*1^2+4*1-8=1+1+2+4-8=0}}} ,
{{{x-1}}} must be a factor of {{{g(x)}}} .
Dividing, we find that
{{{g(x)/(x-1)=(x^4+x^3+2x^2+4x-8)/(x-1)=x^3+2x^2+4x+8}}} .
So {{{g(x)=(x-1)(x^3+2x^2+4x+8)}}} and {{{f(x)=(x-1)g(x)}}} ,
which means that
{{{f(x)=(x-1)(x-1)(x^3+2x^2+4x+8)=(x-1)^2(x^3+2x^2+4x+8)}}}
 
At this point, either we may realize that {{{x^3+2x^2+4x+8}}} is zero for {{{x=-2}}} ,
or we may realize that {{{x^3+2x^2+4x+8}}} is divisible by {{{x+2}}} .
In either case we divide and find that {{{x^3+2x^2+4x+8=(x+2)(x^2+4)}}} .
So, putting it all together,
{{{f(x)=(x-1)^2(x^3+2x^2+4x+8)=(x-1)^2(x+2)(x^2+4)}}} .
 
Alternatively, we may realize that
{{{(x^3+2x^2+4x+8)(x-2)=x^4-16=(x^2+2)(x^2-2)=(x^2+2)(x+2)(x-2)}}}
and since {{{(x^3+2x^2+4x+8)(x-2)=(x^2+2)(x+2)(x-2)}}} ,
dividing both sides by {{{(x-2)}}} we conclude that
{{{(x^3+2x^2+4x+8)=(x^2+2)(x+2)}}}