Question 695861
The factoring of such a cubic polynomial would be
{{{f(x)=(-1)(x-1)(x-2)(x-k)}}}
so that each factor would be zero for each of the roots of {{{f(x)=0}}}.
 
The remainder of {{{f(x)}}} when divided by {{{(x-3)}}} is
{{{f(3)=(-1)(3-1)(3-2)(3-k)=8}}} --> {{{(-1)(2)(1)(3-k)=8}}} --> {{{2k-6=8}}} --> {{{2k=8+6}}} --> {{{2k=14}}} --> {{{highlight(k=7)}}}
and {{{f(x)=(-1)(x-1)(x-2)(x-7)}}}
 
The remainder of {{{f(x)}}} when divided by {{{(x+3)}}} is
{{{f(-3)=(-1)(-3-1)(-3-2)(-3-7)=(-1)(-4)(-5)(-10)=highlight(-200)}}}