Question 1183612
If a polynomial in {{{x}}} has {{{x+1}}} as a factor, it can be written as {{{(x+1)Q(x)}}} where {{{Q(x)}}} is another polynomial,
and the value of the original polynomial is
{{{(-1+1)Q(-1)=0*Q(-1)=0}}} for {{{x=-1}}}
Then, for {{{x=-1}}} we have
{{{3(-1+3)^4-(k+(-1))^2=0}}}
{{{3(2)^4-(k-1)^2=0}}}
{{{3*16-(k-1)^2=0}}}
{{{(k-1)^2=3*16}}}
{{{k-1=" " +- sqrt(3*16)}}}
{{{k=1 +- sqrt(3*16)}}}
{{{highlight(k=1 +- 4sqrt(3))}}}
 
You can also solve the problem the long, cumbersome, treacherous way, but that increases the risk of errors.