Question 821628
Case level is important.  Use either H or h, because each may name different functions.  H is not taken as equal to h.


Synthetic division of H(x) using the zero of 3 (not shown here), gives {{{x^2-4x-1}}}, which has the zeros, {{{2+sqrt(5)}}} and {{{2-sqrt(5)}}}.  In factored form, {{{H(x)=(x-3)(x-(2+sqrt(5)))(x-(2-sqrt(5)))}}}
which you could simplify, carrying out the multiplications and groupings, to 
{{{H(x)=(x-3)((x-2)^2-5)}}}, but this is not linear factors.  The best form of the answer is then {{{highlight(H(x)=(x-3)(x-(2+sqrt(5)))(x-(2-sqrt(5))))}}}