Question 976987
{{{f(x)=3x^4+4x^3+13x^2+16x+4}}}.........write {{{13x^2}}} as {{{x^2+12x^2}}}
{{{f(x)=3x^4+4x^3+x^2+12x^2+16x+4}}}...........group
{{{f(x)=(3x^4+12)+(4x^3+16x)+(x^2+4)}}}
{{{f(x)=3x^2(x^2+4)+4x(x^2+4)+(x^2+4)}}}
{{{f(x)=(3x^2+4x+1) (x^2+4) }}}
 {{{f(x)=(3x^2+x+3x+1) (x^2+4) }}}
{{{f(x)=((3x^2+x)+(3x+1)) (x^2+4) }}}
{{{ f(x)=(x+1) (3x+1) (x^2+4) }}}

zeros:
{{{ 0=(x+1) (3x+1) (x^2+4) }}}
{{{ 0=(x+1)  }}}=>{{{x=-1}}}
{{{ 0= (3x+1) }}}=>{{{3x=-1}}}=>{{{x=-1/3}}}
and, since {{{(x^2+4)=0}}}=>{{{x^2=-4}}}=>{{{x=sqrt(-4)}}}=>and zeros are {{{x=2i}}} or {{{x=-2i}}},

then {{{(x^2+4)=(x-2i)(x+2i)}}}

and our  the polynomial as a product of linear factors is:

 {{{f(x)=(x+1) (3x+1)(x-2i)(x+2i)}}}