Question 1127540

given:

{{{x[1]=5}}}, 
{{{x[2]=-3}}}
 and 
{{{x[3]=-1 + 2i}}}=> complex roots come in pairs, so you also have {{{x[4]=-1 - 2i}}}

{{{f(x)=(x-x[1])(x-x[2])(x-x[3])(x-x[4])}}}

{{{f(x)=(x-5)(x-(-3))(x-(-1 + 2i))(x-(-1 - 2i))}}}

{{{f(x)=(x-5)(x+3)(x+1 -2i)(x+1+ 2i)}}}

{{{f(x)=(x^2-2x-15)(x^2 + 2x + 5)}}}

{{{f(x)=x^4 + 2x^3 + 5x^2-2x^3-4x^2-10x-15x^2-30x-75}}}

{{{f(x)=x^4 + cross(2x^3) + 5x^2-cross(2x^3)-19x^2-40x-75}}}

{{{f(x)=x^4-14x^2-40x-75}}}