Question 930626
Write the equation, in standard form, of a polynomial with real coefficients that has roots at :

{{{2}}},
{{{1-2i}}}, complex roots always come in pairs, so you have {{{1+2i}}} too
{{{1+4i}}}, and {{{1-4i}}} as well

passes through the point ({{{0}}},{{{98}}}). 

{{{f(x)=a(x-2)(x-(1-2i))(x-(1+2i))(x-(1+4i))(x-(1-4i))}}} where {{{a}}} is coefficient

{{{f(x)=a(x^5-6x^4+34x^3-96x^2+173x-170)}}}

since passes through the point ({{{0}}},{{{98}}}) set {{{f(x)=98}}} and {{{x=0}}} and solve for coefficient {{{a}}}


{{{98=a(0^5-6*0^4+34*0^3-96*0^2+173*0-170)}}}

{{{98=a(-170)}}}

{{{a=98/(-170)}}}

{{{a=-0.5764705882352941}}} exact value, and for more accurate result I will use exact value

{{{f(x)=-0.5764705882352941(x^5-6x^4+34x^3-96x^2+173x-170)}}}

{{{f(x)=-0.576470588235294x^5+3.45882352941176x^4-19.6x^3+55.3411764705882x^2-99.7294117647059x+98}}}
 

we can round it and get shorter form, but zeros will not be exact

{{{f(x)=-0.58x^5+3.46x^4-19.6x^3+55.34x^2-99.7x+98}}}