Question 1023290
The roots we know of are the real roots
{{{x[1]=1-sqrt(3)}}} , and {{{x[2]=1+sqrt(3)}}} ,
and the complex root {{{x[3]=6-i}}} .
For the polynomial to have only real coefficients,
it must also have the conjugate complex root {{{x[4]=6+i}}} .
So, the polynomial of least degree, with a leading coefficient of {{{red(1)}}} is
{{{P(x)=red(1)*(x-x[1])*(x-x[2])*(x-x[3])*(x-x[4])}}} .
{{{P(x)=(x-(1-sqrt(3)))*(x-(1+sqrt(3)))*(x-(6-i))*(x-(6+i))}}}
{{{P(x)=(x-1+sqrt(3))*(x-1-sqrt(3))*(x-6+i)*(x-6-i)}}}
{{{P(x)=((x-1)+sqrt(3))*((x-1)-sqrt(3))*((x-6)+i)*((x-6)-i)}}}
{{{P(x)=((x-1)^2-(sqrt(3))^2)*((x-6)^2-i^2)}}}
{{{P(x)=((x^2-2x+1)-3)*((x^2-12x+36)-(-1))}}}
{{{P(x)=(x^2-2x-2)*((x^2-12x+36)+1)}}}
{{{P(x)=(x^2-2x-2)*(x^2-12x+37)}}}
From here on, it is just busywork.
That is the part where I get bored and distracted, and make mistakes, so check my math from this point on.
{{{P(x)=x^2*(x^2-12x+37)-2x*(x^2-12x+37)-2*(x^2-12x+37)}}}
{{{P(x)=(x^4-12x^3+37x^2)-(2x^3-24x^2+74x)-(2x^2-24x+74)}}}
{{{P(x)=x^4-12x^3+37x^2-2x^3+24x^2-74x-2x^2+24x-74}}}
{{{P(x)=x^4-14x^3+59x^2-50x-74}}}