Question 1200976
<br>
If the coefficients of the polynomial are real and 2+5i is a root, then 2-5i is also a root.<br>
So the three roots are
-5
2+5i
2-5i<br>
and the polynomial is<br>
{{{f(x)=a(x-(-5))(x-(2+5i))(x-(2-5i))}}}<br>
where a can be any constant.<br>
{{{(x-(-5))=(x+5)}}}<br>
{{{(x-(2+5i))(x-(2-5i))=((x-2)-5i)((x-2)+5i)=(x^2-4x+4)-25i^2=(x^2-4x+4)+25=x^2-4x+29}}}<br>
{{{(x+5)(x^2-4x+29)=x^3-4x^2+29x+5x^2-20x+145=x^3+x^2+9x+145}}}<br>
ANSWER: {{{f(x)=a(x^3+x^2+9x+145)}}}