Question 889313
If {{{1+i}}} is a root, then {{{1-i}}} is also a root since complex roots only occur in complex conjugate pairs.
{{{(x-(1+i))(x-(1-i))=x^2-(1-i)x-(1+ix)-(1+i)(1-i)}}}
{{{(x-(1+i))(x-(1-i))=x^2-2x+(1-i+i-i^2))}}}
{{{(x-(1+i))(x-(1-i))=x^2-2x+(1+1))}}}
{{{(x-(1+i))(x-(1-i))=x^2-2x+2)}}}
Now use polynomial long division to factor out this polynomial.
*[illustration Capture.JPG].
Symbolically shown here, the divisor turns out to be,
{{{x^2+5}}}
So then,
{{{x^4 - 2x^3 + 7x^2 - 10x + 10=(x^2-2x+2)(x^2+5)}}}
So the other two roots are also complex,
{{{x=0 +- 5i}}}