If one root of bi-quadratic equation x4+2x3−16x2−22x+7=0is2+√3. Find the other three roots.
Since irrational roots occurs in pairs
2−√3 must also be a root
x=2+√3
x−2=√3
x2−4x+4=3
x2−4x+1=0
Divide x4+2x3−16x2−22x+7 by x2−4x+1 we get the quotient is x2+6x+7
∴x2+6x+7=0⇒x=−3±√2