3p2x3+px2+qx+3=0 p=1 and q=−7 ⇒3x3+x2−7x+3=0.....(i) On putting x=1 3×13+12−7×1+3=7−7=0 ∴(x−1) is one factor in the Eq. (i), 3x2+4x−3 ────────── (x−1)3x3+x2−7x+3 +3x3−3x2 −+ ──────────── 4x2−7x+3 4x2∓4x −+ ──────────── −3x+3 −3x+3 +− ───────────── 0 ∴ Eq. (i) can be written as (x−1)(3x2+4x−3)=0 3x2+4x−3=0 gives the irrational roots. ⇒α+β=