Applying R3→(R3−R2);R3→(R2−R1)
∣∣
∣
∣∣(x+2)(2x+3)(3x+4)(x+1)(x+1)(x+1)(x+2)2(x+2)6(x+2)∣∣
∣
∣∣=0
Applying C3→C3−C2;C2→C2−C1
⇒∣∣
∣
∣∣(x+2)(x+1)(x+1)(x+1)00(x+2)(x+2)6(x+2)∣∣
∣
∣∣=0⇒−(x+1){4(x+1)(x+2)−(x+2)(x+2)}=0⇒−3(x+1){(x+1)(x+2)}=0⇒x=−1,−1,−2
Hence sum of roots is −4 its absolute value is 4