3x+4(x−2)(x+3)≡Px+3+Qx−2
3x+4(x−2)(x+3)=P(x−2)+Q(x+3)(x−2)(x+3)
=3x+4=P(x−2)+Q(x+3)
=3x+4=Px−2P+Qx+3Q
=3x+4=Px+Qx−2P+3Q
=3x+4=(P+Q)x−2P+3Q
Equating x and the constants
P+Q=3------(i)
2P+3Q=4-------(ii)
From equation (i)
P=3-Q------(iii)
Substitute (3-Q) for P inequation (ii)
=-2(3-Q)+3Q=4
=-6+2Q+3Q=4
=-6+5Q=4
=5Q=4+6
=5Q=10
∴Q=105=2