Find the L.C.M. of 6x2−x−1, 3x2+7x+2 and 2x2+3x−2
(x+2)(2x−1)(3x+1)
First expression = 6x2−x−1=6x2−3x+2x−1=3x(2x−1)+1(2x−1)=(3x+1)(2x−1)
Second expression = 3x2+7x+2=3x2+6x+x+2=3x(x+2)+1(x+2)=(3x+1)(x+2)
Third expression= 2x2+3x−2=2x2+4x−x−2=2x(x+2)−1(x+2)=(2x−1)(x+2)
Required L.C.M. is (x+2)(2x−1)(3x+1)