Question

factorise 8(2x-3y)^2-10(2x-3y)^7​

Answer 1

7+10(2x−3y)−8(2x−3y)

2

Let 2x−3y=z

Then,7+10(2x−3y)−8(2x−3y)

2

becomes

=7+10z−8z

2

=7+14z−4z−8z

2

=7(1+2z)−4z(1+2z)

=(1+2z)(7−4z)

Now, on substituting z=2x−3y, we get

=[(1+2(2x−3y))][7−4(2x−3y)]

=(1+4x−6y)(7−8x+12y)

Answer 2

Answer:

7+10(2x−3y)−8(2x−3y)

2

Let 2x−3y=z

Then,7+10(2x−3y)−8(2x−3y)

2

becomes

=7+10z−8z

2

=7+14z−4z−8z

2

=7(1+2z)−4z(1+2z)

=(1+2z)(7−4z)

Now, on substituting z=2x−3y, we get

=[(1+2(2x−3y))][7−4(2x−3y)]

=(1+4x−6y)(7−8x+12y)