Question

please solve the above questions.​

Answer 1

Answer:

Plz refer to the attachment also

I got the answer for questions 10 & 11

Hope its helps u..

11(i)

(x−2y)2−6(x−2y)+5

Let x−2y=z

Then,(x−2y)2−6(x−2y)+5 becomes

=z2−6z+5

=z2−5z−z+5

=z(z−5)−1(z−5)

=(z−5)(z−1)

Now, on subtituting z=x−2y,we get

=[(x−2y)−5][(x−2y)−1]

=(x−2y−5)(x−2y−1)

(ii)

7+10(2x−3y)−8(2x−3y)2

Let 2x−3y=z

Then,7+10(2x−3y)−8(2x−3y)2 becomes

=7+10z−8z2

=7+14z−4z−8z2

=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:

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