Factorise 21×2 – 59xy + 40y2

(i) 21x² – 59xy + 40y²

(ii) 4x³y – 44x²y + 112xy

Answer :

(i) 21x² – 59xy + 40y²

By multiplying the first and last term we get, 21 × 40 = 840

Then, (-35) × (-24) = 840

So, 21x² – 35xy – 24xy + 40y²

7x(3x – 5y) – 8y(3x – 5y)

(3x – 5y) (7x – 8y)

(ii) 4x³y – 44x²y + 112xy

4x³y – 44x²y + 112xy

Take out common in all terms,

4xy(x² – 11x + 28)

Then, 4xy (x² – 7x – 4x + 28)

4xy [x(x – 7) – 4(x + 7)]

4xy (x – 7) (x – 4)