**(i) 21x ^{2} – 59xy + 40y^{2}**

**(ii) 4x ^{3}y – 44x^{2}y + 112xy**

**Answer :**

**(i) 21x**^{2} – 59xy + 40y^{2}

^{2}– 59xy + 40y

^{2}

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

Then, (-35) × (-24) = 840

So, 21x^{2} – 35xy – 24xy + 40y^{2}

7x(3x – 5y) – 8y(3x – 5y)

(3x – 5y) (7x – 8y)

**(ii) 4x**^{3}y – 44x^{2}y + 112xy

^{3}y – 44x

^{2}y + 112xy

4x^{3}y – 44x^{2}y + 112xy

Take out common in all terms,

4xy(x^{2} – 11x + 28)

Then, 4xy (x^{2} – 7x – 4x + 28)

4xy [x(x – 7) – 4(x + 7)]

4xy (x – 7) (x – 4)