NCERT Solutions
Class 8 Maths
Factorisation

Ex.14.3 Q.5
Factorize the expressions and divide them as directed:
1. (y2 + 7y + 10) ÷ (y + 5)
2. (m2 – 14m - 32) ÷ (m + 2)
3. (5p2 – 25p + 20) ÷ (p - 1)
4. 4yz (z2 + 6z - 16) ÷ 2y (z + 8)
5. 5pq (p2 – q2) ÷ 2p (p + q)
6. 12xy (9x2 – 16y2) ÷ 4xy (3x + 4y)
7. 39y3(50y2 - 98) ÷ 26y2(5y + 7)
1. (y2 + 7y + 10) ÷ (y + 5) = (y2 + 7y + 10) ÷ (y + 5)
= {y2 + (2 + 5) y + 2 × 5} ÷ (y + 5)
= {y2 + 2y + 5y + 2 × 5} ÷ (y + 5)
= {y (y + 2) + 5(y + 2)} ÷ (y + 5)
= {(y + 2) (y + 5)} ÷ (y + 5)
= y + 2
2. (m2 – 14m - 32) ÷ (m + 2) = (m2 – 14m - 32) ÷ (m + 2)
= {m2 + (-16 + 2) m – 16 × 2)} ÷ (m + 2)
= {m2 – 16m + 2m – 16 × 2)} ÷ (m + 2)}
= {m (m - 16) + 2(m – 16)} ÷ (m + 2)
= {(m - 16) (m + 2)} ÷ (m + 2)
= m + 2
3. (5p2 – 25p + 20) ÷ (p - 1) = (5p2 – 25p + 20) ÷ (p - 1)
= 5(p2 – 5p + 4) ÷ (p - 1)
= 5{p2 – (1 + 4) p + 4} ÷ (p - 1)
= 5{p2 – p - 4p + 4} ÷ (p - 1)
= 5{p (p - 1) – 4(p - 1)} ÷ (p - 1)
= 5{(p - 1) (p - 4)} ÷ (p - 1)
= 5(p - 4)
4. 4yz (z2 + 6z - 16) ÷ 2y (z + 8) = {4yz (z2 + 6z - 16)} ÷ {2y (z + 8)}
= {4yz (z2 + (8 - 2) z – 8 × 2)} ÷ {2y (z + 8)}
= {4yz (z2 + 8z - 2z – 8 × 2)} ÷ {2y (z + 8)}
= [4yz {z (z + 8) – 2(z + 8)}] ÷ {2y (z + 8)}
= [4yz (z + 8) (z - 2)] ÷ {2y (z + 8)}
= 2z (z - 2)
5. 5pq (p2 – q2) ÷ 2p (p + q) = {5pq (p2 – q2)} ÷ {2p (p + q)}
= {5pq (p – q) (p + q)} ÷ {2p (p + q)}
= 5q (p – q) ÷ 2
6. 12xy (9x2 – 16y2) ÷ 4xy (3x + 4y) = 12xy{(3x)2 – (4y)2} ÷ {4xy (3x + 4y)}
= 12xy {(3x – 4y) (3x + 4y)} ÷ {4xy (3x + 4y)}
= 3(3x – 4y)
7. 39y3(50y2 - 98) ÷ 26y2(5y + 7) = {39y3(50y2 - 98)} ÷ {26y2(5y + 7)}
= {39y3 × 2(25y2 - 49)} ÷ {26y2(5y + 7)}
= [39y3 × 2{(5y)2 – 72}] ÷ {26y2(5y + 7)}
= [39y3 × 2{(5y - 7) (5y + 7)}] ÷ {26y2(5y + 7)}
= [13 × 3y3 × 2{(5y - 7) (5y + 7)}] ÷ {13 × 2y2(5y + 7)}
= 3y (5y - 7)