NCERT Class-8 Maths Factorisation Chapter-14.3 Factorise the expressions

Question

Factorise the expressions and divide them as directed

(a) (y² + 7y + 10) ÷ (y + 5)

(b) (m² – 14m – 32) ÷ (m + 2)

(c) 5p² – 25p + 20) ÷ (p – 1)

(d) 4yz(z² + 6z – 16) ÷ 2y (z + 8)

(e) 5pq(p² – q²) ÷ 2p (p + q)

(f) 12xy (9x² – 16y²) ÷ 4xy (3x + 4y)

(g) 39y³(50y² – 98) ÷ 26y² (5y + 7)

Answer 1

(a) (y² + 7y + 10) ÷ (y + 5)

Answer: (y² + 7y + 10) ÷ (y + 5)
Here, dividend can be factorised by splitting the middle term as follows:
= (y² + 5y + 2y + 10) ÷ (y + 5)
= [y (y + 5) + 2 (y + 5) ] ÷ (y + 5)
= (y + 2) (y + 5) ÷ (y + 5)
= y + 2

(b) (m² – 14m – 32) ÷ (m + 2)

Answer: (m² – 14m – 32) ÷ (m + 2)
Here, dividend can be factorised by splitting the middle term as follows:
= (m² – 16m + 2m – 32) ÷ (m + 2)
= [ m(m – 16) + 2(m – 16) ] ÷ (m + 2)
= (m + 2) (m – 16) ÷ (m +2)
= m – 16

(c) 5p² – 25p + 20) ÷ (p – 1)

Answer: 5p² – 25p + 20) ÷ (p – 1)
Here, dividend can be factorised by splitting the middle term as follows:
= (5p² – 5p – 20p + 20) ÷ (p - 1)
= [5p(p – 1) – 20(p – 1) ] ÷ (p – 1 )
= (5p – 20) (p – 1) ÷ (p – 1)
= 5p – 20

(d) 4yz(z² + 6z – 16) ÷ 2y (z + 8)

Answer: 4yz(z² + 6z – 16) ÷ 2y (z + 8)
= 2z(z² + 8z – 2z – 16) ÷ (z + 8)²= 2z [ z(z + 8) – 2(z + 8) ] ÷ (z + 8)
= 2z (z – 2) (z + 8) ÷(z + 8)
= 2z (z – 2)

(e) 5pq(p² – q²) ÷ 2p (p + q)

Answer: 5pq(p² – q²) ÷ 2p (p + q)
Using (a +b) (a – b) = a² – b²; the equation can be written as follows:
= 5pq (p + q) (p – q) ÷ 2p (p + q)
= 5q(p – q) ÷ 2

(f) 12xy (9x² – 16y²) ÷ 4xy (3x + 4y)

Answer: 12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
This can be solved as previous question;
= 3 (3x + 4y) (3x – 4y) ÷ (3x + 4y)
= 3 (3x – 4y)

(g) 39y³(50y² – 98) ÷ 26y² (5y + 7)

Answer: 39y³(50y² – 98) ÷ 26y² (5y + 7)
= 2 x 39y³(25y² – 49) ÷ 26y² (5y + 7)
= 3y (25y² – 49) ÷ (5y + 7)
= 3y (5y + 7) (5y – 7) ÷ (5y + 7)
= 3y (5y – 7)