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Factorize the expressions and divide them as directed.
(i) (y2 + 7y + 10) ÷ (y + 5)
(ii) (m2 -14m - 32) ÷ (m + 2)
(iii) (5p2 - 25p + 20) ÷ (p -1)
(iv) 4yz(z2 + 6z -16) ÷ 2y(z + 8)
(v) 5pq(p2 - q2) ÷ 2p(p + q)
(vi) 12xy(9x2 -16y2) ÷ 4xy(3x + 4y)
(vii) 39y3(50y2 - 98) ÷ 26y2(5y + 7)

Solution:

We will factorize the expressions then cancel out common factors from the numerator and denominator.

(i) (y2 + 7y + 10) ÷ (y + 5)

(y2 + 7y + 10) can be written as ,

y2 + 2 y + 5 y +10 = y(y + 2) + 5(y + 2)

= (y + 2)(y + 5)

Thus, (y2 + 7y + 10) ÷ (y + 5) = (y + 2)(y + 5) / (y + 5)

= y + 2

(ii) (m2 -14m - 32) ÷ (m + 2)

(m2 -14m - 32) can be written as ,

m2 + 2m -16m - 32 = m(m + 2) -16(m + 2)

= (m + 2)(m - 16)

Thus, (m2 -14m - 32) ÷ (m + 2) = (m + 2)(m - 16) / (m + 2)

= m - 16

(iii) (5p2 - 25p + 20) ÷ (p -1)

 (5p2 - 25p + 20) can be written as ,

5(p2 - 5p + 4) = 5(p2 - p - 4p + 4)

= 5[p(p - 1) - 4(p - 1)]

= 5(p - 1)(p - 4)

Thus, (5p2 - 25p + 20) ÷ (p - 1) = 5(p - 1)(p - 4) / (p -1)

= 5(p - 4)

(iv) 4yz(z2 + 6z -16) ÷ 2y(z + 8)

4yz(z2 + 6z -16) can be written as ,

4 yz (z2 - 2z + 8z -16) = 4 yz [z(z - 2) + 8(z - 2)]

= 4 yz(z - 2)(z + 8)

Thus, 4yz(z2 + 6z -16) ÷ 2y(z + 8) = 4 yz(z - 2)(z + 8) / 2y(z + 8)

= 2z(z - 2)

(v) 5pq(p2 - q2) ÷ 2p(p + q)

5pq(p2 - q2) can be written as 5pq(p - q)(p + q) [Using identity a2 - b2 = (a + b)(a - b)]

Thus, 5pq(p2 - q2) ÷ 2p(p + q) = 5pq(p - q)(p + q) / 2p(p + q)

= 5q(p - q) / 2

(vi) 12xy(9x2 -16y2) ÷ 4xy(3x + 4y)

12xy(9x2 -16y2) can be written as

12xy[(3x)2 - (4y)2] = 12xy(3x - 4y)(3x + 4y)  [Using identity a2 - b2 = (a + b)(a - b)]

= 2 × 2 × 3 × x × y × (3x - 4y) × (3x + 4y)

Thus, 12xy(9x2 -16y2) ÷ 4xy(3x + 4y) = 2 × 2× 3 × x × y × (3x - 4y) × (3x + 4y) / 4xy(3x + 4y)

= 3(3x - 4y)

(vii) 39y3(50y2 - 98) ÷ 26y2(5y + 7)

39y3(50y2 - 98) can be written as

3 × 13 × y × y × y × [2 × (25y2 - 49)] = 3 × 13 × 2 × y × y × y × [(5y)2 - (7)2]

= 3 × 13 × 2 × y × y × y(5y - 7)(5y + 7)  [Using identity a2 - b2 = (a + b)(a - b)]

Now,

26y2(5y + 7) can be written as 2 × 13 × y × y × (5y + 7)

Thus, 39y3(50y2 - 98) ÷ 26y2(5y + 7) = 3 × 13 × 2 × y × y × y(5y - 7)(5y + 7) / 2 × 13 × y × y × (5y + 7)

= 3y(5y - 7)

☛ Check: NCERT Solutions for Class 8 Maths Chapter 14

Video Solution:

Factorize the expressions and divide them as directed. (i) (y² + 7y + 10) ÷ (y + 5) (ii) (m² -14m - 32) ÷ (m + 2) (iii) (5p² - 25p + 20) ÷ (p -1) (iv) 4yz(z² + 6z -16) ÷ 2y(z + 8) (v) 5pq(p² - q²) ÷ 2p(p + q) (vi) 12xy(9x² -16y²) ÷ 4xy(3x + 4y) (vii) 39y³(50y² - 98) ÷ 26y²(5y + 7)

Class 8 Maths NCERT Solutions Chapter 14 Exercise 14.3 Question 5

Summary:

The following expressions are factorised and divided. (i) (y² + 7y + 10) ÷ (y + 5) (ii) (m² -14m - 32) ÷ (m + 2) (iii) (5p² - 25p + 20) ÷ (p -1) (iv) 4yz(z² + 6z -16) ÷ 2y(z + 8) (v) 5pq(p² - q²) ÷ 2p(p + q) (vi) 12xy(9x² -16y²) ÷ 4xy(3x + 4y) (vii) 39y³(50y² - 98) ÷ 26y²(5y + 7) and the results are (i) y + 2 (ii)m - 16 (iii) 5(p - 4) (iv) 2z(z - 2) (v) 5q(p - q) / 2 (vi) 3(3x - 4y) (vii) 3y(5y - 7)

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