Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1

Question 1.

Find the common factors of the given terms?

- 12x, 36

- 2y, 22xy

- 4pq, 28p
^{2}q^{2}

- 2x, 3x
^{2}, 4

- 6abc, 24ab
^{2}, 12a^{2}b

- 6x
^{3}, -4x^{2}, 32x

- 10pq, 20qr, 30rp

- 3x
^{2}y^{3}, 10x^{3}y^{2}, 6x^{2}y^{2}z

Solution:

1. ∵ 12x = 2 × 2 × 3 × x = (2 × 2 × 3) × x

and 36 = 2 × 2 × 3 × 3 = (2 × 2 × 3) × 3

∴ the common factor = 2 × 2 × 3 = 12

2. ∵ 2y = 2 × y = (2 × y)

and 22y = 2 × 11 × y

= (2 × y) × 11

∴ the common factor = 2 × y = 2y

3. ∵ 14pq = 2 × 7 × p × q = (2 × 7 × p × q)

and 28 p^{2}q^{2} = 2 × 2 × 1 × p × p × q × q

= (2 × 7 × p × q) × 2 × p × q

∴ the common factor = (2 × 7 × p × q) =14pq

4. ∵ 2x = 1 × 2 × x = 1 × 2 × x

3x^{2} = 1 × 3 × x × x = 1 × 3 × x and

4 = 1 × 2 × 2 = 1 × 2 × 2

∴ the common factor = 1

Note: 1 is a factor of every term.

5. ∵ 6 abc = 2 × 3 × a × b × c

= (2 × 3 × a × b) × c

24 ab^{2} = 2 × 2 × 2 × 3 × a × b × b

= (2 × 3 × a × b) × 2 × 2 × b

12 a^{2}b = 2 × 2 × 3 × a × a × b

= (2 × 3 × a × b) × 2 × a

∴ the common factor

6. ∵ 10pq = 2 × 5 × p × q

= (2 × 5) × p × q

20qr = 2 × 2 × 5 × q × r

= (2 × 5) × 2 × q × r

30rp = 2 × 3 × 5 × r × p

∴ the common factor = 2 × 5 = 10

7. ∵ 3x^{2}y^{3} = 3 × x × x × y × y × y

= (x × x × y × y) × 3 × y

10x^{3}y^{2} = 2 × 5 × x × x × x × y × y

= (x × x × y × y) × 2 × 5 × x

6x^{2}y^{2}z = 2 × 3 × x × x × y × y × z

= (x × x × y × y) × 2 × 3 × z

∴ the common factor = (x × x × y × y) = x^{2}y^{2}

Question 2.

Factorise the following expressions:

- 7x – 42
- 6p – 12q
- 7a
^{2}+ 14a - -16z + 20z
^{3} - 20l
^{2}m + 30alm - 5x
^{2}y – 15xy^{2} - 10 a
^{2}– 15 b^{2}+ 20 c^{2} - -4a
^{2}+ 4ab – 4ca - x
^{2}yz + xy^{2}z + xyz^{2} - ax
^{2}y + bxy^{2}+ cxyz

Solution:

1. ∵ 7x = 7 × x = (7) × x

42 = 2 × 7 × 3 = (7) × 2 × 3

∴ 7x – 42 = 7[(x)] – (2 × 3)] = 7[x – 6]

2. ∵ 6p = 2 × 3 × p

= (2) × (3) × p = (2 × 3) × p

12q = 2 × 2 × 3 × q

= (2) × 2 × (3) × q

= (2 × 3) × 2 × q

∴ 6p – 12q = (2 × 3) [(p) – (2 × q)]

= 6[p – 2q]

3. ∵ 7a^{2} = 7 × a × a = (7 × a) × a

14a = 2 × 7 × a = (7 × a) × 2

∴ 7a^{2} – 14a = (7 × a)[a – 2] = 7a(a – 2)

4. ∵ -16z = (-1) × 2 × 2 × 2 × 2 × z

= (2 × 2 × z) × (-1) × 2 × 2

20z^{3} = 2 × 2 × 5 × z × z × z

= (2 × 2 × z) × 5 × z × z

∴ -16z + 20z^{3} = (2 × 2 × z) [(-1) × 4 + 5 × z × z] = 4z[-4 + 5z^{2}]

5. ∵ 20l^{2}m = 2 × 2 × 5 × l × l × m

= (2 × 5 × l × m) × 2 × l

30alm = 2 × 3 × 5 × a × l × m

= (2 × 5 × l × m) × 3 × a

∴ 2ol^{2}m + 30alm = (2 × 5 × l × m) [2 × l + 3 × a] = 10lm [2l + 3a]

6. ∵ 5x^{2}y = 5 × x × x × y

= (5 × x × y) [x]

15x^{2}y = 5 × 3 × x × y × y

= (5 × x × y) [3 × y]

∴ 5x^{2}y – 15xy^{2} = (5 × x × y)[x – 3 × y]

= 5xy(x – 3y)

7. ∵ 10a^{2} = 2 × 5 × a × a

= (5) [2 × a × a]

15b^{2} = 3 × 5 × b × b

= (5) [3 × b × b]

20c^{2} = 2 × 2 × 2 × 5 × c × c

= (5) [2 × 2 × c × c]

∴ 10a^{2} – 15b^{2} + 20c^{2} = (5)[2 × a × a – 3 × b × b + 2 × 2 × c × c]

= 5[2a^{2} – 3b^{2} + 4c^{2}]

8. ∵ -4a^{2} = (-1) × 2 × 2 × a × a

= (2 × 2 × a) [(-1) × a]

4ab = 2 × 2 × a × b

= (2 × 2 × a)[b]

-4ca = (-1) × 2 × 2 × c × a

= (2 × 2 × a)[-1) × c]

∴ -4a^{2} + 4ab – 4ca = (2 × 2 × a) [(-1)] × a + b – c]

= 4a[-a + b – c]

9. ∵ x^{2}yz = x × x × y × z = (xyz)[x]

xy^{2}z = x × y × y × z = (xyz)[y]

xyz^{2} = x × y × z × z = (xyz)[z]

∴ x^{2}yz + xy^{2}z + xyz^{2} = (xyz)[x + y + z]

= xyz(x + y + z)

10. ∵ ax^{2}y = a × x × x × y = (x × y)[a × x]

bxy^{2} = b × x × y × y

= (x × y)[a × x + b × y + c × z]

= xy(ax + by + cz)

Question 3.

Factorise:

- x
^{2}+ xy + 8x + 8y - 15xy – 6x + 5y – 2
- ax + bx – ay – by
- 15pq + 15 + 9q + 25p
- z – 7 + 7xy – xyz

Solution:

1. x^{2} + xy + 8x + 8y

= x[x + y] + 8[x + y]

= (x + y)(x + 8)

2. 15xy – 6x + 5y – 2

= 3x[5y – 2] + 1[5y – 2]

= (5y – 2)(3x + 1)

3. ax + bx – ay – by

= x[a + b] + (-y)[a + b]

= (x – y)[a + b]

4. Regrouping the terms, we have

15pq + 15 + 9q + 25p

= 15pq + 9q + 25p + 15

= 3q[5p + 3] + 5[5p + 3]

= (5p + 3)[3q + 5]

5. Regrouping the terms, we have

z – 7 + 7xy – xyz = z – 7 – xyz + 7xy

= 1[z – 7] – xy[z – 7] = (z – 7)(1 – xy)