## MP Board Class 8th Maths Solutions Chapter 14 Factorization Ex 14.1

The prime factorization of 84 is 2 x 2 x 3 x 7 x 1 = 22 x 3 x 7 x 1.

Question 1.

Find the common factors of the given terms,

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28p^{2}q^{2}

(iv) 2x, 3x^{2}, 4

(v) 6abc, 24ab^{2},12a^{2}b

(vi) 16x^{3}, – 4x^{2}, 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x^{2}y^{3}, 10x^{3}y^{2}, 6x^{2}y^{2}z

Solution:

(i) The given terms are 12x and 36.

12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

Thus, the common factor of the given terms is 2 × 2 × 3 = 12

(ii) The given terms are 2y and 22xy

2y = 2 xy

22xy = 2 × 11 × X × y

Thus, the common factor of the given terms is 2 × y = 2y.

(iii) The given terms are 14pq and 18p^{2}q^{2}

14 pq = 2 × 7 × p × q

18p2q2 = 2 × 2 × 7 × p × p × q × q

Thus, the common factors of the given terms is 2 × 7 × p × q = 14pq

(iv) The given terms are 2x, 3x^{2} and 4

2x = 2 × x

3x^{2} = 3 × x × x

4 = 2 × 2

Hence, the given three terms have no factor in common except 1.

(v) The given terms are 6abc, 24ab^{2} and 12a^{2}b

6abc= 2 × 3 × a × b × c

24ab^{2} = 2 × 2 × 2 × 3 × a × b × b.

12 a^{2}b = 2 × 2 × 3 × a × a × b

Thus, the common factor of the given terms is 2 × 3 × a × b = 6ab.

(vi) The given terms are 16x^{3}, -4x^{2} and 32x

16x^{3} = 2 × 2 × 2 × 2 × x × x × x

– 4x^{2} = -1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

Thus, the common factor of the given terms is 2 × 2 × x = 4x.

(vii) The given terms are 10pq, 20qr and 30rp

10pq = 2 × 5 × p × q

20 qr = 2 × 2 × 5 × q × r

30 rp.= 2 × 3 × 5 × r × p

Thus, the common factor of the given terms is 2 × 5 = 10.

(viii)The given terms are 3x^{2}y^{3}, 10x^{3}y^{2} and 6x^{2}y^{2}z

3x^{2}y^{2} = 3 × x × x × y × y × y

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

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

Thus, the common factor of the given terms is x × x y × y = x^{2}y^{2}.

Factors of 70 are any integer that can be multiplied by another integer to make exactly 70.

Question 2.

Factorise the following expressions.

(i) 7x – 42

(ii) 6p – 12g

(iii) 7a^{2} + 14a

(iv) -16z + 20x^{3}

(v) 20l^{2}m + 30alm

(vi) 5x^{2}y – 15xy^{2}

(vii) 10a^{2} – 15b^{2} + 20c^{2}

(viii) – 4a^{2} + 4ab – 4ca

(ix) x^{2}yz + xy^{2}z + xyz^{2}

(x) ax^{2}y + bxy^{2} + cxyz

Solution:

(i) The expression is 7x – 42

Factors of 7x = 7 × x and 42 = 2 × 3 × 7

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

(ii) The expression is 6p – 12q

Factors of 6p = 2 × 3 × p and

12q = 2 × 2 × 3 × q

∴ 6p – 12 q = 2 × 3 × p – 2 × 2 × 3 × q

= 2 × 3(p – 2 × q) = 6(p – 2 q)

(iii) The expression is 7a^{2} + 14a

Factors of 7a^{2} = 7 × a × a and

14a = 2 × 7 × a

∴ 7a^{2} + 14a = 7 × a × a + 2 × 7 × a

= 7 × a(a + 2) = 7a (a + 2)

(iv) The expression is -16z + 20z^{3}

Factors of -16z = – 1 × 2 × 2 × 2 × 2 × z and

20z^{3} = 2 × 2 × 5 × z × z × z

∴ -16z + 20z^{3} = -1 × 2 × 2 × 2 × 2 × z + 2 × 2 × 5 × z × z × z

= 2 × 2 × z(-1 × 2 × 2 + 5 × z × z)

= 4z (- 4 + 5z^{2})

(v) The expression is 20l^{2}m + 30alm

Factors of 20l^{2}m = 2 × 2 × 5 × l × l × m and

30alm = 2 × 3 × 5 × a × l × m;

∴ 20l^{2}m + 30alm = 2 × 2 × 5 × l × l × m + 2 × 3 × 5 × a × l × m

= 2 × 5 × l × m (2 × l + 3 × a)

= 10lm (2l + 3a)

(vi) The expression is 5x^{2}y – 15xy^{2}

Factors of 5x^{2}y = 5 × x × x × y and

15xy^{2} = 3 × 5 × x × y × y

∴ 5x^{2}y – 15xy^{2} = 5 × x × x × y – 3 × 5 × x × y × y

= 5 × x × y (x – 3 × y) = 5xy (x – 3y).

(vii) The expression is 10a^{2} – 15b^{2} + 20c^{2}

Factors of 10a^{2} = 2 × 5 × a × a

-15b^{2} = (-1) × 3 × 5 × b × b

20c^{2} = 2 × 2 × 5 × c × c

∴ 10a^{2} – 15b^{2} + 20c^{2}

= 2 × 5 × a × a – 3 × 5 × b × b + 2 × 2 × 5 × c × c

= 5(2a^{2} – 3b^{2} + 4c^{2})

(viii) The expression is – 4a^{2} + 4ab – 4ca

Factors of – 4a^{2} = (-1) × 2 × 2 × a × a

4ab = 2 × 2 × a × b and

-4ca = -1 × 2 × 2 × c × a

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

= 2 × 2 × a (-1 × a + b – 1 × c)

= 4a(-a + b – c)

(ix) The expression is x^{2}yz + xy^{2}z + xyz^{2}

Factors of x^{2}yz = x × x × y × z and

xy^{2}z = x × y × y × z and xyz^{2} = x × y × z × z

∴ x^{2} yz + xy^{2}z + xyz^{2} = x × x × y × z + x × y × y × z + x × y × z × z

= x × y × z(x + y + z) = xyz (x + y + z)

(x) The expression is ax^{2}y + bxy^{2} + cxyz

Factors of ax^{2}y = a × x × x × y,

bxy^{2} = b × x × y × y and cxyz = c × x × y × z

∴ ax^{2}y + bxy^{2} + cxyz = a × x × x × y + b × x × y × y + c × x × y × z

= xy (a × x + b × y + c × z) = xy (ax + by + cz)

Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. There are 8 integers that are factors of 24.

Question 3.

Factorise.

(i) x^{2} + xy + 8x + 8y

(ii) 15xy – 6x + 5y – 2

(iii) ax + bx – ay-by

(iv) 15pq + 15 + 9q + 25p

(v) z – 7 + 7xy – xyz

Solution:

(i) The given expression is

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

= x × x + x × y + 8 × x + 8xy

= x(x + y) + 8(x + y) = (x + 8) (x + y).

(ii) The given expression is 15xy – 6x + 5y – 2

= 3x(5y – 2) + 1 × (5y – 2)

= (3x + 1) (5y – 2)

(iii) The given expression is ax + bx – ay – by

= x (a + b) – y (a + b) = (x – y) (a + b)

(iv) The given expression is

15 pq + 15 + 9 q + 25 p

= 15pq + 25p + 15 + 9q = 5p(3q + 5) + 3(5 + 3q)

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

(v) The given expression is z – 7 + 7xy – xyz

= z – 7 + 7 × x × y – x × y × z

= z – 7 + xy (7 – z) = 1 (z – 7) – xy (z – 7)

= (z – 7)(1 – xy)

Factors of 144 are any whole number that can be multiplied by another whole number to make exactly 144.