NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.3

Question 1.
Carry out the following divisions.
(i) 28x4 ÷ 56x
(ii) – 36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (- 6a6b4)
(i) 28x4 ÷ 56x
= $$\frac{2 \times 2 \times 7 \times x^{4}}{2 \times 2 \times 2 \times 7 \times x}$$
= $$\frac{x^{4-1}}{2}=\frac{x^{3}}{2}=\frac{1}{2} x^{2}$$

(ii) – 36y3 ÷ 9y2
= $$\frac{-2 \times 2 \times 3 \times 3 \times y^{3}}{3 \times 3 \times y^{2}}$$
= – 4y3-2
= -4y

(iii) 66pq2r3 ÷ 11qr2
= $$\frac{2 \times 3 \times 11 \times p \times q^{2} \times r^{3}}{11 \times q \times r^{2}}$$
= 2 x 3 x p x q2-1 x r3-2
= 6 x p x q x r = 6pqr

(iv) 34x2y2z2 ÷ 51xy2z3
= $$\frac{2 \times 17 \times x^{3} \times y^{3} \times z^{3}}{3 \times 17 \times x \times y^{2} \times z^{3}}$$
= $$\frac{2}{3}$$ x x3-1 x y3-2 x z3-3
= $$\frac{2}{3}$$ x x2 x y x z0
= $$\frac{2 x^{2} y}{3}$$

(v) 12a8b8 ÷ (-6 a6 b4)
= $$\frac{2 \times 2 \times 3 \times a^{8} \times b^{8}}{-2 \times 3 \times a^{6} \times b^{4}}$$
= -2 x a8-6 x b8-4
= – 2 x a2 x b4
= – 2a2b4

Question 2.
Divide the given polynomial by the given monomial.
(i) (5x2 – 6x) ÷ 3x
(ii) (3y8 – 4y6 + 5y4) ÷ y4
(iii) 8 (x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
(iv) (x3 + 2x2 + 3x) ÷ 2x
(v) (p3q6– p6q3) ÷ p3q3
(i) (5x2 – 6x) ÷ 3x

(ii) (3y8 – 4y6 + 5y4) ÷ y4

=3y4 – 4y6-4 + 5y4-4
=3y4 – 4y2 – 5y°
= 3y4 – 4y2 + 5

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) 4x2y2z2

= [x3-2 x y2-2 x z2-2 + x2-2 x y3-2 x z2-2 + x2-2 x y2-2 x z3-2]
= 2 (x x y° x z° + x° x y x z° + x° x y° x z)
= 2 (x + y + z)
[using a° = 1]

(iv) x3 + 2x2 + 3x ÷ 2x

(v) (p3q6– p6q3) ÷ p3q3

= p3-3 x q6-3 – p6-3 x q3-3
= p0 x q3 – p3 x q0
= p3 x q3

Question 3.
Work out the following divisions:
(i) (10x – 25) ÷ 5
(ii) (10x-25) ÷ (2x-5)
(iii) 10y(6y + 21) ÷ 5 (2y + 7)
(iv) 9x2y2(3z – 24) ÷ 27xy (z – 8)
(v) 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
(i) (10x – 25) ÷ 5

= 2x – 5

(ii) (10x – 25) ÷ (2x – 5)

(iii) 10y (6y + 21) ÷ 5 (2y + 7)

[Taking 3 as common from 6y + 21]
= 2 x y x 3
= 6y

(iv) 9x2y2(3z – 24) ÷ 27xy (z – 8)

= x2-1 x y2-1 = xy.

(v) 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)

Question 4.
Divide as directed.
(i) 5(2x+1) (3x +5) ÷ (2x+1)
(ii) 26xy(x + 5) (y-4) ÷ 13x(y-4)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
(iv) 20 (y + 4) (y2 + 5y + 3) ÷ 5(y + 4)
(v) x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
= $$\frac{5(2 x+1)(3 x+5)}{(2 x+1)}$$
= 5 (3x + 5)

(ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)
= $$\frac{2 \times 13 \times x \times y \times(x+5) \times(y-4)}{13 \times x \times(y-4)}$$
= 2 x y x (x + 5) = 2y (x + 5)

(iii) 52pqr (p +q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
= $$\frac{2 \times 2 \times 13 \times p \times q \times r \times(p+q) \times(q+r) \times(r+9)}{2 \times 2 \times 2 \times 13 \times p \times q \times(q+r) \times(r+p)}$$
_ 2x2xl3xpxqxrx(p + q)x(q + r)x(r + 9) 2x2x2xl3xpxqx(q + r)x(r + p)
= $$\frac{r(p+q)}{2}=\frac{1}{2}$$r(p+q)

(iv) 20(y + 4) (y2 + 5y + 3) ÷ 5 (y + 4)
= $$\frac{4 \times 5(y+4)\left(y^{2}+5 y+3\right)}{5 \times(y+4)}$$
= 4 (y2 + 5y + 3)

(v) x(x+ 1) (x + 2) (x + 3) ÷ x(x + 1)
= $$\frac{x(x+1)(x+2)(x+3)}{x(x}$$
= (x + 2)(x + 3)

Question 5.
Factorise the expressions and divide them as directed.
(i) (y2 + 7y + 10) – (y + 5)
(ii) (m2 – 14m – 32) + (m + 2)
(iii) (5p2 – 25p + 20) 4- (p – 1)
(iv) 4yz(z2 + 6z – 16) -f- 2y (z + 8)
(v) 5pq (p2 – q2) -f- 2p (p + q)
(vi) 12xy (9x2 – 16y2) -J- 4xy (3x + 4y)
(vii) 39y3 (50y2 – 98) + 26y2 (5y + 7)
(i) y2 + 7y + 10
[10 = 2 x 5
[7 = 2 + 5]
= y2 + 5y + 2y + 10
= y( y + 5) + 2 (y + 5) = (y + 5) (y + 2)
∴ (y2 + 7y + 10) ÷ (y + 5)
$$\frac{(y+5)(y+2)}{(y+5)}$$ = y + 2

(ii) m2 – 14m – 32
– 32 = -16 x 2
-14 = -16 + 2
= m (m – 16) + 2(m – 16)
= (m – 16) (m + 2)
(m2 – 14m – 32) ÷ (m + 2)
= $$\frac{(m-16)(m+2)}{m+2}$$
= m – 16

(iii) 5p2 – 25p + 20
= 5[p2 – 5p + 4]
= 5 [p2 – 4p – p + 4]
= 5[p(p – 4)- 1 (p – 4)
= 5(p – 4) (p – 1)
∴ 5p2 – 25 p + 20 ÷ (p – 1)
= $$\frac{5(p-4)(p-1)}{(p-1)}$$
= 5 (p -4)

(iv) z2 + 6z – 16 = z2 + 8z – 2z – 16
-16 = 8 x – 2
6 = 8 + (-2)
= z(z + 8) – 2 (z – 2)
= (z + 8) (z – 2)
∴ 4yz (z2 + 6z – 16) 2y (z + 8)
= $$\frac{4 \times y \times z \times(z+8) \times(z-2)}{2 \times y \times(z+8)}$$
4xyxzx(z + 8)x(z-2)

(v) 5pq(p2 – q2) ÷ 2p(p + q)
= $$\frac{5 \times p \times q \times(p+q)(p-q)}{2 \times p \times(p+q)}$$
= $$\frac{5 q(p-q)}{2}$$
= $$\frac { 5 }{ 2 }$$q(p-q)

(vi) 12xy (9x2 – 16y2)
= 2 x 2 x 3 x x x y x [(3x)2 – (4y)2]
= 2 x 2 x 3 x x x y x (3x + 4y) (3x – 4y)
12xy (9x2 – 16y2) ÷ 4xy (3x + 4y)
=$$\frac{2 \times 2 \times 3 \times x \times y \times(3 x+4 y) \times(3 x-4 y)}{2 \times 2 \times x \times y \times(3 x+4 y)}$$
= 3 (3x – 4y)

(vii) 39y3 (50y2 – 98)
= 3 x 13 x y3 x 2 (25y2 – 49)
= 3 x 13 x 2 x y3 [(5y)2 – 72]
= 3 x 13 x 2 x y3 x (5y + 7) (5y – 7)
39y3 (50y2 – 98) + 26y2 (5y + 7)
= $$\frac{3 \times 13 \times 2 \times y^{3} \times(5 y+7) \times(5 y-7)}{2 \times 13 \times y^{2}(5 y+7)}$$
= 3 x y3-2 x (5y – 7)
= 3y(5y – 7)

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