# PSEB 8th Class Maths Solutions Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Ex 14.3

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Ex 14.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 8 Maths Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Exercise 14.3

1. ਹੇਠਾਂ ਲਿਖਿਆਂ ਦੀ ਵੰਡ ਕਰੋ :

ਪ੍ਰਸ਼ਨ (i).
28x4 ÷ 56x
ਹੱਲ:
28x4 ÷ 56x = $$\frac{28 x^{4}}{56 x}$$
= $$\frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x}$$
= $$\frac{1}{2}$$x3

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

ਪ੍ਰਸ਼ਨ (iii).
66pq2r3 ÷ 11qr2
ਹੱਲ:
66pq2r3 ÷ 11qr2 = $$\frac{66 p q^{2} r^{3}}{11 q r^{2}}$$
= $$\frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r}$$
= 2 × 3 × p × q × r
= 6pqr

ਪ੍ਰਸ਼ਨ (iv).
34x3y3z3 ÷ 51xy2z3
ਹੱਲ:
34x3y3z3 ÷ 51xy2z3 = $$\frac{34 x^{3} y^{3} z^{3}}{51 x y^{2} z^{3}}$$
= $$\frac{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{3 \times 17 \times x \times y \times y \times z \times z \times z}$$
= $$\frac{2 \times x \times x \times y}{3}$$ = $$\frac{2}{3}$$x2y

ਪ੍ਰਸ਼ਨ (v).
12a8b8 ÷ (-6a6b4).
ਹੱਲ:
12a8b8 ÷ (-6a6b4) = $$\frac{12 a^{8} b^{8}}{-6 a^{6} b^{4}}$$
= $$\begin{gathered} 2 \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times \\ \frac{a \times a \times b \times b \times b \times b \times b \times b \times b \times b}{-2 \times 3 \times a \times a \times a \times a} \\ \times a \times a \times b \times b \times b \times b \end{gathered}$$
= -2 × a × a × b × b × b × b
= -2a2b4

2. ਦਿੱਤੇ ਹੋਏ ਬਹੁਪਦ ਨੂੰ ਦਿੱਤੀ ਹੋਈ ਇਕ ਪਦੀ ਨਾਲ ਭਾਗ ਦਿਉ :

ਪ੍ਰਸ਼ਨ (i).
(5x2 – 6x) ÷ 3
ਹੱਲ:
5x2 – 6x = 5 × x × x – 2 × 3 × x
= x × (5 × x – 2 × 3)
= x × (5x – 6)
ਇਸ ਲਈ, (5x2 – 6x) ÷ 3x = $$\frac{x \times(5 x-6)}{3 \times x}$$
= $$\frac{1}{3}$$(5x – 6)

ਪ੍ਰਸ਼ਨ (ii).
(3y8 – 4y6 + 5y4) ÷ y4
ਹੱਲ:
3y8 – 4y6 + 5y4 = y4(3y4 – 4y2 + 5)
ਇਸ ਲਈ, (3y8 – 4y6 + 5y4) ÷ y4
= $$\frac{\left(3 y^{8}-4 y^{6}+5 y^{4}\right)}{y^{4}}$$ = $$\frac{y^{4}\left(3 y^{4}-4 y^{2}+5\right)}{y^{4}}$$
= 3y4 – 4y2 + 5

ਪ੍ਰਸ਼ਨ (iii).
8 (x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
ਹੱਲ:
8 (x3y2z2 + x2y3z2 + x2y2z3) .
= 2 × 2 × 2 × x2 × y2 × z2 (x+y + z)
ਇਸ ਲਈ, 8 (x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
= $$\frac{2 \times 2 \times 2 \times x^{2} \times y^{2} \times z^{2} \times(x+y+z)}{2 \times 2 \times x^{2} \times y^{2} \times z^{2}}$$
= 2 (x + y + z)

ਪ੍ਰਸ਼ਨ (iv).
(x3 + 2x2 + 3x) ÷ 2x
ਹੱਲ:
x3 + 2x2 + 3x = x × x × x + 2 × x × x + 3 × x
= x × (x × x + 2 × x + 3)
= x × (x2 + 2x + 3)
ਇਸ ਲਈ, (x3 + 2x2 + 3x) ÷ 2x
= $$\frac{x \times\left(x^{2}+2 x+3\right)}{2 x}$$ = $$\frac{1}{2}$$(x2 + 2x + 3)

ਪ੍ਰਸ਼ਨ (v).
(p3q6 – p6q3) ÷ p3q3.
ਹੱਲ:
p3q6 – p6q3 = p3 × q3 × q3 – p3 × p3 × q3 = p3 × q3 × (q3 – p3)
ਇਸ ਲਈ, (p3q6 – p6q3) ÷ p3q3
= $$\frac{p^{3} \times q^{3} \times\left(q^{3}-p^{3}\right)}{p^{3} \times q^{3}}$$ = q3 – p3

3. ਹੇਠਾਂ ਲਿਖਿਆ ਦੀ ਵੰਡ ਕਰੇ :

ਪ੍ਰਸ਼ਨ (i).
(10x – 25) ÷ 5
ਹੱਲ:
10x – 25 = 2 × 5 × x -5 × 5
= 5 (2 × x – 5)
= 5 × (2x – 5)
ਇਸ ਲਈ, (10x – 25) ÷ 5 = $$\frac{10 x-25}{5}$$
= $$\frac{5×(2x-5)}{5}$$ = 2x – 5

ਪ੍ਰਸ਼ਨ (ii).
(10x – 25) ÷ (2x – 5)
ਹੱਲ:
10x – 25 = 2 × 5 × x – 5 × 5
= 5 × (2 × x – 5) = 5 × (2x – 5)
ਇਸ ਲਈ, (10x – 25) ÷ 2x – 5
$$\frac{10 x-25}{2x-5}$$ = $$\frac{5×(2x-5)}{(2x-5)}$$

ਪ੍ਰਸ਼ਨ (iii).
10y (6y + 21) ÷ 5 (2y + 7)
ਹੱਲ:
10y (6y + 21) = 2 × 5 × y × (2 × 3 × y + 3 × 7)
= 2 × 5 × y × 3 × (2 × y + 7)
= 2 × 5 × 3 × y (2y + 7)
ਇਸ ਲਈ, 10y (6y + 21) ÷ 5 (2y + 7)
= $$\frac{10 y(6 y+21)}{5(2 y+7)}$$
= $$\frac{2 \times 5 \times 3 \times y \times(2 y+7)}{5 \times(2 y+7)}$$
= 2 × 3 × y = 6y

ਪ੍ਰਸ਼ਨ (iv).
9x2y2 (3z – 24) ÷ 27xy (z – 8)
ਹੱਲ:
9x2y2 (3z – 24)
= 3 × 3 × x × x × y × y × (3 × 2 = 3 × 8)
= 3 × 3 × 3 × x × x × y × y × (z – 8)
ਇਸ ਲਈ, 9x2y2(3z – 24) = 27xy (z – 8)
= $$\frac{9 x^{2} y^{2}(3 z-24)}{27 x y(z-8)}$$
= $$\frac{3 \times 3 \times 3 \times x \times x \times y \times y \times(z-8)}{3 \times 3 \times 3 \times x \times y \times(z-8)}$$
= x × y = xy

ਪ੍ਰਸ਼ਨ (v).
96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6).
ਹੱਲ:
96abc (3a – 12) (5b – 30)
= 2 × 2 × 2 × 2 × 2 × 3 × a × b × c × (3 × a – 3 × 4) (5 × b – 5 × 6)
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × a × b × c × (a – 4) × 5 × (b – 6)
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × a × b × c (a – 4) (b – 6)
ਇਸ ਲਈ, 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
= $$\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}$$
= $$\begin{gathered} 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \\ \times a \times b \times c \times(a-4) \times(b-6) \\ \hline 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4)(b-6) \end{gathered}$$
= 2 × 5 × a × b × c.
= 10abc

4. ਨਿਜਦੋਸ ਆਨੁਮਾਰ ਭਾਗ ਕਰੇ :

ਪ੍ਰਸ਼ਨ (i).
5 (2x + 1) (3x + 5) ÷ (2x + 1)
ਹੱਲ:
5 (2x + 1) (3x + 5) ÷ (2x + 1)
= $$\frac{5(2x+1)(3x+5)}{2x+1}$$
= 5 (3x + 5)

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

ਪ੍ਰਸ਼ਨ (iii).
52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
ਹੱਲ:
52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)
= $$\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r) \times(r+p)}$$
= $$\frac{2 \times 2 \times 13 \times p \times q \times r \times(p+q) \times(q+r) \times(r \times p)}{2 \times 2 \times 2 \times 13 \times p \times q \times(q+r) \times(r+p)}$$
= $$\frac{1}{2}$$ × r × (p + q) = $$\frac{1}{2}$$r(p + q)

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

ਪ੍ਰਸ਼ਨ (v).
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
ਹੱਲ:
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
= $$\frac{x \times(x+1) \times(x+2) \times(x+3)}{x \times(x+1)}$$
= (x + 2) (x + 3)

5. ਵਿਅੰਜਕ ਦੇ ਗੁਣਨਖੰਡ ਕਰੋ ਅਤੇ ਨਿਰਦੇਸ਼ ਅਨੁਸਾਰ ਭਾਗ ਕਰੇ :

ਪ੍ਰਸ਼ਨ (i).
(y2 + 7y + 10) ÷ (y + 5)
ਹੱਲ:
(i) y2 + 7y + 10
= y2 + 2y + 5y + 10
= y(y + 2) + 5 (y + 2)
= (y + 2) (y + 5)
ਇਸ ਲਈ, (y2 + 7y + 10) ÷ (y + 5)
= $$\frac{(y+2)(y+5)}{(y+5)}$$ = y + 2

ਪ੍ਰਸ਼ਨ (ii).
(m2 – 14m – 32) ÷ (m + 2)
ਹੱਲ:
m2 – 14m – 32
= m2 – 16m + 2m = 32
= m (m – 16) + 2 (m – 16)
= (m – 16) (m + 2)
ਇਸ ਲਈ, (m2 – 14m – 32) ÷ (m + 2)
= $$\frac{m^{2}-14 m-32}{m+2}$$
= $$\frac{(m-16)(m+2)}{(m+2)}$$
= m – 16

ਪ੍ਰਸ਼ਨ (iii).
(5p2 – 25p + 20) ÷ (p – 1)
ਹੱਲ:
5p2 – 25p + 20 = 5 × (p2 – 5p + 4)
= 5 × [p2 – 4p – p + 4]
= 5 × [p (p – 4) – 1 (p – 4)]
= 5 × (p – 4) (p – 1)
ਇਸ ਲਈ, 5p2 – 25p + 20 ÷ (p – 1)
= $$\frac{5 p^{2}-20 p+20}{p-1}$$
= $$\frac{5 \times(p-4)(p-1)}{(p-1)}$$
= 5 × (p – 4) = 5(p – 4)

ਪ੍ਰਸ਼ਨ (iv).
4yz (z2 + 6z – 16) ÷ 2y (z + 8)
ਹੱਲ:
4yz (z2 + 6z – 16) = 2 × 2 × y × z × (z2 + 8z – 27 – 16) = 2 × 2 × y × z × [z (z + 8) – 2 (z + 8)]
= 2 × 2 × y × z × (z + 8) (z – 2)
ਇਸ ਲਈ, 4yz (z2 + 6z – 16) ÷ 2y (2 + 8)
= $$\frac{4 y z\left(z^{2}+6 z-16\right)}{2 y(z+8)}$$
= $$\frac{2 \times 2 \times y \times z \times(z+8)(z-2)}{2 \times y \times(z+8)}$$
= 2 × z × (z – 2)
= 2z (z – 2)

ਪ੍ਰਸ਼ਨ (v).
5pq (p2 – q2) ÷ 2p (p + q)
ਹੱਲ:
5pq (p2 – q2) = 5 × p × q × (p + q) (p – q)
ਇਸ ਲਈ, 5pq (p2 – q2) ÷ 2p (p + q)
= $$\frac{5 p q\left(p^{2}-q^{2}\right)}{2 p(p+q)}$$
= $$\frac{5 \times p \times q \times(p+q)(p-q)}{2 \times p \times(q+q)}$$
= $$\frac{5}{2}$$q (p – q)

ਪ੍ਰਸ਼ਨ (vi).
12xy (9x2 – 16y2) ÷ 4xy (3x + 4y)
ਹੱਲ:
12xy (9x2 – 16y2) = 2 × 2 × 3 × x × y × [(3x)2 – (4y)2]
= 2 × 2 × 3 × x × y × (3x + 4y) (3x – 4y)
ਇਸ ਲਈ, 12xy (9x2 – 16y2) ÷ 4y (3x + 4y)
= $$\frac{12 x y\left(9 x^{2}-16 y^{2}\right)}{4 x y(3 x+4 y)}$$
= $$\frac{2 \times 2 \times 3 \times x \times y \times(3 x+4 y)(3 x-4 y)}{2 \times 2 \times x \times y \times(3 x+4 y)}$$
= 3 (3x – 4y)

ਪ੍ਰਸ਼ਨ (vii).
39y3(50y2 – 98) ÷ 26y2(5y + 7)
ਹੱਲ:
39y3(50y2 – 98)
= 3 × 13 × y × y × y × (2 × 5 × 5 × y × y – 2 × 7 × 7)
= 3 × 13 × y × y × y × 2 × (5y × 5y – 7 × 7)
= 2 × 3 × 13 × y × y × y × (5y + 7) (5y – 7)
ਇਸ ਲਈ, 39y (50y3 – 98) ÷ 26y2(5y + 7)
= $$\frac{39 y^{3}\left(50 y^{2}-98\right)}{26 y^{2}(5 y+7)}$$
= $$\frac{2 \times 3 \times 13 \times y \times y \times y \times(5 y+7)(5 y-7)}{2 \times 13 \times y \times y \times(5 y+7)}$$
= 3 × y × (5y – 7) = 3y (5y – 7)