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)