HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.3

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.3

प्रश्न 1.
निम्नलिखित विभाजन कीजिए
(i) 28x4 ÷ 56x
(ii) -36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z2 ÷ 51xy2z3
(v) 12a8b8 ÷ (-6a6b4)
हल:
(i) 28x4 ÷ 56x
= $$\frac{28 x^{4}}{56 x}$$
= $$\frac{7 \times 2 \times 2 \times x \times x \times x \times x}{7 \times 2 \times 2 \times 2 \times x}$$
= $$\frac{x^{3}}{2}$$
अतः 28x4 ÷ 56x = $$\frac{x^{3}}{2}$$

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

(iii) 66pq2r3 ÷ 11qr2
= $$\frac{66 p q^{2} r^{3}}{11 q r^{2}}$$
= $$\frac{6 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r}$$
= 6 pqr
∴ 66pq2r3 ÷ 11qr2 = 6 pqr

(iv) 34x3y3z2 ÷ 51xy2z3
= $$\frac{34 x^{3} y^{3} z^{3}}{51 x y^{2} z^{3}}$$
= $$\frac{17 \times 2 \times x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot z z z}{17 \times 3 \times x \cdot y \cdot y \cdot z Z z}$$
= $$\frac{2 x^{2} y}{3}$$
∴ 34x3y3z2 ÷ 51xy2z3 = $$\frac{2 x^{2} y}{3}$$

(v) 12a8b8 ÷ (-6a6b4)
= $$\frac{12 a^{8} b^{6}}{-6 a^{6} b^{6}}$$
= $$\frac{3 \times 4 a^{6} \times a^{2} b^{4} \times b^{4}}{-3 \times 2 a^{6} b^{4}}$$
= -2a2b4
∴ 12a8b8 ÷ (-6a6b4) = -2a2b4

प्रश्न 2.
दिए हुए बहुपद को दिए हुए एकपदी से भाग दीजिए
(i) (5x2 – 6x) ÷ 3x
(ii) (3y8 – 4y6 + 5y4), y4
(iii) 8 (x4y2z2 + x2y3z2 + x2y2z2)
(iv) 9x2y2(3z – 24) ÷ 27xy(z – 8)
(v) 96abc(3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
हल:
(i) (5x2 – 6x) ÷ 3x

(ii) (3y8 – 4y6 + 5y4), y4

(iii) 8 (x4y2z2 + x2y3z2 + x2y2z2)

(iv) 9x2y2(3z – 24) ÷ 27xy(z – 8)

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

प्रश्न 3.
निम्नलिखित विभाजन कीजिए
(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)
हल:

प्रश्न 4.
निर्देशानुसार भाग दीजिए
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
(ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)
(iii) 52pqr (p + q) (q + r) (q + 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)
हल:
(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)
= $$\frac{26 x y(x+5)(y-4)}{13 x(y-4)}$$
= $$\frac{26 x y(x+5)}{13 x}$$
= 2y(x+5)

(iii) 52pqr (p + q) (q + r) (q + p) ÷ 104pq (q + r) (r + p)

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

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

प्रश्न 5.
व्यंजक के गुणनखंड कीजिए और निर्देशानुसार भाग दीजिए
(i) (y2 + 7y + 10) ÷ (y + 5)
(ii) (m2 – 14m – 32) ÷ (m + 2)
(iii) (5p2 – 25p + 20) ÷ (p – 1)
(iv) 4yz (z2 + 6z – 16) ÷ 2y (z + 8)
(v) 5pq(p2 – q2) ÷ 2p(p + q)
(vi) 12xy(9x2 – 16y2) ÷ 4xy(3x + 4y)
(vii) 39y3(50y2 – 98) ÷ 26y2(5y + 7)
हल:
(i) (y2 + 7y + 10) ÷ (y + 5)
भाज्य = y2 + 7y + 10
= y2 + 5y + 2y + 10
= (y2 + 5y) + (2y + 10)
=y(y + 5) + 2(y + 5)
= (y + 5) + 2(y + 5)
= (y + 5) (y + 2)
इस प्रकार, $$\frac{y^{2} + 7y + 10}{y+5}$$ = $$\frac{(y + 5) (y + 2)}{(y + 5)}$$
= (y + 2)

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

(iii) (5p2 – 25p + 20) ÷ (p – 1)
भाज्य = 5p2 – 20p – 5p + 20
= 5p2 – 20p) – (5p – 20)
= 5p(p – 4) – 5(p – 4)
= (p – 4)(5p – 5)
= 5(p – 4)(p – 1)
∴ $$\frac{5 p^{2}-25 p+20}{p-1}$$ = $$\frac{5(p-4)(p-1)}{(p-1)}$$
= 5 (p – 4)

(iv) 4yz(z2 + 6z – 16) ÷ 2y (z + 8)
भाज्य = 4yz(z2 + 6z – 16)
= 4yz(z2 + 8z – 2z – 16)
= 4yz [(z + 8z) – (2z + 16)]
= 4yz [z (z +8) – 2 (z + 8)]
= 4yz(z + 8) (z – 2)
∴ $$\frac{4 y z\left(z^{2}+6 z-16\right)}{2 y(z+8)}$$ = $$\frac{4 y z(z+8)(z-2)}{2 y(z+8)}$$
= $$\frac{4 y z(z-2)}{2 y}$$
= 2z(z – 2)

(v) 5pq(p2 – q2) ÷ 2p(p + q)

(vi) 12xy (9x2 – 16y2) + 4xy (3x + 4y)

(vii) 39y3(50y2 – 98) + 26y2(5y + 7)