# HBSE 8th Class Maths Solutions Chapter 9 बीजीय व्यंजक एवं सर्वसमिकाएँ Ex 9.4

Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 बीजीय व्यंजक एवं सर्वसमिकाएँ Ex 9.4 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 9 बीजीय व्यंजक एवं सर्वसमिकाएँ Ex 9.4

प्रश्न 1.
द्विपदों को गुणा कीजिए
(i) (2x + 5) और (4x – 3)
(ii) (y – 8) और (3y – 4)
(iii) (2.5l – 0.5 m) और (2.5l + 0.5 m)
(iv) (a + 3b) और (x + 5)
(v) (2pq + 3q2) और 3(pq – 2q2)
(vi) ($$\frac{3}{4}$$a2 + 3b2) और (a2 – $$\frac{2}{3}$$b2)
हल:
(i) (2x + 5) × (4x – 3)
= 2x(4x – 3) + 5(4x – 3)
= 8x2 – 6x + 20x – 15
= 8x2 + 14x – 15

(ii) (y – 8) × (3y – 4)
= y(3y – 4) – 8(3y – 4)
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32

(iii) (2.5l – 0.5m) × (2.51 + 0.5m)
= 2.5l (2.51 + 0.5m) – 0.5m(2.5l + 0.5m)
= 6.25l2 + 1.25lm – 1.25ml – 0.25m2
= 6.25l2 – 0.25m2

(iv) (a + 3b) × (x + 5)
= a (x + 5) + 3b(x + 5)
= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) × (3pq – 2q2)
= 2pq(3pq – 2q2) + 3q2(3pq – 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4

(vi)

प्रश्न 2.
गुणनफल ज्ञात कीजिए
(i) (5 – 2x)(3 + x)
(ii) (x + 7y)(7x – y)
(iii) (a2 + b) (a + b2)
(iv) (p2 – q2) (2p + q).
हल:
(i) (5 – 2x) × (3 + x)
= 5 × (3 + x) – 2x(3 + x)
= 15 + 5x – 6x – 2x2
= – 2x2 – x + 15

(ii) (x + 7y) × (7x – y)
= x(7x – y) + 7y(7x – y)
= 7x2 – xy + 49xy – 7y2
= 7x2 – 7y2 + 48xy.

(iii) (a2 + b) × (a + b2)
= a2(a + b2) + b(a + b2)
= a3 + a2b2 + ab + b3

(iv) (p2 – q2) × (2p + q)
= p2× (2p + q) – q2 × (2p + q)
= 2p3 + p2q – 2pq2 – q3

प्रश्न 3.
सरल कीजिए-
(i) (x2 – 5) (x + 5) + 25
(ii) (a2 + 5) (b3 + 3) + 5
(iii) (t + s2) (t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x +y) (2x +y) + (x + 2y) (x – y)
(vi) (x + y) (x2 – xy + y2)
(vii) (1.5x – 4y) (1.5x + 4y + 3)- 4.5x + 12y
(viii) (a + b + c) (a + b – c).
हल:
(i) (x2 – 5) (x + 5) + 25
= (x + 5) (x2 – 5) + 25
= x(x2 – 5) + 5(x2 – 5) + 25
= x3 – 5x + 5x2 – 25 + 25
= x3 + 5x2 – 5x

(ii) (a2 + 5) (b3 + 3) + 5
= a2(b3 + 3) + 5(b3 + 3) + 5
= a2b3 + 3a2 + 5b2 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20.

(iii) (t + s2) (t2 – s)
= t(t2 – s) + s2(t2 – s)
= t3 – ts + s2t2 – s3

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac +bd)
= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= 4ac – 2bd + 2bd
= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= 3x2 – y2 + 4xy.

(vi) (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3.

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x(1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y
= 2.25x2 – 16y2

(viii) (a + b + c) (a + b – c)
= a (a + b – c) + b(a + b – c) + c (a + b – c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + b2 – c2 + 2ab.