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.