RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4

Question 1.
Multiply the binomials
(i) (2x + 5) and (4x - 3)
(ii) (y - 8) and (3y - 4)
(iii) (2.5l - 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q²) and (3pq - 2q²)
(vi) (\(\frac{3}{4}\)a² + 3b²) and 4(a² - \(\frac{2}{3}\)b²)
Answer:
(i) (2x + 5 ) x (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) × (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) - 0.5m (2.5l + 0.5m)
= 6.25l² + 1.25lm - 1.25ml - 0.25m²
= 6.25l^{2</sup. - 0.25m2 Ans. [∵ lm = ml]}

(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) (\(\frac{3}{4}\)a² + 3b²) × 4(a² - \(\frac{2}{3}\)b²)
= (\(\frac{3}{4}\)a² + 3b²) (4a² - \(\frac{8}{3}\)b²)
= \(\frac{3}{4}\)a²(4a² - \(\frac{8}{3}\)b²) + 3b²(4a² - \(\frac{8}{3}\)b²)
= 3a⁴ - 2a²b² + 12oa²b² - 8b⁴
= 3a⁴ + 10a²b² - 8b⁴

Question 2.
Find the product
(i) (5 - 2x) (3 + x)
Answer:
(5 - 2x) (3 + x) = 5(3 + x) - 2x(3 + x)
= 15 + 5x - 6x - 2x2
= 15 - x - 2x²

(ii) (x + 7y) (7x - y)
Answer:
(x + 7y) (7x - y) = x(7x - y) + 7y(7x - y)
= 7x² - xy + 49xy - 7y²
= 7x² + 48xy - 7y²

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

(iv) (p² - q²) (2p + q)
Answer:
(p² - q²) (2p + q) = p²(2p + q) - q²(2p + q)
= 2p³ + p²q - 2pq² - q³

Question 3.
Simplify
(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)
Answer:
(i) (x² - 5) (x + 5) + 25
= x² (x + 5) - 5 (x + 5) + 25
= x³ + 5x² - 5x - 25 + 25
= x² + 5x² - 5x

(ii) (a² + 5) (b³ + 3) + 5
= a²(b³ + 3) + 5 (b³ + 3) + 5
= a²b³ + 3a² + 5b³ +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
= (1 + 1 + 2) ac + (- 1 + 1 )ad + (1 - l)bc + (- 1 - 1 + 2) bd
= 4ac + (0) ad + (0) (bc) + (0) bd
= 4ac + 0 + 0 = 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
= (2 + 1)x² + (1 + 2 - 1 + 2)xy + (1 - 2)y²
= 3x² + 4xy - y²

(vi) (x + y) (x² - xy + y²)
= x(x² - xy + y²) + y(x² - xy + y²)
= x³ - x²y + xy² + x²y - xy² + y³
= x³ + (-1 + 1 )x²y + (1 - 1) xy² + y³
= x³ + (0)x²y + (0) (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
= 1.5x × 1.5x + 1.5x × 4y + 1.5x × 3 - 4y × 1.5x - 4y × 4y - 4y × 3 - 4.5x + 12y
= 2.25x² + 6xy + 4.5x - 6xy - 16y² - 12y - 4.5x + 12y
= 2.25x² + (6 - 6)xy + (4.5 - 4.5)x - 16y² + (- 12 + 12 )y
= 2.25x² + (0)xy + (0)x - 16y² + (0)y
= 2.25x² + 0 + 0 - 16y² + 0
= 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