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

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KSEEB Solutions for Class 8 Maths 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 (2pq – 2q²)
(vi) \(\left(\frac{3}{4} a^{2}+3 b^{2}\right)\) and \(4\left(a^{2}-\frac{2}{3} b^{2}\right)\)
Solution:
(i) (2x + 5) and (4x – 3)
= (2x + 5) (4x – 3)
= 2x(4x – 3) + 5(4x – 3)
= 2x × 4x – 3 × 2x + 5 × 4x – 5 × 3
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15

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

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

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

(v) (2pq + 3q²) and (2pq – 3q²)
= (2pq + 3q²) (2pq – 3q²)
= 2pq(3pq – 2q²) + 3q²(3pq – 2q²)
= 2pq × 3pq – 2q² × 2q² + 3q² × 3pq – 3q² × 2q²
= 6p²q² – 4pq³ + 9pq³ – 6pq⁴
= 6p²q² + 5pq³ – 6q⁴

(vi) \(\left(\frac{3}{4} a^{2}+3 b^{2}\right)\) and \(4\left(a^{2}-\frac{2}{3} b^{2}\right)\)
= \(\left(\frac{3}{4} a^{2}+3 b^{2}\right)\left(4 a^{2}-\frac{8}{3} b^{2}\right)\)
= \(\frac{3}{4} a^{2}\left(4 a^{2}-\frac{8}{3} b^{2}\right)\) + \(3 b^{2}\left(4 a^{2}-\frac{8}{3} b^{2}\right)\)
= \(\frac{3}{4} a^{2}+4 a^{2}-\frac{3}{4} \times \frac{8}{3} a^{2} b^{2}\) + \(3 \times 4 b^{2} a^{2}-\frac{3 \times 8}{3} b^{2} b^{2}\)
= \(\frac{3}{4} \times 4 a^{4}\) – 2b²a² + 12b²a² – 8b⁴
= 3a⁴ + 10a²b² – 8b⁴

Question 2.
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a² + b) (a + b²)
(iv) (p² + q²) (2p + q)
Solution:
(i) (5 – 2x) (3 + x)
= 5(3 + x) – 2x(3 + x)
= 5 × 3 + 5 × x – 2x × 3 – 2x × x
= 15 + 5x – 6x – 2x²
= 15 – x – 2x²

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

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

(iv) (p² + q²) (2p + q)
= p²(2p + q) – q²(2p + q)
= 2p³ + p²q – q² × 2p – q² × 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)
Solution:
(i) (x² – 5) (x + 5) + 25
= x²(x + 5) – 5(x + 5) + 25
= x² × x + x² × 5 – 5 × x – 5 × 5 + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x

(ii) (a² + 5) (b³ + 3) + 5
= a²(b² + 3) + 5(b² + 3) + 5
= a² × b² + a² × 3 + 5b² + 5 × 3 + 5
= a²b² + 3a² + 5b² + 15 + 5
= a²b² + 3a² + 5b² + 20

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

(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
= a(c – b) + b(c – d) + a(c + d) – b(c + d) + 2 × ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac – ad + ad + bc – bc – bd – bd + 2bd
= 4ac + 0 + 0 + 0 – 2bd + 2bd
= 4ac + 0 + 0 + 0
= 4ac

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

(vi) (x + y) + (x² + xy + y²)
= x(x² – xy + y²) + y(x² + xy + y²)
= x × x² – x × xy + xy² + yx² – y × xy + y × y²
= x³ – x²y + xy² + yx² – xy² + y³
= x³ – x²y + yx² + xy² – 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
= 1.5 × 1.5x × x + 1.5 × 4xy + 1.5 × 3x – 4 × 1.5xy – 4 × 4 × y × y – 4 × 3y – 4.5x + 12y
= 2.25x² + 6xy – 6xy + 4.5x – 12y – 4.5x – 16y² + 12y – 4.5x + 12y
= 2.25x² + 6xy – 6xy + 4.5x – 4.5x – 12y + 12y – 16y²
= 2.25x² + 0 + 0 + 0 – 16y²
= 2.25x² – 16y²

(viii) (a + b + c) (a + b – c)
= a(a + b – c) + b(a + b – c) + c(a + b – c)
= a × a + a × b – a × c + b × a + b × b – b × c + c × a + c × b – c × c
= a² + ab – ac + ab + b² – bc + ca + cb – c²
= a² + ab + ab + ac + ac + b² – bc + bc – c²
= a² + 2ab + 0 + b² + 0 – c²
= a² + 2ab + b² – c²
= a² + b² – c² + 2ab