KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4
KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4
KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4
Question 1.
Multiply the binomials.
(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
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
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