NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities Ex 8.4
Ex 8.4 Class 8 Maths 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) (3/4 a2 + 3b2) and 4(a2 – 2/3 b2)
Solution:
(i) (2x
+ 5) × (4x – 3)
= 2x × (4x – 3) + 5 × (4x – 3)
= (2x × 4x) – (2x × 3) + (5 × 4x) – (5 × 3)
= 8x2 –
6x + 20x – 15
= 8x2 +
14x – 15
(ii) (y – 8) × (3y – 4)
= y × (3y – 4) – 8 × (3y – 4)
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4)
= 3y2 –
4y – 24y + 32
= 3y2 –
28y + 32
(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
= 6.25l2 +
1.25ml – 1.25ml – 0.25m2
= 6.25l2 +
0 – 0.25m2
= 6.25l2 –
0.25m2
(iv) (a + 3b) × (x + 5)
= a × (x + 5) + 3b × (x + 5)
= (a × x) + (a × 5) + (3b × x) + (3b × 5)
= ax + 5a + 3bx + 15b
(v) (2pq + 3q2)
× (3pq – 2q2)
= 2pq × (3pq – 2q2)
+ 3q2 (3pq
– 2q2)
= (2pq × 3pq) – (2pq × 2q2)
+ (3q2 ×
3pq) – (3q2 ×
2q2)
= 6p2q2 –
4pq3 +
9pq3 –
6q4
= 6p2q2 +
5pq3 –
6q4
Ex 8.4 Class 8 Maths 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) – (x × y) + (7y × 7x) – (7y × y)
= 7x2 –
xy + 49xy – 7y2
= 7x2 +
48xy – 7y2
(iii) (a2 +
b) (a + b2)
= a2(a
+ b2)
+ b(a + b2)
= (a2 ×
a) + (a2 ×
b2)
+ (b × a) + (b × b2)
= a3 +
a2b2 +
ab + b3
(iv) (p2 –
q2)
(2p + q)
= p2(2p
+ q) – q2(2p
+ q)
= (p2 ×
2p) + (p2 ×
q) – (q2 ×
2p) – (q2 ×
q)
= 2p3 +
p2q
– 2pq2 –
q3
Ex 8.4 Class 8 Maths 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
= x3 +
5x2 –
5x – 25 + 25
= x3 +
5x2 –
5x + 0
= x3 +
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 –
st + s2t2 –
s3
= t3 +
s2t2 –
st – 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
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 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)
= 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)
= x3 –
x2y
+ xy2
+ x2y –
xy2 +
y3
= x3 –
0 + 0 + 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 +
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)
= a2 +
ab – ac + ab + b2 –
bc + ac + bc – c2
= a2 +
ab + ab – bc + bc – ac + ac + b2 –
c2
= a2 +
2ab + b2 –
c2 +
0 + 0
= a2 +
2ab + b2 –
c2
Related Links:
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