NCERT Solution Class 8th Maths Chapter – 8 Algebraic Expressions and Identities
| Textbook | NCERT |
| Class | 8th |
| Subject | Mathematics |
| Chapter | 8th |
| Chapter Name | Algebraic Expressions and Identities |
| Category | Class 8th Maths |
| Medium | English |
| Source | Last Doubt |
NCERT Solution Class 8th Maths Chapter – 8 Algebraic Expressions and Identities
Chapter – 8
Algebraic Expressions and Identities
Exercise 8.3
1. Carry out the multiplication of the expressions in each of the following pairs. (i) 4p, q + r
(ii) ab, a − b
(iii) a2 + b, 7a2b2
(iv) a2 − 9, 4a
(v) pq + qr + rp, 0 Solution: (i) 4p(q + r) = 4pq + 4pr
(ii) ab(a – b) = a2b – ab2
(iii) (a + b) (7a2b2) = 7a3b2 + 7a2b3
(iv) (a2 – 9 )(4a) = 4a3 – 36a
(v) (pq + qr + rp) × 0 = 0 ( Anything multiplied by zero is zero ) |
2. Complete the table. | First expression | Second expression | Product | | (i) | a | b + c + d | ______ | | (ii) | x + y – 5 | 5xy | ______ | | (iii) | p | 6p2 – 7p + 5 | ______ | | (iv) | 4 p2 q2 | P2 – q2 | ______ | | (v) | a + b + c | abc | ______ |
Solution: | First expression | Second expression | Product | | (i) | a | b + c + d | a(b + c + d) = a × b + a × c + a × d = ab + ac + ad | | (ii) | x + y – 5 | 5xy | p(6p2– 7p + 5) = p × 6p2 – p × 7p + p × 5 = 6p3 − 7p2 + 5p | | (iii) | p | 6p2 – 7p + 5 | p (6 p 2-7 p +5)= p× 6 p2 – p× 7 p + p×5 = 6 p3 – 7 p2 + 5 p | | (iv) | 4p2q2 | P2 – q2 | 4p2q2(p2 – q2) = 4p4q2 − 4p2q4 | | (v) | a + b + c | abc | abc(a + b + c ) = abc × a + abc × b + abc × c = a2bc + ab2c + abc2 |
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3. Find the product. (i) (a2) × (2a22) × (4a26) Solution:
(i) a2 x (2a22) x (4a26)
= (2 × 4) (a2 × a22 × a26)
= 8 × a2 + 22 + 26
= 8a50 (ii) (2/3xy) × (-9/10x2y2) Solution:
(ii) (2/3xy) × (-9/10x2y2)
=(2/3 × -9/10 ) ( x × x2 × y × y2 )
= -3/5x3y3 (iii) (-10/3pq3/) × (6/5p3q)
Solution:
(iii) (-10/3pq3) × (6/5p3q)
= ( -10/3 × 6/5 ) (p × p3× q3 × q)
= -4p4q4 (iv) (x) × (x2) × (x3) × (x4) Solution:
(iv) ( x) × (x2) × (x3) × (x4)
= x1 + 2 + 3 + 4
= x10 |
4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =1/2 Solution: (a) 3x(4x – 5) + 3
= 3x(4x) – 3x(5) + 3
=12x2 – 15x + 3 (i) Putting x = 3 in the equation we gets
= 12x2 – 15x + 3
= 12(32) – 15(3) +3
= 108 – 45 + 3
= 66 (ii) Putting x = 1/2 in the equation we get
12x2 – 15x + 3
= 12(1/2)2 – 15(1/2) + 3
= 12(1/4) – 15/2 + 3
= 3 – 15/2 + 3
= 6 – 15/2
= (12 – 15 )/2
= -3/2 (b) Simplify a (a2+ a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = – 1 Solution:
(b) a(a2 + a + 1) + 5 = (a × a2 + a × a + a × 1) + 5 = a3 + a2 + a + 5 (i) putting a = 0 in the equation we get
= 03 + 02 + 0 + 5 = 5
= 5 (ii) putting a=1 in the equation we get
= 13 + 12 + 1 + 5
= 1 + 1 + 1 + 5
= 8 (iii) Putting a = -1 in the equation we get
= (-1)3 + (-1)2 + (-1) + 5
= -1 + 1 – 1 + 5
= 4 |
5. (a) Add: p(p – q), q (q – r) and r ( r – p) Solution:
(a) p(p – q) + q(q – r) + r( r – p)
= (p2 – pq) + (q2 – qr) + (r2 – pr)
= p2 + q2 + r2 – pq – qr – pr (b) Add: 2x(z – x – y) and 2y(z – y – x) Solution:
(b) 2x(z – x – y) + 2y(z – y – x)
= (2xz – 2x2 – 2xy) + (2yz – 2y2 – 2xy)
= 2xz – 4xy + 2yz – 2x2 – 2y2 (c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l ) Solution:
(c) 4l(10n – 3m + 2l ) – 3l(l – 4m + 5n)
= (40ln – 12lm + 8l2) – (3l2 – 12lm + 15ln)
= 40ln – 12lm + 8l2 – 3l2 + 12lm -15ln
= 5l2 + 25ln (d) Subtract: 3a(a + b + c ) – 2b(a – b + c) from 4c(– a + b + c ) Solution:
(d) 4c(– a + b + c ) – (3a(a + b + c ) – 2b(a – b + c))
= (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – ( 2ab – 2b2 + 2bc ))
=-4ac + 4bc + 4c2 – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)
= -4ac + 4bc + 4c2 – 3a2 – 3ab – 3ac + 2ab – 2b2 + 2bc
= -7ac + 6bc + 4c2 – 3a2 – ab – 2b2 |
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