NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions
and Identities Ex 9.3 are the part of NCERT Solutions for Class 8 Maths. Here
you can find the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic
Expressions and Identities Ex 9.3.
Ex 9.3 Class 8 Maths Question 1.
Carry out the multiplication of the expressions in each of the following pairs.(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a2b2
(iv) a2– 9, 4a
(v) pq + qr + rp, 0
Solution:
(i) 4p × (q +
r) = 4p × q + 4p × r
= 4pq + 4pr
(ii) ab × (a –
b) = ab × a – ab × b
= a2b
– ab2
(iii) (a + b) ×
7a2b2 = a × 7a2b2 + b ×
7a2b2
= 7a3b2 +
7a2b3
(iv) (a2 –
9) × 4a = a2 × 4a – 9 × 4a
= 4a3 –
36a
(v) (pq + qr +
rp) × 0 = 0
Ex 9.3 Class 8 Maths Question 2.
Complete the table.|
S. No. |
First Expression |
Second Expression |
Product |
|
(i) |
a |
b + c + d |
– |
|
(ii) |
x + y – 5 |
5xy |
– |
|
(iii) |
p |
6p2 – 7p + 5 |
– |
|
(iv) |
4p2q2 |
p2 – q2 |
– |
|
(v) |
a + b + c |
abc |
– |
Solution:
(i) a × (b + c + d) = (a × b) + (a × c) + (a × d) = ab + ac + ad
(ii) (x + y – 5) (5xy) = (x × 5xy) + (y × 5xy) – (5 × 5xy) = 5x2y +
5xy2 – 25xy
(iii) p × (6p2 – 7p + 5) = (p × 6p2) – (p × 7p) + (p
× 5) = 6p3 – 7p2 + 5p
(iv) 4p2q2 × (p2 – q2) =
4p2q2 × p2 – 4p2q2 ×
q2 = 4p4q2 – 4p2q4
(v) (a + b + c) × (abc) = (a × abc) + (b × abc) + (c × abc) = a2bc +
ab2c + abc2
Completed table is as follows:
|
S. No. |
First Expression |
Second Expression |
Product |
|
(i) |
a |
b + c + d |
ab + ac + ad |
|
(ii) |
x + y – 5 |
5xy |
5x2y + 5xy2 –
25xy |
|
(iii) |
p |
6p2 – 7p + 5 |
6p3 – 7p2 +
5p |
|
(iv) |
4p2q2 |
p2 – q2 |
4p4q2 – 4p2q4 |
|
(v) |
a + b + c |
abc |
a2bc + ab2c + abc2 |
Ex 9.3 Class 8 Maths Question 3.
Find the products.
Solution:
Ex 9.3 Class 8 Maths Question 4.
(a) Simplify: 3x(4x – 5) + 3 and find its values for (i) x = 3 (ii) x = ½.(b) Simplify: a(a2 + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1
Solution:
(a) We have 3x(4x – 5) + 3 = 3x × 4x – 3x × 5 + 3 = 12x2 –
15x + 3
(i) For x = 3, 12x2 – 15x + 3
= 12 × (3)2 – 15 × 3 + 3 = 12 × 9 – 45 + 3 = 108 – 42 = 66
= (a2 × a) + (a × a) + (1 × a) + 5
= a3 + a2 + a + 5
(i) For a = 0, a3 + a2 + a + 5
= (0)3 + (0)2 + (0) + 5 = 5
(ii) For a = 1, a3 + a2 + a + 5
= (1)3 + (1)2 + (1) + 5 = 1 + 1 + 1 + 5 = 8
(iii) For a = -1, a3 + a2 + a + 5
= (-1)3 + (-1)2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4
Ex 9.3 Class 8 Maths Question 5.
(a) Add: p(p – q), q(q – r) and r(r – p)(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
Solution:
(a) p(p – q) + q(q – r) + r(r – p)
= (p × p) –
(p × q) + (q × q) – (q × r) + (r × r) – (r × p)
= p2 –
pq + q2 –
qr + r2 –
rp
= p2 +
q2 +
r2 –
pq – qr – rp
(b) 2x(z –
x – y) + 2y(z – y – x)
= (2x × z)
– (2x × x) – (2x × y) + (2y × z) – (2y × y) – (2y × x)
= 2xz – 2x2 –
2xy + 2yz – 2y2 –
2xy
= -2x2 –
2y2 –
4xy + 2yz + 2xz
(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
= (4l ×
10n) – (4l × 3m) + (4l × 2l) – (3l × l) – (3l × -4m) – (3l × 5n)
= 40ln –
12lm + 8l2 –
3l2 +
12lm – 15ln
= (40ln –
15ln) + (-12lm + 12lm) + (8l2 –
3l2)
= 25ln + 0
+ 5l2
= 5l2 +
25ln
(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
= -3a2 –
2b2 +
4c2 –
ab + 6bc – 7ac
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