NCERT solutions for Exercise 9.3 Class 8 Maths Chapter 9 Algebraic Expressions And Identities

1.Carry out the multiplication of the expressions in each of the following pairs:
(i) \(4p, q + r\)

(ii) \(ab, a – b\)

(iii) \(a + b, 7a^2b^2\)

(iv) \(a^2 – 9, 4a\)

(v) \(pq + qr + rp, 0\)


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Answer :

Multiplication of the expressions are as follows:
(i) \(4p \times (q + r) = (4p \times q) + (4p \times r) = 4pq + 4pr\)

(ii) \(ab, a – b = ab \times (a – b) = (ab \times a) – (ab \times b) = a^2b – ab^2\)

(iii) \((a + b) \times 7a^2b^2 = (a \times 7a^2b^2) + (b \times 7a^2b^2) = 7a^3b^2 + 7a^2b^3\)

(iv) \((a2 – 9) \times 4a = (a^2 \times 4a) – (9 \times 4a) = 4a^3 – 36a\)

(v) \((pq + qr + rp) \times 0 = 0\)

2. Complete the table.


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Answer :

Products of the given expressions in the table are done as follows:
.(i) \(a \times (b + c + d) = (a \times b) + (a \times c) + (a \times d) = ab + ac + ad\)

(ii) \((x + y – 5) (5xy) = (x \times 5xy) + (y \times 5xy) – (5 \times 5xy) = 5x^2y + 5xy^2 – 25xy\)

(iii) \(p \times (6p^2 – 7p + 5) = (p \times 6p^2) – (p \times 7p) + (p \times 5) = 6p^3 – 7p^2 + 5p\)

(iv) \(4p^2q^2 \times (p^2 – q^2) = 4p^2q^2 \times p^2 – 4p^2q^2 \times q^2 = 4p^4q^2 – 4p^2q^4\)

(v) \((a + b + c) \times (abc) = (a \times abc) + (b \times abc) + (c \times abc) = a^2bc + ab^2c + abc^2\)

Therefore, the completed table is as follows:


3.Find the products
(i)\((a^2)\times(2a^{22})\times(4a^{26})\)

(ii)\((\frac{2}3xy)\times(-\frac{9}{10}x^2y^2)\)

(iii)\((-\frac{10}3pq^3)\times(\frac{6}{5}p^3q)\)

(iv)\(x\times x^2\times x^3\times x^4\)


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Answer :

(i)\((a^2)\times(2a^{22})\times(4a^{26})\)

\(=(1\times 2\times 4)\times a^{2+22+26}=8a^{50}\)

(ii)\((\frac{2}3xy)\times(-\frac{9}{10}x^2y^2)\)

\(=(\frac{2}3\times (-\frac{9}{10}))\times x^{(1+2})y^{(1+2)}\)

(iii)\((-\frac{10}3pq^3)\times(\frac{6}{5}p^3q)\)

\(=(-\frac{10}3\times\frac{6}{5})\times p^{(1+3)}q^{(1+3)}\)

\(=-4p^4q^4\)

(iv)\(x\times x^2\times x^3\times x^4\)

\(= x^{(1+2+3+4)}=x^{10\)

4.(a) Simplify: \(3x(4x – 5) + 3\) and find its values for (i) x = 3 (ii) x = \(\frac { 1 }{ 2 }\).
(b) Simplify: \(a(a^2 + a + 1) + 5\) and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1


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Answer :

(a) Given the expression: \(3x(4x – 5) + 3 = 4x \times 3x – 5 \times 3x + 3 = 12x^2 – 15x + 3\)
(i) So for x = 3, we have

\(12 \times (3)2 – 15 \times 3 + 3 = 12 \times 9 – 45 + 3 = 108 – 42 = 66\)

(ii)For x=\(\frac{1}2\), we have:

\(\Rightarrow 12(\frac{1}2)^2 - 15(\frac{1}2)+3\)

\(\Rightarrow 12\times(\frac{1}4) - \frac{15}2+3\)

\(\Rightarrow 3 - \frac{15}2+3\)

\(\Rightarrow \frac{6-15+6}2=\frac{12-15}2=-\frac{3}2\)

(b) We have \(a(a^2 + a + 1) + 5\)

\(= (a^2 \times a) + (a \times a) + (1 \times a) + 5\)

\(= a^3 + a^2 + a + 5\)

(i) For a = 0, we have:

\(= (0)^3 + (0)^2 + (0) + 5 = 5\)

(ii) For a = 1, we have:

\(= (1)^3 + (1)^2 + (1) + 5 = 1 + 1 + 1 + 5 = 8\)

(iii) For a = -1, we have:

\(= (-1)^3 + (-1)^2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4\)

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) \(Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)\)


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Answer :

(a) \(p(p – q) + q(q – r) + r(r – p)\)

=\( (p \times p) – (p \times q) + (q \times q) – (q \times r) + (r \times r) – (r \times p)\)[Opening the brackets]

= \(p^2 – pq + q^2 – qr + r^2 – rp\)

=\( p^2 + q^2 + r^2 – pq – qr – rp\)

(b) \(2x(z – x – y) + 2y(z – y – x)\)

= \((2x \times z) – (2x \times x) – (2x \times y) + (2y \times z) – (2y \times y) – (2y \times x)\)[Opening the brackets]

= \(2xz – 2x^2 – 2xy + 2yz – 2y^2 – 2xy\)

\(= -2x^2 – 2y^2 + 2xz + 2yz – 4xy\)

\(= -2x^2 – 2y^2 – 4xy + 2yz + 2xz\)

(c) \(4l(10n – 3m + 2l) – 3l(l – 4m + 5n)\)

\(= (4l \times 10n) – (4l \times 3m) + (4l \times 2l) – (3l \times l) – (3l \times -4m) – (3l ×\times 5n)\)[Opening the brackets]

\(= 40ln – 12lm + 8l^2 – 3l^2 + 12lm – 15ln\)

\(= (40ln – 15ln) + (-12lm + 12lm) + (8l^2 – 3l^2)\)

\(= 25ln + 0 + 5l^2\)

\(= 25ln + 5l^2\)

\(= 5l^2 + 25ln\)

(d)\( [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]\)

\(= (-4ac + 4bc + 4c^2) – (3a^2 + 3ab + 3ac – 2ab + 2b^2 – 2bc)\)[Opening the brackets]

\(= -4ac + 4bc + 4c^2 – 3a^2 – 3ab – 3ac + 2ab – 2b^2 + 2bc\)

\(= -3a^2 – 2b^2 + 4c^2 – ab + 6bc – 7ac\)




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