# 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) $$pq + qr + rp, 0$$

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.

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)$$x\times x^2\times x^3\times x^4$$

(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

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

(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]

$$= -4ac + 4bc + 4c^2 – 3a^2 – 3ab – 3ac + 2ab – 2b^2 + 2bc$$

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