• # question_answer 10)                 Carry out the multiplication of the expressions in each of the following pairs:                 (i) $4p+q+r$                     (ii) $ab,\,a-b$                  (iii) $a+b,\,7{{a}^{2}}{{b}^{2}}$                                (iv) ${{a}^{2}}-9,\,4a$                 (v) $pq+qr+2p,\,0$

(i) $4p+q+r$                     $(4p)\times (q+r)=(4p)\times (q)+(4p)\times (r)$ $=4pq+4pr$     8959781310 (ii) $ab,\,a-b$ $(ab)\times (a-b)=(ab)\times (a)-(ab)\times (b)$ $={{a}^{2}}b-a{{b}^{2}}$                 (iii) $a+b,\,7{{a}^{2}}{{b}^{2}}$                                $(a+b)\times (7{{a}^{2}}{{b}^{2}})=(7{{a}^{2}}{{b}^{2}})\times (a+b)$  |by commutative law $(7{{a}^{2}}{{b}^{2}})\times (a)+(7{{a}^{2}}{{b}^{2}})\times (b)$ $=7{{a}^{3}}{{b}^{2}}+7{{a}^{2}}{{b}^{3}}$ (iv) ${{a}^{2}}-9,\,4a$ $({{a}^{2}}-9)\,\times (4a)=(4a)\times ({{a}^{2}}-9)$                     | by commutative law $=(4a)\times ({{a}^{2}})-(4a)\times (9)$ $=4{{a}^{3}}-36a$                 (v) $pq+qr+2p,\,0$ $(pq+qr+2p)\times (0)$ $=(0)\times (pq+qr+2p)$                           |by commutative law $=(0)\times (pq)+(0)\times (qr)+(0)$$\times (2p)$ $=0+0+0$                 = 0.