By Xavier Roche, oth.

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**Sample text**

6 + 1 = 7. Thus 32 X 24 = 768 We can write it as follows 32 24 ¯¯¯¯ 668 1 ¯¯¯¯ 768. , 1 is placed under the previous digit 3 X 2 = 6 and added. After sufficient practice, you feel no necessity of writing in this way and simply operate or perform mentally. 5 28 X 35. Step (i) : 8 X 5 = 40. 0 is retained as the first digit of the answer and 4 is carried over. Step (ii) : 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried over 4 to 34. Now the result is 34 + 4 = 38. Now 8 is retained as the second digit of the answer and 3 is carried over.

15x3 v) 3 X 0 = 0 Hence the product is 15x3 + 38x2 + 59x + 42 Find the products using urdhva tiryagbhyam process. 1) 25 X 16 2) 32 X 48 3) 56 X 56 4) 137 X 214 5) 321 X 213 6) 452 X 348 7) (2x + 3y) (4x + 5y) 9) (6x2 + 5x + 2 ) (3x2 + 4x +7) 8) (5a2 + 1) (3a2 + 4) 10) (4x2 + 3) (5x + 6) Urdhva – tiryak in converse for division process: As per the statement it an used as a simple argumentation for division process particularly in algebra. Consider the division of (x3 + 5x2 + 3x + 7) by (x – 2) process by converse of urdhva – tiryak : i) x3 divided by x gives x2 .

Hence second term of Q is 7x. x3 + 5x2 + 3x + 7 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x–2 gives Q = x2 + 7x + - - - - - - - - iii)We now have – 2 X 7x = -14x. But the 3rd term in the dividend is 3x for which ‘17x more’ is required since 17x – 14x =3x. Now multiplication of x by 17 gives 17x. , 17 multiplied by –2 gives 17X–2 = -34 but the relevant term in dividend is 7. So 7 + 34 = 41 ‘more’ is required. As there no more terms left in dividend, 41 remains as the remainder. x3 + 5x2 + 3x + 7 ________________ x–2 gives Q= x2 + 7x +17 and R = 41.