MSC #27: Transpose and Apply

“Given that |x|< 2, what are the first three terms, in ascending powers of x, for the expansion of 4 /(2+ x)²?

A) 1− 4x +5x² + …; B) 1− x + 3/4 x² +…; C) 1 + x – 4/5 x² +…; D)1− x + 5/4 x² +…;  E) 1+ x + 1/4 x² + …”

This question, given in the senior level of the 2^nd International Vedic Mathematics Olympiad can be easily solved using the technique discussed in out 27^th of 30 MATH-Inic Specials for Christmas series.

The Vedic Mathematics Sutra or word formula “Transpose and Apply” has a much wider application than the “transposition” which we know in Algebra as a short cut of the “golden rule’ of Algebra: do unto one side of the equation, what you do on the other side.

We first introduced this Sutra in MSC #12, Base Division. We will now apply this technique in polynomial division where the divisor is a quadratic expression.

2x³ + 5x² + 12x + 12 ÷ x² + 2x + 3. 

In using this technique, except for the first term, all others are transposed or expressed in their negative form. To keep the solution simple, only the coefficients are used.

 x² + 2x + 3    | 2x³ + 5x² +| 12x + 12

         -2   -3      |      – 4     | –  6 

   |_________________|__- 2_____-3_

                        | 2x  +  1      | 4x  + 9

1. Transpose all terms in the divisor except the first, in this case, the last two terms and write down their coefficients in the next row.
2. Place a remainder bar to separate the last two terms of the dividend. This is because there are two terms after the first in the divisor.
3. Divide the first term of the dividend by the first term of the divisor (the First by the First), 2x³ ÷ x² = 2x; write this down in the answer line.
4. Multiply 2x  by the transposed digits and place the results (coefficients only) under the 2^nd and 3^rd terms of the dividend.
5. Add 5x² and -4(x²) to get x² which when divided by x² will give 1, which we will write in the answer row.
6. Multiply 1 by the transposed digits and write the results under the 3^rd and 4^th terms of the dividend.
7.  Add all the figures to the left of the remainder bar to get the remainder:
8. (12 – 6 – 2) = 4x and (12 – 3) = 9.  

In the IVMO problem, since it is given that |x| < 2, (2 + x) cannot be 0 or negative. We can then expand (2 +x)² as (4 + 4x + x² )and use this form as the divisor. The dividend can be expressed as 4   + 0x   + 0x²   +  0x³  +  …

4 + 4x + x²  |   4   + 0x   + 0x²   +  0x³  +  …

      -4    – 1  |          – 4        -1

                   |                       4          1

_____________________________                       

1  –   x   + 3/4 x² …