# 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 +5x^{2} + …; B) 1− x + 3/4 x^{2} +…; C) 1 + x – 4/5 x^{2} +…; D)1− x + 5/4 x^{2} +…; E) 1+ x + 1/4 x^{2} + …”

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.

2x^{3} + 5x^{2} + 12x + 12 ÷ x^{2} + 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.

x^{2} + 2x + 3 | 2x^{3} + 5x^{2} +| 12x + 12

-2 -3 | – 4 | – 6

|_________________|__- 2_____-3_

| 2x + 1 | 4x + 9

- 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.
- 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.
- Divide the first term of the dividend by the first term of the divisor (
), 2x*the First by the First*^{3}÷ x^{2}= 2x; write this down in the answer line. - Multiply
**2x**by the transposed digits and place the results (coefficients only) under the 2^{nd}and 3^{rd}terms of the dividend. - Add 5x
^{2}and -4(x^{2}) to get x^{2}which when divided by x^{2}will give 1, which we will write in the answer row. - Multiply 1 by the transposed digits and write the results under the 3
^{rd}and 4^{th}terms of the dividend. - Add all the figures to the left of the remainder bar to get the remainder:
- (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 + x^{2} )and use this form as the divisor. The dividend can be expressed as 4 + 0x + 0x^{2} + 0x^{3 }+ …

4 + 4x + x^{2} | 4 + 0x + 0x^{2} + 0x^{3 }+ …

-4 – 1 | – 4 -1

| 4 1

_____________________________

1 – x + 3/4 x^{2} …