“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 2nd International Vedic Mathematics Olympiad can be easily solved using the technique discussed in out 27th 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
| 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 (the First by the First), 2x3 ÷ x2 = 2x; write this down in the answer line.
- Multiply 2x by the transposed digits and place the results (coefficients only) under the 2nd and 3rd terms of the dividend.
- Add 5x2 and -4(x2) to get x2 which when divided by x2 will give 1, which we will write in the answer row.
- Multiply 1 by the transposed digits and write the results under the 3rd and 4th 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 + 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 …