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# MSC # 27 – Transpose and Apply

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.

In our example, the divisor is **x ^{2} – 2x + 3** and we transpose (reverse the signs of) the coefficients of all the terms of the divisor after the leading term. We also separate by a remainder bar the last two terms of the dividend. We write the transposed figures, 2 and -3, below the original coefficients.

We obtain the first term of the quotient by dividing the first term of the dividend by the first term of the divisor **(the First by the First), 2x ^{4}/ x^{2} = 2x^{2}**.

We then multiply the coefficient of the first quotient term, **2**, by the transposed figures to get **4** and **-6**, and place it under the 2^{nd} and 3^{rd} terms of the dividend as shown in the illustration.

By adding the coefficient of the **x ^{3}** term of the dividend,

**-1**, and

**4**, we get

**3**, the coefficient of the x term in the answer.

Again, we multiply this **3** by the transposed figures to give **6** and **-9 **which we will place in the 3^{rd} and 4^{th} columns of our solution.

We the add **1** which is the coefficient of the **x ^{2}** term in the dividend,

**-6**and

**6**to get

**1**as the last answer figure. This is actually

**1x**

^{2}^{ }but we have to divide it by

**x**to get

^{2 }**1**.

Since the last answer figure is **1**, we need only to place the transposed figures in the fourth row as shown.

The first figure in the remainder is the sum of **9(x), – 9** and** 2** which is **2x,** while the last figure is **8 + – 3 = 5**.

So we have **(2x ^{4} – x^{3} + x^{2} + 9x + 8) ÷ (x^{2} – 2x + 3) = 2x^{2 } + 3x + 1 rem 2 x + 5**