What is the quotient when (x^4 + 3x^3 + 2x^2 + 5x + 5) is divided by (x + 1)?
- x^3 + 3x^2+ 5x + 5
- x^3 + 2x^2 + 5x + 5
- x^3 + 3x^2 + 5
- x^3 + 2x^2+ 5
- x^3 + 5
I’m not sure if this type of question has been given in any of the previous Vedic Math competitions but this would be a good one and is related to another post in this issue about division by 11.
Just by looking at the coefficients of the alternating terms of the polynomial dividend, we know that it is exactly divisible by (x + 1): (1 + 2 + 5) = (3 + 5).
Using the First by the First (x^4 ÷ x = x^3) and the Last by the Last (5 ÷ 1 = 5) is not much help because all the choices begin with x^3 and end in 5.
We could, of course, perform the actual division using the special technique we discussed in Chapter 28 of our book Algebra Made Easy as Arithmetic (pp 73-74) as follows:
- The first term of the quotient is the first term of the dividend with the exponent decreased by 1: (x^4 ÷ x = x^3)
(x^4 + 3x^3 + 2x^2 + 5x + 5) ÷ (x + 1) = x^3 +
- The next term of the quotient (x2 term) is obtained by subtracting the first term of the quotient from the second term of the dividend, (3x^3 – x^3 = 2x^3) and decreasing the exponent of the result by 1.
(x^4 + 3x^3 + 2x^2 + 5x + 5) ÷ (x + 1) = x^3 + 2x^2 +
- Apply step 2 to the succeeding terms of the quotient and the dividend, (2x^2 – 2x^2 = 0x^2).
(x^4 + 3x^3 + 2x^2 + 5x + 5) ÷ (x + 1) = x^3 + 2x^2 + 0x +
- Repeat step 3: (5x – 0x = 5x)
(x^4 + 3x^3 + 2x^2 + 5x + 5) ÷ (x + 1) = x^3 + 2x^2 + 0x + 5
- Finally, we have 5 – 5 = 0, which means no remainder and confirms that that the polynomial is exactly the divisible by (x + 1).
We can see that choice d is the correct quotient.
The easiest solution here, however, will be to use the Sutra, the Product of the Sums is the Sum of the Product. The product of the digit sum of the coefficients of the quotient and the digit sum of (x + 1) which is 2 is equal to the digit sum of the coefficients of the dividend which is 8 + 8 = 16.
We can also see that the digit sum of the coefficients of the quotient is equal to the coefficients of the odd placed or even placed terms which in this case is 8.
Only case d satisfies this condition.