 # Multiplying by (x + 1)

“2x^3 + x^2 + 3x + k is exactly divisible by (x + 1). What is the value of k?”

Less than half of the competitors in the Seniors age group in the 3rd MATH-Inic Vedic Mathematics National Challenge held last April were able to get the correct answer to this question which can be easily solved in less than 3 seconds!

In this question, if x = 10, then (x + 1) = 11. And now let us recall the divisibility rule for 11 which states that if the difference of the sum of the alternating digits of a number is equal to zero or a multiple of 11, then the number is divisible by 11. Applying this rule to polynomials, if the difference between the sums of the coefficients of alternating terms of a polynomial is zero, then the polynomial is exactly divisible by (x + 1).

If 2x^3 + x^2 + 3x + k is exactly divisible by (x + 1) then (2 + 3) – (1 + k) = 0.

This is equivalent to saying that the sum of the coefficients of even and odd placed terms are equal for all multiples of (x + 1) or (2 + 3) = (1 + k). Then k = 4.

If the question is modified as: “What is the quotient when 2x^3 + x^2 + 3x + 4 is divided by (x + 1)?

From the previous question, we know that the quadrinomial is exactly divisible by (x + 1). How can we quickly determine the quotient?

By using the First by the First (2x^3/x = 2x^2) and the Last by the Last, (4/1 = 4), we can say that the quotient has the form (2x^2 + bx + 4).  Knowing that the digit sum of the coefficients of the quadrinomial is (2 + 1 + 3 + 4 = 10) and that of (x + 1) is (1 + 1 = 2), we can now apply the sub-Sutra the Product of the Sums is equal to the Sum the Product: the digit sum of the coefficients of the quotient x the digit sum of the coefficients of the divisor = digit sum of the coefficients of the dividend.  That is, 2 x DS(2x^2 + bx + 4) = 10.

Thus DS(2x^2 + bx + 4) = 5 and b = -1. Therefore, the quotient is (2x^2 – x + 4).

Is (x^3 + 3x + 4) divisible by (x + 1)?

The 3rd degree polynomial is missing an x^2 term so we must place a placeholder and write it as (x^3 + 0x^2 + 3x + 4). Here we can see that (1 + 3) = (0 + 4) which make the dividend exactly divisible by (x = 1).

Multiplication by (x +1) is discussed in Chapter 25 while division by (x + 1) is explained in Chapter 28 of our book Algebra Made Easy as Arithmetic.

Applications of The First by the First and the Last by the Last (FFLL) and the Product of the Sums is equal to the Sum of the Products (PSSP) are given in chapters 2, 3, 20 and 21 of the same book.