# 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 3^{rd} 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 3^{rd} 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.