# Division by (x + 1) Part II

**2x^3 + 5x^2 + 7x + k is exactly divisible by (x + 1). What is the value of k?**

Only a few got the correct answer to this question given in the **3 ^{rd} Math-Inic Vedic Mathematics National Challenge** last April 2023.

But as we have noted in the past articles in this newsletter, if we assign the value of 10 to x, then (x + 1) = 11 and the arithmetic rules applicable to 11 may be applied to (x + 1).

Therefore, if a polynomial is exactly divisible by (x + 1) then the alternating sum of the coefficients of the polynomial is zero. Note that k here is the coefficient of x^0.

So, we have (2 + 7 ) = (5 + k) or k = 4.

An interesting question which may be be asked here is: **“If 2x^3 + 5x^2 + 7x + k is exactly divisible by (x + 1). What is the quotient?**”

Here we have determined that k = 4.

Thus, by applying the** First by the First and the Last by the Last, **we have the first and last terms of the quotient as (2x^3/x = 2x^2) and (4/1 = 4).

The quotient then must be in the form (2x^2 + bx + 4).

Applying the **Product of the Sums is the Sum of the Product **to(2x^2 + bx + 4) ( x = 1) = (2x^3 + 5x^2 + 7x + 4), we have

(2 + b + 4) ( 1 + 1) = (2 + 5 + 7 + 4)

Or

(6 + b)(2) = 18;

6 + b = 9

b = 3

We can check that (2x^2 + 3 x + 4) (x + 1) = (2x^3 + 5x^2 + 7x + 4)

What if we have not determined the value of k first? Can we get the quotient?

Yes, of course.

By applying division by (x + 1) and the **First by the First**, we can immediately say that the first term of the quotient is **2x^2**,

Next, we subtract 2x^2 from the next term of the polynomial 5x^2 to get 3x^2. If we decrease the exponent of this difference, we will get **3x**, the next term of the quotient.

Then, if we subtract 3x from the third term of the polynomial which is 7x, we will get 4x and if we decrease its exponent by 1 the result is **4**, the last term of the quotient.

Now if we subtract 4 from k, we should get zero since the polynomial is exactly divisible by (x + 1). Therefore **k = 4**.