# Degree of the Remainder Polynomial

Why should the degree of the remainder polynomial be *less than* that of the divisor polynomial? Think of a simple analogy: the division of two numbers. Suppose that you divide 27 by 7. You have

\[27 = 7\left( 3 \right) + 6\]

That is, the remainder is 6, which is less than the divisor 7. It does not make sense for the remainder to be more than 7, because if the remainder were more than 7, the remainder itself could have *accommodated* additional 7s in itself. For example, suppose that someone writes

\[27 = 7\left( 2 \right) + 13\]

and says that the remainder in this case is 13. Even though this relation is correct, we want the quotient to be *as high as it can go*. In this case, the quotient is 2, while as we saw above, it should actually be 3:

\[27 = 7\left( 2 \right) + 7 + 6\\\,\qquad = 7\left( 3 \right) + 6\]

An analogous reasoning applies in the case of division of polynomials. Intuitively speaking, we want the quotient to be of as high a degree as possible, so *the remainder will always have a degree less than the divisor*. Let us state this formally. We denote the dividend and the divisor polynomials by \(a\left( x \right)\) and \(b\left( x \right)\), and the quotient and remainder polynomials by \(q\left( x \right)\) and \(r\left( x \right)\). Thus, we have

\[a\left( x \right) = b\left( x \right)q\left( x \right) + r\left( x \right)\]

And,

\({\rm{Deg}}\left( {r\left( x \right)} \right) < {\rm{Deg}}\left( {b\left( x \right)} \right)\)

In the case of linear polynomials as divisors, we can therefore conclude that the remainders will always be polynomials of degree 0, that is, *constants*.

If you are still not convinced that the degree of the remainder should be less than that of the divisor, try to think of a scenario where the remainder has a degree greater than that of the divisor, and then observe that the remainder itself can be further divided by the divisor, which defeats the whole purpose of it being a remainder.