# Remainder Theorem

## Remainder Theorem or Little Bézout’s theorem

### Divisor is linear

**Theorem:** When the polynomial \( p(x) \) is divided by the linear polynomial \( ax + b \), the remainder is equal to \( p\left(\frac{-b}{a}\right) \).

**Proof:** By Euclid’s Lemma for polynomials, when \( p(x) \) is divided by \( ax + b \), there exist polynomials \( q(x) \) and \( r(x) \) such that \( p(x) = q(x)(ax + b) + r(x) \).

Since the degree of the divisor is 1, the degree of \( r(x) \) must be zero, i.e., \( r(x) \) must be a constant. Let \( r(x) = R \), then \( p(x) = q(x)(ax + b) + R \) .....[1]

Since equation (1) is an identity with respect to \( x \), let us put \( x = \frac{-b}{a} \) (which is nothing but the root of \( ax + b = 0 \)) into equation 1. We get,

Thus the remainder \( R = p\left(\frac{-b}{a}\right) \).

Example: If \( p(x) = x^3 + 2x^2 - 3x + 1 \), then the remainder when \( p(x) \) is divided by \( x - 1 \) is \( p(1) = 1 \).

### Divisor is quadratic

Consider the following two problems to understand how to obtain the remainder when the divisor is quadratic. Only thing is that it should be in factorized form. We handle the cases of distinct factors and repeated factors in different ways.

Distinct Factors

Find the remainder when the polynomial \( p(x) \) is divided by \( (x - \alpha)(x - \beta) \) where \( \alpha \neq \beta \).

**Solution:** The divisor has a degree equal to 2. The remainder may have a degree equal to 0 or 1 (the degree of the remainder is at least one unit less than the degree of the divisor). Let us assume that the remainder is \( ax + b \) which has degree 1 when \( a \neq 0 \) but degree 0 when \( a = 0 \).

Thus \( p(x) = q(x)(x - \alpha)(x - \beta) + ax + b \) .....(1)

Since equation 1 is an identity, plugging \( x = \alpha \) and \( x = \beta \) we get

Solving (2) and (3) for \( a \) and \( b \) we get,

Hence the remainder \( r(x) = ax + b \) is

Which is clearly true only when \( \alpha \neq \beta \).

Repeated Factors

Find the remainder when polynomial \( p(x) \) is divided by \( (x - \alpha)^2 \).

**Solution:** The degree of the divisor is 2, thus let the remainder be \( ax + b \). We cannot solve this problem in the same way we solved the previous problem, as both factors of the divisor have the same root. By Euclid’s Lemma,

By putting \( x = \alpha \) we get \( p(\alpha) = a\alpha + b \) .....(1)

To find \( a \) and \( b \) we need one more equation. Using differentiation we may obtain a second equation.

Differentiating both sides with respect to \( x \) we get

Now put \( x = \alpha \) in this equation to obtain \( p'(\alpha) = a \) .....(2)

Thus from equation (2) and (1) we get \( b = p(\alpha) - p'(\alpha)\alpha \) and \( a = p'(\alpha) \)

Therefore the remainder is \( p'(\alpha)x + p(\alpha) - p'(\alpha)\alpha \)