Remainder Theorem

Let us consider two numbers 15 and 6. You know that when we divide 15 by 6, we get the quotient 2 and remainder 3. Do you remember how this fact is expressed? We write 15 as

                                       15 = (6 × 2) + 3
We observe that the remainder 3 is less than the divisor 6. Similarly, if we divide12 by 6, we get

                                      12 = (6 × 2) + 0
What is the remainder here? Here the remainder is 0, and we say that 6 is a factor of 12 or 12 is a multiple of 6.
Now, the question is: can we divide one polynomial by another? To start with, let us try and do this when the divisor is a monomial. So, let us divide the polynomial 2x3 + x2   + x by the monomial x.
We have (2x3 + x2 + x) ÷ x =

                                           = 2x2 + x + 1
In fact, you may have noticed that x is common to each term of 2x3 + x2 + x. So we can write

                                    (2x3 + x2 + x as x(2x2 + x + 1).
We say that x and 2x2 + x + 1 are factors of 2x3 + x2 + x, and 2x3 + x2 + x is a multiple of x as well as a multiple of 2x2 + x + 1.
Consider another pair of polynomials 3x2 + x + 1 and x.
Here,                          (3x2 + x + 1) ÷ x = (3x2 ÷ x) + (x ÷ x) + (1 + x).
We see that we cannot divide 1 by x to get a polynomial term. So in this case we stop here, and note that 1 is the remainder. Therefore, we have

                                   3x2 + x + 1 = {x × (3x + 1) } + 1
In this case, 3x + 1 is the quotient and 1 is the remainder. Do you think that x is a factor of 3x2 + x + 1? Since the remainder is not zero, it is not a factor.
Now let us consider an example to see how we can divide a polynomial by any non-zero polynomial.
i.e.,    Dividend = (Divisor × Quotient) + Remainder
In general, if p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that:

                                        p(x) = g(x)q(x) + r(x),
Where r (x) = 0 or degree of r(x) < degree of g(x). Here we say that p(x) divided by g(x), gives q(x) as quotient and r(x) as remainder.
In the example above, the divisor was a linear polynomial. In such a situation, let us see if there is any link between the remainder and certain values of the dividend.
Remainder Theorem: Let p(x ) be any polynomial of degree greater than or equal  to one and let a be any real number. If p (x) is divided by the linear polynomial x – a, then the remainder is p (a).
Proof: Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r (x), i.e.,

                                         p(x) = (x – a) q(x) + r (x)
Since the degree of x – a is 1 and the degree of r(x) is less than the degree of x – a, the degree of r(x) = 0. This means that r(x) is a constant, say r.
So, for every value of x, r(x) = r.
Therefore,   p(x) = (x – a) q(x) + r
In particular, if x = a, this equation gives us

                                     p(a) = (a – a) q(a) + r   = r,

Which proves the theorem?
Let us use this result in another example.