# Remainder Theorem

Before tackling Remainder theorem, you might want to revisit long division (also known
as synthetic division) and
. Just like numbers, polynomials can be divided through
by numbers.

The Remainder Theorem is a useful mathematical theorem that can be used to factorize
polynomials of any degree in a neat and fast manner.

The Remainder Theorem states that when you divide a polynomial P(x) by any
factor (x – a); which is not necessarily a factor of the polynomial; you’ll
obtain a new smaller polynomial and a remainder, and this remainder is the value
of P(x) at x = a, i.e P(a)

Remainder Theorem operates on the fact that a polynomial is completely divisible
once by its factor to obtain a smaller polynomial and a remainder of zero. This
provides an easy way to test whether a value a is a root of the polynomial
P(x).

For example, given a polynomial P(x), and also given that a is a root
of the polynomial, then when P(x) is divided by the factor (x – a),
the result should be a smaller polynomial P1(x) and a remainder
zero.

Below is an example that serves to prove the remainder theorem prove that x = 1 is a root of P(x),

solution:    which implies that x = 1 is a root of the polynomial P(x), and (x –
1)
is a factor of P(x)

Therefore if we were to synthetically divide through P(x) by (x – 1),
we should get a new smaller polynomial and a remainder of zero:

## Remainder Theorem Examples

The first step of syntheic division is to arrange the polynomial and the factor
in the format below. The factor is the divisor and is on the outside, while the
polynomial is the dividend and goes under the division bar. The next step is to divide the first term of the polynomial (the first term should
be the one with the highest power) by the x part of the factor. In this case, you
divide x3 by x, to get x2 which you then
write at the top of the division bar. Next you multiply the term that you wrote at the top by the entire divisor, in this
case you multiply x2 by (x – 1) to get (x3
x2)
which you then subtract from the polynomial under
the division bar as shown below. By subtracting, you get rid of the term with the largest exponent to reduce the
size of the polynomial which is a remainder as shown below: Now we have a new polynomial for which we repeat the second step. In this example,
we divide -5x2 by x and add the result which is -5xto
the term already at the top of the division bar Next we repeat the third step, we multiply this result which is on top of the division
bar (only the most recently added) by the entire divisor and then subtract
what you get from the divident polynomial, as shown in this step below: Once again we have a new polynomial under the division bar. As you have probably guessed, we repeat the second step by dividing the first term
of the polynomial by the x part of the divisor and add the result to whatever
is already on top of the division bar. Next we multiply the most recently added term by the entire divisor and again subtract
the result from the dividend polynomial. Now the subtraction step is repeated to obtain a new polynomial. The remainder here is zero because we have completely divided through P(x)
by the factor (x – 1). When you reach this step, you’re done, dividing any
further by zero would result in zero.

Now lets try to use the remainder theorem to find a value of P(-1), to see
what we result with. We already know that x = -1 is not a root of the polynomial
P(x)

The first step is the same as in the previous example; arrange the divisor and dividend
on the outside and inside of the division bar respectively. Next we perform the first division; the first term of the dividend by the x
term of the divisor Then comes the multiplication; the most recent result by the entire divisor and
subtract the result from the dividend The subtraction gets rid of the term with the largest power and yields a new dividend
under the division bar. We once again divide the first term of the dividend by the x term of the
divisor and add the result to whatever is already on top of the division bar. Next we repeat the multiplication step Then we subtract the result from the dividend to obtain a new polynomial dividend. Once again we divide the first term of the dividend by the x term of the
divisor. Multiplication once again: Then subtraction: and we’re left with a remainder. We can’t continue division any further since the
x term in the divisor has a higher exponent (x1) than the
x in the dividend polynomial (which in this case happens to be x0)
and we don’t want to end up with negative exponents so we stop there and say that
whatever is left is the remainder which in this case is -24.

Next we check to see if the value of P(-1) is the same as the remainder obtained
above   Thus we have proved the remainder theorem. You may try a few more values of x
as a way of practicing synthetic division of polynomials.

## Solving Polynomials Using the Remainder Theorem

From the section on polynomials, we know that the root of a polynomial P(x)
is defined as the value of x for which the polynomial is equal to zero.

Using Remainder Theorem, we can redefine a root as a value a for which the factor
(x – a) divides through the polynomial P(x) to get a remainder of
zero. In other words it divides through the polynomial completely.

The remainder theorem is especially useful in finding the roots of polynomials of
a large degree (4 or more). By finding the one root and using it to synthetically
divide through the polynomial, we’re able to obtain a smaller polynomial for which
we repeat the process until we find the last root.

The example below better explains the steps involved;

Given the polynomial P(x) below, find all its roots. Solution:

From the degree of the polynomial P(x) which is 5, we can tell that P(x)
will have 5 roots.

The first step is to use small numbers values of x to find which ones will
give a value of zero ie we need to find So we first attempt x = 0, we substitute for x = 0 into P(x)  since P(0) is not equal to zero, we can safely conclude that 0 is not a root
of P(x)

Next we move on to another number, let’s try substituting for x = -3  P(-3) = 0 which means that -3 is a root of the polynomial and that (x + 3)
is a factor of the polynomial. So now we can use synthetic division to get a smaller
polynomial We obtain a new polynomial, lets call this polynomial f(x) We then repeat the process for f(x). We try to find a root of f(x)
by trial and error

First we try x = 1  f(1) is not equal to zero, so we move on to another value of x

Try x = 3  f(3) = 0 which implies that x = 3 is a root of the polynomial and (x-3) is a factor.

Next we divide f(x) by (x – 3) We obtain a new polynomial g(x) and a remainder of zero We repeat the process one last time.

Since we can see that g(x) is a polynomial of degree 2 and that there only
addition operators in the polynomial, we can conclude the that one of the roots
of g(x) must be a negative.

So we try to substitute for x = -4   since g(4) is equal to zero, we conclude that x = -4 is a root of
the polynomial and (x+4) is a factor, which allows us to perform synthetic
division on g(x) from which we get the last factor of the P(x) as (x + 3) and thus
the root as x = -3.

So now we can rewrite P(x) as and the roots of P(x) are x = {4,-3,-3,3}

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