The Remainder Theorem

Remainder Theorem: The Remainder Theorem is useful for evaluating polynomials at a given value of x, though it might not seem so, at least at first blush. This is because the tool is presented as a theorem with a proof, and you probably don’t feel ready for proofs at this stage in your studies. Fortunately, you don’t “have” to understand the proof of the Theorem; you just need to understand how to use the Theorem.

Everyone loves to find a shortcut whether it involves driving directions or some other type of long task. Discovering a quicker and more efficient way to arrive at the same endpoint makes you feel good since you’ve most likely saved time, effort, and/or money. Math is filled with these types of shortcuts and one of the more useful ones is the remainder theorem.

The remainder theorem states that when a polynomial, f(x), is divided by a linear polynomial, x – a the remainder of that division will be equivalent to f(a). In other words, if you want to evaluate the function f(x) for a given number, a, you can divide that function by x – a, and your remainder will be equal to f(a).

It should be noted that the remainder theorem only works when a function is divided by a linear polynomial, which is of the form x + number or x – number. How does the remainder theorem save you time? Let’s find out.

Chinese Remainder Theorem

The remainder theorem is especially useful when it is paired with the synthetic division. If you remember, the synthetic division is an alternate method to quickly and easily divide polynomials instead of using long division. Also, remember that in synthetic division, the number in the bottom row in the last column on the right is the remainder. Thus, rather than plugging a value in and using order of operations, you can use synthetic division as a way to evaluate a polynomial for a given value.

Additionally, synthetic division and the remainder theorem can be used to determine if a value is a zero of a function. Hopefully, you remember that a zero of a function, by definition, is any point c, where f(c) = 0. Therefore, if you find a remainder of zero after performing synthetic division, the number listed out front, referred to as an in the definition above, evaluates to zero, or f(a) = 0.

Note, that you can use long division instead of synthetic division, but it’s almost always faster and easier to use synthetic division.

Remainder Theorem Calculator

Chinese Remainder Theorem Example

The Remainder Theorem starts with an unnamed polynomial p(x), where “p(x)” just means “some polynomial p whose variable is x“. Then the Theorem talks about dividing that polynomial by some linear factor x – a, where a is just some number. Then, as a result of the long polynomial division, you end up with some polynomial answer q(x) (the “q” standing for “the quotient polynomial”) and some polynomial remainder r(x).

As a concrete example of p, a, q, and r, let’s look at the polynomial p(x) = x³ – 7x – 6, and let’s divide by the linear factor x – 4 (so a = 4):

So we get a quotient of q(x) = x² + 4x + 9 on top, with a remainder of r(x) = 30.

You know, from the long division of regular numbers, that your remainder (if there is one) has to be smaller than whatever you divided by. In polynomial terms, since we’re dividing by a linear factor (that is, a factor in which the degree on x is just an understood “1”), then the remainder must be a constant value. That is, when you divide by “x – a“, your remainder will just be some number.

The Remainder Theorem then points out the connection between division and multiplication. For instance, since 12 ÷ 3 = 4, then 4 × 3 = 12. If you get a remainder, you do the multiplication and then add the remainder back in. For instance, since 13 ÷ 5 = 2 R 3, then 13 = 5 × 2 + 3. This process works the same way with polynomials. That is:

If   ^p(x) / _(x – a)  =  q(x)  with remainder  r(x),

then  p(x)  =  (x – a) q(x)  +  r(x).

(Technically, this “if-then” statement is the “Division Algorithm for Polynomials”. But the Algorithm is the basis for the Remainder Theorem.)

In terms of our concrete example: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Since  ^{(x^3 – 7x – 6)} / _{(x – 4)}   =   x² + 4x + 9   with remainder 30,

then   x³ – 7x – 6  =   (x – 4) (x² + 4x + 9)  +  30.

The Remainder Theorem says that we can restate the polynomial in terms of the divisor, and then evaluate the polynomial at x = a. But when x = a, the factor “x – a” is just zero! Then evaluating the polynomial at x = a gives us:

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p(a) = (a – a)q(a) + r(a)
= (0)q(a) + r(a)
= 0 + r(a)
= r(a)

But remember that the remainder term r(a) is just a number! So the value of the polynomial p(x) at
x = a is the same as the remainder you get when you divide that polynomial p(x) by x – a. In terms of our concrete example:

p(4) = (4 – 4)((4)² + 4(4) + 9) + 30
= (0)(16 + 16 + 9) + 30
= 0 + 30
= 30

But you gotta think: Okay, fine; the value of the polynomial p(x) at x = a is the remainder r(a) when you divide by x – a, but who wants to do the long division each time you have to evaluate a polynomial at a given value of x?!? You’re right; this would be overkill. Fortunately, that’s not what they really want you to do.

When you are dividing by a linear factor, you don’t “have” to use long polynomial division; instead, you can use synthetic division, which is much quicker. In our example, we would get:

Note that the last entry in the bottom row is 30, the remainder from the long division (as expected), and also the value of p(x) = x³ – 7x – 6 at x = 4. And that is the point of the Remainder Theorem: There is a simpler, quicker way to evaluate a polynomial p(x) at a given value of x, and this simpler way is not to evaluate p(x) at all but to instead do the synthetic division at that same value of x. Here are some examples:

• Use the Remainder Theorem to evaluate  f (x) = 6x³ – 5x² + 4x – 17 at x = 3.

First off, even though the Remainder Theorem refers to the polynomial and to long division and to restating the polynomial in terms of a quotient, a divisor, and a remainder, that’s not actually what I’m meant to be doing. Instead, I’m supposed to be doing synthetic division, using “3” as the divisor:

Since the remainder (the last entry in the bottom row) is 112, then the Remainder Theorem says that:

f (3) = 112.

• Using the Remainder Theorem, find the value of f (–5), for   f (x) = 3x⁴ + 2x³ + 4x.

I need to do the synthetic division, remembering to put zeroes in for the powers of x that are not included in the polynomial:

Since the remainder is 1605, then, thanks to the Remainder Theorem, I know that:

f (–5) = 1605.

• Use the Remainder Theorem to determine whether x = 2 is a zero of
  f (x) = 3x⁷ – x⁴ + 2x³ – 5x² – 4

For x = 2 to be a zero of f (x), then f (2) must evaluate to zero. In the context of the Remainder Theorem, this means that my remainder, when dividing by x = 2, must be zero:

The remainder is not zero. Then  x = 2 is not a zero of f (x).

• Use the Remainder Theorem to determine whether x = –4 is a solution of
  x⁶ + 5x⁵ + 5x⁴ + 5x³ + 2x² – 10x – 8 = 0

For x = –4 to be a solution of f (x) = x⁶ + 5x⁵ + 5x⁴ + 5x³ + 2x² – 10x – 8 = 0, it must be that  f (–4) = 0. In the context of the Remainder Theorem, this means that the remainder, when dividing by x = –4, must be zero:

The remainder is zero. Then x = –4 is a solution of the given equation.

Polynomial Remainder Theorem

Divide 15 by 6. What answer do you get? By using the simple division process, we find that the quotient is 2 and the remainder is 3. Hence, we write 15 = (2 x 6) + 3. Note: The remainder ‘3’ is less than the divisor ‘6’. On the other hand, when we divide 12 by 6, we get a quotient of 2 and remainder 0. In this case, we say that 6 is a factor of 12 OR 12 is a multiple of 6.

Dividing One Polynomial by Another

Let’s start by dividing a polynomial with a monomial as follows:

Dividend Polynomial: 2x³ + x² + x
Divisor Monomial: x
We have, (2x³ + x² + x) / x = (2x³)/x + x²/x + x/x = 2x² + x + 1

Observe that ‘x’ is common to each term of the dividend polynomial. We can also write the dividend as, x(2x² + x + 1). Hence, ‘x’ and ‘2x² + x + 1’ are factors of ‘2x³ + x² + x’.

Let’s look at another example.

Dividend polynomial: 3x² + x + 1
Divisor Monomial: x
We have, (3x² + x + 1) / x = (3x²)/x + x/x + 1/x = 3x + 1 + 1/x

Here, 1 is not divisible by x. So, we stop the division here and note that 1 is the remainder. Hence, we have 3x² + x + 1 = {x(3x + 1)} + 1. So, the result of the division is,

Quotient: 3x + 1
Remainder: 1

Note that since the remainder is not zero, x is not a factor of 3x² + x + 1.

Dividing a Polynomial by Any Non-zero Polynomial

Let’s do the following division: Dividend: x + 3x² – 1 and Divisor: 1 + x

Step 1.

We arrange the terms on the descending order of their degrees. Hence, we have dividend: 3x² + x – 1 and divisor: x

Step 2.

Divide the first term of the dividend by the first term of the divisor: 3x²/x = 3x. This is the first term of the quotient.

Step 3.

Multiply the divisor by the first term of the quotient and subtract this product from the dividend,
{3x² + x – 1} – {3x(x + 1)}
= {3x² + x – 1} – {3x² + 3x}
= – 2x – 1
‘– 2x – 1’ is the remainder.

Step 4.

Now, the new dividend is ‘– 2x – 1’ and the divisor is still the same. Repeat step 2 to get the next term of the quotient. Divide the first term of the new dividend by the first term of the divisor: (– 2x)/x = – 2 = the second term of the quotient

Step 5.

Multiply the divisor by the second term of the quotient and subtract this product from the new dividend,
(– 2x – 1) – {– 2(x + 1)} = (– 2x – 1) – (– 2x – 2)
= – 2x – 1 + 2x + 2 = 1 = Remainder

Remember: This process continues until the degree of the new dividend is less than the degree of the divisor.

Step 6.

Hence, we have quotient: 3x – 2 and remainder: 1. It is important to note here that, 3x² + x – 1 = (x + 1) (3x – 2) + 1 or Dividend = (Divisor × Quotient) + Remainder

Therefore, we can conclude that, 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.

Next, let’s try to find a link between the remainder and the dividend. Let’s find the value of the dividend polynomial at x = -1

p(x) = 3x² + x – 1
p(-1) = 3(- 1)² + (- 1) – 1
= 3 – 1 – 1
= 1 … which is the remainder!

So, the remainder of (3×2 + x – 1) / (x + 1) = Value of (3×2 + x – 1) at x = – 1 (or the zero of the divisor [x + 1]). In other words, the remainder obtained on dividing a polynomial by another is the same as the value of the dividend polynomial at the zero of the divisor polynomial. This brings us to the first theorem of this article.

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: p(x) is a polynomial with a degree greater than or equal to one. It is divided by a polynomial (x – a), where ‘a’ is a real number. Let’s assume that the quotient is q(x) and the remainder is r(x). So, we can write,

p(x) = (x – a)q(x) + r(x)

Now, the degree of (x – a) is 1. Also, since r(x) is the remainder, its degree is less than the degree of the divisor: (x – a). Therefore, the degree of r(x) = 0. In other words, r(x) I constant. Let’s call the constant ‘r’. Hence, for all values of ‘x’, r(x) = r. Therefore,

p(x) = (x – a) q(x) + r

Now, let’s find p(a) or the value of p(x) at x = a.

p(a) = (a – a) q(a) + r
= (0)q(a) + r
= r

We see that when a polynomial p(x) of a degree greater than or equal to one is divided by a linear polynomial (x – a), where a is a real number, then the remainder is r which is also equal to p(a). This proves the Remainder Theorem.

For example, check whether the polynomial q(t) = 4t³ + 4t² – t – 1 is a multiple of 2t+1.
Solution: q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t) with remainder zero. Let’s find the zero of the divisor polynomial:

2t + 1 = 0 Or  t = – ½

Next, let’s find the value of q(t) at t= – ½
q(- ½) = 4(- ½)³ + 4(- ½)² – (- ½) – 1
= 4(- 1/8) + 4(- ¼) + ½ – 1
= – ½ + 1 + ½ – 1 = 0.

Hence, we can conclude that the remainder obtained on dividing q(t) by 2t + 1 is 0. And, (2t + 1) is a factor of ‘4t³ + 4t² – t – 1’.