Explainer: Polynomial Long Division with Remainder

In this explainer, we will learn how to find the quotient and remainder when polynomials are divided, including the case when the divisor is irreducible.

As with integers, dividing a polynomial ๐‘(๐‘ฅ) (the dividend) by a divisor ๐‘‘(๐‘ฅ) gives a quotient ๐‘ž(๐‘ฅ) and a remainder ๐‘Ÿ(๐‘ฅ).

Recall that a polynomial is a finite sum of monomials which has nonnegative exponents. Hence, expressions of the forms 2๐‘ฅ+2, ๐‘ฅ๐‘ฆโˆ’10๐‘ฅ๐‘ฆ+๐‘ฅ๏Šจ๏Šจ๏Šจ, and 8 are all examples of polynomials, whereas expressions such as โˆš๐‘ฅ, 3๐‘ฅ๏Šฑ๏Šฏ, and 3๐‘ฅ๏Šจ are not polynomial expressions. In this explainer, we will focus on dividing polynomials of one variable.

Usually when considering the division of polynomials, we write ๐‘(๐‘ฅ)๐‘‘(๐‘ฅ) rather than ๐‘(๐‘ฅ)รท๐‘‘(๐‘ฅ). We can think of long division as finding polynomials ๐‘ž and ๐‘Ÿ such that ๐‘(๐‘ฅ)๐‘‘(๐‘ฅ)=๐‘ž(๐‘ฅ)+๐‘Ÿ(๐‘ฅ)๐‘‘(๐‘ฅ) and we say that that the division yields a quotient ๐‘ž(๐‘ฅ) and a remainder ๐‘Ÿ(๐‘ฅ).

We can write this equivalently as a multiplication equation as follows:

However, not all equations in this form are division equations. For example, consider the equation 2๐‘ฅ+7๐‘ฅโˆ’4=(๐‘ฅโˆ’3)ร—(๐‘ฅโˆ’1)+๏€น๐‘ฅ+11๐‘ฅโˆ’7๏….๏Šจ๏Šจ

This can be written as 2๐‘ฅ+7๐‘ฅโˆ’4๐‘ฅโˆ’3=(๐‘ฅโˆ’1)+๐‘ฅ+11๐‘ฅโˆ’7๐‘ฅโˆ’3๏Šจ๏Šจ but it does not qualify as division by ๐‘ฅโˆ’3 because, just as with integer division, the remainder must always have a lower degree than the divisor.

A correct division equation, in this case, would be 2๐‘ฅ+7๐‘ฅโˆ’4๐‘ฅโˆ’3=(2๐‘ฅ+13)+35๐‘ฅโˆ’3.๏Šจ

The remainder is 35 which has degree 0, which is less than the degree of ๐‘ฅโˆ’3 which is 1.

When we use the division algorithm to get an ๐‘Ÿ of degree less than ๐‘‘, the quotient ๐‘ž and the remainder ๐‘Ÿ are uniquely determined. We will now outline the division algorithm we can use to find ๐‘ž and ๐‘‘.

Long division of polynomials is much the same as long division for integers: at each step, we compare the leading coefficient of the divisor with the current remainder, which starts off being the dividend itself. The objective at each step is to remove this leading term. Let us look at an example of how to do this.

We will use the example of dividing 2๐‘ฅ+7๐‘ฅโˆ’4๏Šจ by ๐‘ฅโˆ’3 to demonstrate the method.

In the first step, we divide the term of the highest degree in the dividend by the term of the highest degree in the divisor. Hence, we divide 2๐‘ฅ๏Šจ by ๐‘ฅ to get 2๐‘ฅ.

We write the result of this division above the line.

We now multiply this term by the divisor and write the result below the dividend so that the terms of equal degree align.

We now subtract the resulting expression from the dividend.

This should result in us eliminating the term with the highest degree. We can then bring down the terms from the dividend to get an expression for our first remainder. If this is of equal or higher degree than the divisor, as is the case here, we repeat this process again.

Hence, we divide the terms of highest degree. That is we divide 13๐‘ฅ by ๐‘ฅ to get 13.

We write this above the line next to our last term.

We now multiply this term by the divisor and write the result below the dividend so that the terms of equal degree align.

We now subtract the resulting expression from the first remainder.

This should result in us eliminating the term with the highest degree. At this point, we are left with a term of lower degree than the divisor, so we stop. The quotient ๐‘ž(๐‘ฅ) is the expression above the line, and the remainder is the expression at the bottom.

Usually, we write this concisely as follows:

The conventions used when preforming long division this way regarding the placement of the terms of the polynomials vary. However, the technique is the same.

Example 1: Polynomial Long Division with a First-Degree Divisor

Use polynomial division to simplify 2๐‘ฅ+5๐‘ฅ+7๐‘ฅ+4๐‘ฅ+1๏Šฉ๏Šจ.

Answer

In this example, we expect a zero remainder:

So the simplification is 2๐‘ฅ+5๐‘ฅ+7๐‘ฅ+4๐‘ฅ+1=2๐‘ฅ+3๐‘ฅ+4.๏Šฉ๏Šจ๏Šจ

A consequence of a zero remainder is that we get a factorization. In the special case of a linear divisor, we get the following.

The Factor Theorem

The polynomial ๐‘(๐‘ฅ) is divisible by (๐‘ฅโˆ’๐‘Ž) (with zero remainder) if and only if ๐‘(๐‘Ž)=0.

In other words, when ๐‘Ž is a zero of the polynomial.

So ๐‘(๐‘ฅ)=(๐‘ฅโˆ’๐‘Ž)๐‘ž(๐‘ฅ) precisely when ๐‘(๐‘Ž)=0.

Example 2: The Factor Theorem and Long Division

By factoring, find all the solutions to ๐‘ฅโˆ’๐‘ฅโˆ’14๐‘ฅ+24=0๏Šฉ๏Šจ, given that (๐‘ฅ+4) is a factor of ๐‘ฅโˆ’๐‘ฅโˆ’14๐‘ฅ+24๏Šฉ๏Šจ.

Answer

Since (๐‘ฅ+4) is a factor of this polynomial, we can use the factor theorem to conclude that โˆ’4 is a zero of the polynomial. We can use polynomial division to find the other factors.

So ๐‘ฅโˆ’๐‘ฅโˆ’14๐‘ฅ+24=(๐‘ฅ+4)๏€น๐‘ฅโˆ’5๐‘ฅ+6๏…๏Šฉ๏Šจ๏Šจ and we can factorize this quadratic, for example, by inspection: ๐‘ฅโˆ’5๐‘ฅ+6=(๐‘ฅโˆ’2)(๐‘ฅโˆ’3)๏Šจ and therefore ๐‘ฅโˆ’๐‘ฅโˆ’14๐‘ฅ+24=(๐‘ฅ+4)(๐‘ฅโˆ’2)(๐‘ฅโˆ’3).๏Šฉ๏Šจ

The factor (๐‘ฅโˆ’2) corresponds to zero ๐‘ฅ=2, the factor (๐‘ฅโˆ’3) gives the zero ๐‘ฅ=3. So the zeros are ๐‘ฅ=2,๐‘ฅ=3,๐‘ฅ=โˆ’4.

Using the same method, we can perform polynomial long division when the divisor is of degree greater than one. In the next example, we will demonstrate this.

Example 3: Polynomial Long Division with Higher-Degree Divisors

Use polynomial long division to find the quotient ๐‘ž(๐‘ฅ) and the remainder ๐‘Ÿ(๐‘ฅ) for ๐‘(๐‘ฅ)๐‘‘(๐‘ฅ), where ๐‘(๐‘ฅ)=๐‘ฅ+๐‘ฅ+๐‘ฅ+๐‘ฅ+๐‘ฅ+1๏Šญ๏Šฌ๏Šช๏Šจ and ๐‘‘(๐‘ฅ)=๐‘ฅ+๐‘ฅ+1๏Šฉ.

Answer

Applying the long division algorithm, we get the following division:

Hence, the quotient ๐‘ž(๐‘ฅ)=๐‘ฅ+๐‘ฅโˆ’๐‘ฅโˆ’๐‘ฅ๏Šช๏Šฉ๏Šจ and remainder ๐‘Ÿ(๐‘ฅ)=3๐‘ฅ+2๐‘ฅ+1๏Šจ.

Of course, we should not always expect the resulting polynomials ๐‘ž(๐‘ฅ) and ๐‘Ÿ(๐‘ฅ) to have integer coefficients, even when ๐‘(๐‘ฅ) and ๐‘‘(๐‘ฅ) do. The next example demonstrates this.

Example 4: Polynomial Long Division

Express the division ๐‘(๐‘ฅ)๐‘‘(๐‘ฅ)=2๐‘ฅโˆ’๐‘ฅ+52๐‘ฅโˆ’5๐‘ฅ+8๏Šฉ๏Šจ in the form ๐‘ž(๐‘ฅ)+๐‘Ÿ(๐‘ฅ)๐‘‘(๐‘ฅ).

Answer

Using the long division algorithm, we get the following long division:

Hence, 2๐‘ฅโˆ’๐‘ฅ+52๐‘ฅโˆ’5๐‘ฅ+8=๏€ผ๐‘ฅ+52๏ˆ+๐‘ฅโˆ’152๐‘ฅโˆ’5๐‘ฅ+8.๏Šฉ๏Šจ๏Šญ๏Šจ๏Šจ

The factor theorem is a special case of the remainder theorem.

The Remainder Theorem

When the polynomial ๐‘(๐‘ฅ) is divided by (๐‘ฅโˆ’๐‘Ž), the remainder is the constant ๐‘(๐‘Ž).

Example 5: The Remainder Theorem

Find the remainder when 4๐‘ฅ+4๐‘ฅ+3๏Šจ is divided by 2๐‘ฅโˆ’3.

Answer

Although this can be done by long division, we can also use the remainder theorem. We do have to be careful about the application, because (2๐‘ฅโˆ’3) is not (๐‘ฅโˆ’๐‘Ž) for any ๐‘Ž. However, suppose that ๐‘(๐‘ฅ)=4๐‘ฅ+4๐‘ฅ+3=(2๐‘ฅโˆ’3)๐‘ž(๐‘ฅ)+๐‘Ÿ๏Šจ with remainder the constant ๐‘Ÿ and quotient ๐‘ž(๐‘ฅ). Since 2๐‘ฅโˆ’3=2๏€ผ๐‘ฅโˆ’32๏ˆ, we can rewrite the above as 4๐‘ฅ+4๐‘ฅ+3=2๏€ผ๐‘ฅโˆ’32๏ˆ๐‘ž(๐‘ฅ)+๐‘Ÿ.๏Šจ

This says that the remainder when 4๐‘ฅ+4๐‘ฅ+3๏Šจ is divided by 2๐‘ฅโˆ’3 is the same as the remainder on division by ๐‘ฅโˆ’32. Since this has the correct form, the remainder theorem applies and ๐‘Ÿ=๐‘ƒ๏€ผ32๏ˆ=4๏€ผ32๏ˆ+4๏€ผ32๏ˆ+3=44(9)+42(3)+3=9+6+3=18.๏Šจ

Example 6: Using Polynomial Long Division

Find the value of ๐‘˜ that makes the expression 30๐‘ฅ+57๐‘ฅโˆ’48๐‘ฅโˆ’20๐‘ฅ+๐‘˜๏Šซ๏Šจ๏Šฉ๏Šช divisible by 5๐‘ฅโˆ’8๏Šจ.

Answer

We can do this by polynomial division. We should expect a remainder of degree 1 or less which will involve the constant ๐‘˜ and setting that to zero will determine the required ๐‘˜.

The first step is to ensure that the dividend is written correctly in descending powers of ๐‘ฅ: ๐‘(๐‘ฅ)=30๐‘ฅโˆ’20๐‘ฅโˆ’48๐‘ฅ+57๐‘ฅ+๐‘˜.๏Šซ๏Šช๏Šฉ๏Šจ

Using the algorithm:

we find the remainder has degree 0 and is ๐‘˜+40.

Since 5๐‘ฅโˆ’8๏Šจ is a factor only if the division gives a zero remainder, the condition on ๐‘˜ is that ๐‘˜+40=0; in other words ๐‘˜=โˆ’40.

Observe that the method used above will always work. An alternative (which is applicable here) is to use the remainder theorem. Notice that 5๐‘ฅโˆ’8๏Šจ has zeros ยฑ๏„ž85. If ๐‘(๐‘ฅ)=๏€น5๐‘ฅโˆ’8๏…๐‘ž(๐‘ฅ)๏Šจ with some quotient ๐‘ž(๐‘ฅ), then evaluating ๐‘(๐‘ฅ) at, say, ๐‘Ž=๏„ž85 should give zero. Indeed, we find ๐‘๏€ฟ๏„ž85๏‹=๐‘˜+40.

Key Points

  1. Using a similar algorithm for integer long division, we can divide polynomials.
  2. If we divide a polynomial by a factor, we get no remainder. Otherwise, we will be left with a remainder of degree less than the degree of the divisor.
  3. For simple linear factors of the form ๐‘ฅโˆ’๐‘Ž, we can find the remainder by applying the the remainder theorem which states that when the polynomial ๐‘(๐‘ฅ) is divided by (๐‘ฅโˆ’๐‘Ž), the remainder is the constant ๐‘(๐‘Ž).

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