Lesson Explainer: Polynomial Long Division With Remainder - Nagwa

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 𝑝(𝑥) (the dividend) by a divisor 𝑑(𝑥) gives a quotient 𝑞(𝑥) and a remainder 𝑟(𝑥).

Recall that a polynomial is a finite sum of monomials which has nonnegative exponents. Hence, expressions of the forms 2𝑥+2, 𝑥𝑦−10𝑥𝑦+𝑥, and 8 are all examples of polynomials, whereas expressions such as √𝑥, 3𝑥, and 3𝑥 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 𝑝(𝑥)𝑑(𝑥) rather than 𝑝(𝑥)÷𝑑(𝑥). We can think of long division as finding polynomials 𝑞 and 𝑟 such that 𝑝(𝑥)𝑑(𝑥)=𝑞(𝑥)+𝑟(𝑥)𝑑(𝑥) and we say that that the division yields a quotient 𝑞(𝑥) and a remainder 𝑟(𝑥).

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 2𝑥+7𝑥−4=(𝑥−3)×(𝑥−1)+𝑥+11𝑥−7.

This can be written as 2𝑥+7𝑥−4𝑥−3=(𝑥−1)+𝑥+11𝑥−7𝑥−3 but it does not qualify as division by 𝑥−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 2𝑥+7𝑥−4𝑥−3=(2𝑥+13)+35𝑥−3.

The remainder is 35 which has degree 0, which is less than the degree of 𝑥−3 which is 1.

When we use the division algorithm to get an 𝑟 of degree less than 𝑑, the quotient 𝑞 and the remainder 𝑟 are uniquely determined. We will now outline the division algorithm we can use to find 𝑞 and 𝑑.

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 2𝑥+7𝑥−4 by 𝑥−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 2𝑥 by 𝑥 to get 2𝑥.

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 13𝑥 by 𝑥 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 𝑞(𝑥) 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 2𝑥+5𝑥+7𝑥+4𝑥+1.

Answer

In this example, we expect a zero remainder:

So the simplification is 2𝑥+5𝑥+7𝑥+4𝑥+1=2𝑥+3𝑥+4.

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 𝑝(𝑥) is divisible by (𝑥−𝑎) (with zero remainder) if and only if 𝑝(𝑎)=0.

In other words, when 𝑎 is a zero of the polynomial.

So 𝑝(𝑥)=(𝑥−𝑎)𝑞(𝑥) precisely when 𝑝(𝑎)=0.

Example 2: The Factor Theorem and Long Division

By factoring, find all the solutions to 𝑥−𝑥−14𝑥+24=0, given that (𝑥+4) is a factor of 𝑥−𝑥−14𝑥+24.

Answer

Since (𝑥+4) is a factor of this polynomial, we can use the factor theorem to conclude that −4 is a zero of the polynomial. We can use polynomial division to find the other factors.

So 𝑥−𝑥−14𝑥+24=(𝑥+4)𝑥−5𝑥+6 and we can factorize this quadratic, for example, by inspection: 𝑥−5𝑥+6=(𝑥−2)(𝑥−3) and therefore 𝑥−𝑥−14𝑥+24=(𝑥+4)(𝑥−2)(𝑥−3).

The factor (𝑥−2) corresponds to zero 𝑥=2, the factor (𝑥−3) gives the zero 𝑥=3. So the zeros are 𝑥=2,𝑥=3,𝑥=−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 𝑞(𝑥) and the remainder 𝑟(𝑥) for 𝑝(𝑥)𝑑(𝑥), where 𝑝(𝑥)=𝑥+𝑥+𝑥+𝑥+𝑥+1 and 𝑑(𝑥)=𝑥+𝑥+1.

Answer

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

Hence, the quotient 𝑞(𝑥)=𝑥+𝑥−𝑥−𝑥 and remainder 𝑟(𝑥)=3𝑥+2𝑥+1.

Of course, we should not always expect the resulting polynomials 𝑞(𝑥) and 𝑟(𝑥) to have integer coefficients, even when 𝑝(𝑥) and 𝑑(𝑥) do. The next example demonstrates this.

Example 4: Polynomial Long Division

Express the division 𝑝(𝑥)𝑑(𝑥)=2𝑥−𝑥+52𝑥−5𝑥+8 in the form 𝑞(𝑥)+𝑟(𝑥)𝑑(𝑥).

Answer

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

Hence, 2𝑥−𝑥+52𝑥−5𝑥+8=𝑥+52+𝑥−152𝑥−5𝑥+8.

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

The Remainder Theorem

When the polynomial 𝑝(𝑥) is divided by (𝑥−𝑎), the remainder is the constant 𝑝(𝑎).

Example 5: The Remainder Theorem

Find the remainder when 4𝑥+4𝑥+3 is divided by 2𝑥−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 (2𝑥−3) is not (𝑥−𝑎) for any 𝑎. However, suppose that 𝑝(𝑥)=4𝑥+4𝑥+3=(2𝑥−3)𝑞(𝑥)+𝑟 with remainder the constant 𝑟 and quotient 𝑞(𝑥). Since 2𝑥−3=2𝑥−32, we can rewrite the above as 4𝑥+4𝑥+3=2𝑥−32𝑞(𝑥)+𝑟.

This says that the remainder when 4𝑥+4𝑥+3 is divided by 2𝑥−3 is the same as the remainder on division by 𝑥−32. Since this has the correct form, the remainder theorem applies and 𝑟=𝑃32=432+432+3=44(9)+42(3)+3=9+6+3=18.

Example 6: Using Polynomial Long Division

Find the value of 𝑘 that makes the expression 30𝑥+57𝑥−48𝑥−20𝑥+𝑘 divisible by 5𝑥−8.

Answer

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

The first step is to ensure that the dividend is written correctly in descending powers of 𝑥: 𝑝(𝑥)=30𝑥−20𝑥−48𝑥+57𝑥+𝑘.

Using the algorithm:

we find the remainder has degree 0 and is 𝑘+40.

Since 5𝑥−8 is a factor only if the division gives a zero remainder, the condition on 𝑘 is that 𝑘+40=0; in other words 𝑘=−40.

Observe that the method used above will always work. An alternative (which is applicable here) is to use the remainder theorem. Notice that 5𝑥−8 has zeros ±85. If 𝑝(𝑥)=5𝑥−8𝑞(𝑥) with some quotient 𝑞(𝑥), then evaluating 𝑝(𝑥) at, say, 𝑎=85 should give zero. Indeed, we find 𝑝85=𝑘+40.

Key Points

• Using a similar algorithm for integer long division, we can divide polynomials.
• 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.
• For simple linear factors of the form 𝑥−𝑎, we can find the remainder by applying the the remainder theorem which states that when the polynomial 𝑝(𝑥) is divided by (𝑥−𝑎), the remainder is the constant 𝑝(𝑎).