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Lesson Explainer: Polynomial Long Division without Remainder Mathematics • 10th Grade

In this explainer, we will learn how to perform long division on polynomials.

Before we begin explaining the process of long division of polynomials, let’s first recall what it means to divide integers and how we apply long division to integers.

If we have two integers π‘Ž and 𝑏, where 𝑏 is nonzero, then π‘Žπ‘ is the number that when multiplied by 𝑏, gives π‘Ž. Hence, π‘Žπ‘Γ—π‘=π‘Ž. We define the division of polynomials in the same way.

Definition: Division of Polynomials

We define the quotient of two polynomials by finding the polynomial whose product with the divisor yields the dividend.

For example, if we want to find π‘₯+π‘₯π‘₯+1, we can note that we want to find a polynomial that when multiplied by π‘₯+1, gives π‘₯+π‘₯. One way of finding this value is to expand the product (π‘₯+1)π‘₯. We have (π‘₯+1)π‘₯=π‘₯+π‘₯.

Thus, π‘₯+π‘₯π‘₯+1=π‘₯.

We can then ask, β€œHow do we find an expression for a quotient of polynomials in general?”

Let’s first consider an example involving integers. If we want to determine the value of 175, we first note that 5 does not go exactly into 17. We can instead employ long division to evaluate this quotient:

We first see that three 5s make 15 and that we cannot have any more 5s. Removing this from 17, we are left with a remainder of 2. We call 17 the dividend since it is being divided by the divisor 5, we call 3 the quotient, and we call 2 the remainder.

It is also worth noting that we say that 𝑏 divides π‘Ž if its remainder is 0 after division by 𝑏. If we call the remainder after dividing π‘Ž by π‘Β π‘Ÿ, then we can guarantee that 𝑏>π‘Ÿ>0. If this is not the case, then we can increase the value of the quotient.

The reason for this is that the process for long division rewrites the fraction. Let’s say that π‘Žπ‘ has a quotient π‘ž and a remainder π‘Ÿ. Then 𝑏 goes into π‘Žπ‘ž times and has a remainder π‘Ÿ. This means that π‘Γ—π‘ž+π‘Ÿ=π‘Ž. Hence, π‘Žπ‘=π‘Γ—π‘ž+π‘Ÿπ‘π‘Žπ‘=π‘Γ—π‘žπ‘+π‘Ÿπ‘π‘Žπ‘=π‘ž+π‘Ÿπ‘.

In our numerical example, this reads 175=3+25.

We are almost ready to apply this process to polynomials. Let’s say we want to divide a polynomial called the dividend by another polynomial called the divisor. We want to do this by subtracting multiples of the divisor from the dividend until we can no longer do so. To see how to do this, let’s consider an example.

Suppose we want to divide π‘₯+5π‘₯+6 by π‘₯+2. We need to determine how many multiples of π‘₯+2 we can remove from π‘₯+5π‘₯+6. Each multiple will be a polynomial since the products of polynomials are polynomials. We will call the polynomial that tells us the number of multiples of π‘₯+2 we remove the quotient. This means that we want to find a polynomial whose product with π‘₯+2 is as close to π‘₯+5π‘₯+6 as possible.

Let’s write this division using the same long division notation we use for integers:

We note that the leading term of the dividend is π‘₯ and that the leading term of the divisor is π‘₯. We then see that π‘₯π‘₯=π‘₯. We will write this in our quotient above the π‘₯-term:

Remember that we want to subtract the product of the quotient and divisor from the dividend. We can do this term by term. If we multiply the divisor by π‘₯, we get π‘₯(π‘₯+2)=π‘₯+2π‘₯.

Subtracting this from the dividend yields ο€Ήπ‘₯+5π‘₯+6ο…βˆ’π‘₯(π‘₯+2)=π‘₯βˆ’π‘₯+5π‘₯βˆ’2π‘₯+6=3π‘₯+6.

We write this in the long division notation as follows:

Before we continue with the division, it may be useful to consider what we have actually shown. We have shown that π‘₯+5π‘₯+6=π‘₯(π‘₯+2)+(3π‘₯+6). We can then divide through by π‘₯+2, which gives π‘₯+5π‘₯+6π‘₯+2=π‘₯(π‘₯+2)+(3π‘₯+6)π‘₯+2=π‘₯+3π‘₯+6π‘₯+2.

Thus, we have reduced the degree of the dividend. We can apply this process again, this time with 3π‘₯+6 as the dividend. The leading term of this new dividend is 3π‘₯, and the leading term of the divisor is π‘₯. So, 3π‘₯π‘₯=3, and we add this to the quotient as follows:

We want to remove the divisor from the dividend 3 times. This gives us (3π‘₯+6)βˆ’3(π‘₯+2)=3π‘₯+6βˆ’3π‘₯βˆ’6=0.

We can represent this using long division notation as follows:

In this explainer, we will only deal with cases where the remainder is the zero polynomial. However, it is worth noting that each time we perform this process, we are lowering the degree of the dividend by the degree of the divisor. We keep this process going until the remainder has a lower degree than the divisor. At this point when we divide their leading terms, we would get either zero or a variable raised to a negative exponent, which is not a monomial.

Using polynomial long division, we have shown that π‘₯+5π‘₯+6=π‘₯(π‘₯+2)+3(π‘₯+2). So, π‘₯+5π‘₯+6π‘₯+2=π‘₯(π‘₯+2)+3(π‘₯+2)π‘₯+2=π‘₯+3.

We can verify this by multiplying the quotient π‘₯+3 by the divisor π‘₯+2. We have (π‘₯+2)(π‘₯+3)=π‘₯+2π‘₯+3π‘₯+6=π‘₯+5π‘₯+6.

This works as a nice check to make sure that the answer is correct.

We can summarize the working in the above example as follows:

Let’s now see an example of how to apply this process to divide two polynomials.

Example 1: Dividing Polynomials Using Polynomial Long Division

Find the quotient of π‘₯+2π‘₯βˆ’3 divided by π‘₯βˆ’1.

Answer

We can find an expression for π‘₯+2π‘₯βˆ’3π‘₯βˆ’1 using long division. The first step is to find the quotient of the leading terms of the dividend and the divisor. We note that π‘₯π‘₯=π‘₯. So, we write π‘₯ in the quotient, and we subtract π‘₯(π‘₯βˆ’1) from the dividend. We have that π‘₯(π‘₯βˆ’1)=π‘₯βˆ’π‘₯, so we see that

We calculate that ο€Ήπ‘₯+2π‘₯βˆ’3ο…βˆ’ο€Ήπ‘₯βˆ’π‘₯=π‘₯+2π‘₯βˆ’3βˆ’π‘₯+π‘₯=3π‘₯βˆ’3.

We can then add this onto our division:

We want to apply this process again, this time with 3π‘₯βˆ’3 as the dividend. We need to see how many times the divisor π‘₯βˆ’1 goes into the dividend 3π‘₯βˆ’3. We do this by first dividing their leading terms. We see that 3π‘₯π‘₯=3. So, we add 3 to the quotient, and we then need to subtract 3(π‘₯βˆ’1) from the dividend. We find that 3(π‘₯βˆ’1)=3π‘₯βˆ’3. Subtracting this gives us the following:

We find that (3π‘₯βˆ’3)βˆ’(3π‘₯βˆ’3)=0.

Hence, π‘₯+2π‘₯βˆ’3π‘₯βˆ’1=π‘₯+3.

We can verify this answer by calculating (π‘₯+3)(π‘₯βˆ’1)=π‘₯+2π‘₯βˆ’3.

In our next example, we will find the quotient of two polynomials where both of the polynomials are nonmonic.

Example 2: Dividing a Polynomial by a First-Degree Divisor to Find the Quotient

Find the quotient when 2π‘₯+7π‘₯βˆ’8π‘₯βˆ’21 is divided by 2π‘₯+3.

Answer

We can divide two polynomials using polynomial long division. First, we need to divide their leading terms. We have 2π‘₯2π‘₯=π‘₯. We add this into our quotient, and this means we need to subtract π‘₯(2π‘₯+3) from our dividend. Since π‘₯(2π‘₯+3)=2π‘₯+3π‘₯, we can do this as follows:

We calculate that ο€Ή2π‘₯+7π‘₯βˆ’8π‘₯βˆ’21ο…βˆ’ο€Ή2π‘₯+3π‘₯=4π‘₯βˆ’8π‘₯βˆ’21. We have shown that 2π‘₯+7π‘₯βˆ’8π‘₯βˆ’212π‘₯+3=(2π‘₯+3)π‘₯+ο€Ή4π‘₯βˆ’8π‘₯βˆ’212π‘₯+3=π‘₯+ο€Ή4π‘₯βˆ’8π‘₯βˆ’212π‘₯+3.

We can apply this process again to divide 4π‘₯βˆ’8π‘₯βˆ’21 by 2π‘₯+3. We divide the leading terms to get 4π‘₯2π‘₯=2π‘₯. Thus, we add 2π‘₯ to the quotient and subtract 2π‘₯(2π‘₯+3)=4π‘₯+6π‘₯ from the dividend as follows:

We then calculate that ο€Ή4π‘₯βˆ’8π‘₯βˆ’21ο…βˆ’ο€Ή4π‘₯+6π‘₯=βˆ’14π‘₯βˆ’21.

We need to apply this process one final time to fully divide the expression.

This time, the division of the leading terms gives us βˆ’14π‘₯2π‘₯=βˆ’7. We add this to the quotient. We then need to subtract βˆ’7(2π‘₯+3)=βˆ’14π‘₯βˆ’21 from this new dividend. Since this is equal to the dividend, we get a value of 0:

Hence, 2π‘₯+7π‘₯βˆ’8π‘₯βˆ’212π‘₯+3=π‘₯+2π‘₯βˆ’7.

In our next example, we will divide a nonmonic cubic polynomial by a linear polynomial.

Example 3: Dividing a Cubic Polynomial by a First-Degree Divisor to Find the Quotient

Find the quotient when βˆ’6π‘₯+11π‘₯+π‘₯βˆ’6 is divided by π‘₯βˆ’1.

Answer

We can divide two polynomials using polynomial long division. First, we need to divide their leading terms. We have βˆ’6π‘₯π‘₯=βˆ’6π‘₯. We add this to our quotient, and this means we need to subtract βˆ’6π‘₯(π‘₯βˆ’1) from our dividend. Since βˆ’6π‘₯(π‘₯βˆ’1)=βˆ’6π‘₯+6π‘₯, we can do this as follows:

We calculate that ο€Ήβˆ’6π‘₯+11π‘₯+π‘₯βˆ’6ο…βˆ’ο€Ήβˆ’6π‘₯+6π‘₯=5π‘₯+π‘₯βˆ’6.

Since the degree of 5π‘₯+π‘₯βˆ’6 is higher than or equal to the degree of the divisor, we now need to apply this process again, with 5π‘₯+π‘₯βˆ’6 as our new dividend.

We divide their leading terms to get 5π‘₯π‘₯=5π‘₯, which we add to the quotient. We then need to subtract 5π‘₯(π‘₯βˆ’1)=5π‘₯βˆ’5π‘₯ from the dividend. We can do this as follows:

We calculate that ο€Ή5π‘₯+π‘₯βˆ’6ο…βˆ’ο€Ή5π‘₯βˆ’5π‘₯=6π‘₯βˆ’6.

Since the degree of this polynomial is equal to that of the divisor, we need to apply this process one final time, with 6π‘₯βˆ’6 as the dividend.

We divide their leading terms to get 6π‘₯π‘₯=6, which we then add to the quotient. We then need to subtract 6(π‘₯βˆ’1)=6π‘₯βˆ’6 from the dividend. Since this is equal to the dividend, we will end up with a remainder of 0, as shown:

Hence, we have shown that the quotient that results from dividing βˆ’6π‘₯+11π‘₯+π‘₯βˆ’6 by π‘₯βˆ’1 is βˆ’6π‘₯+5π‘₯+6. We could check this answer by calculating (π‘₯βˆ’1)Γ—ο€Ήβˆ’6π‘₯+5π‘₯+6ο…οŠ¨ to ensure that we get our original cubic dividend.

In our next example, we will see how to apply polynomial long division to find a missing dimension in a given diagram.

Example 4: Determining a Dimension Using Polynomial Long Division

Given that the area of the rectangle in the diagram is 2π‘₯+π‘₯βˆ’5π‘₯βˆ’3, find an expression for the width of the rectangle.

Answer

We first recall that the area of rectangle is given by the product of its length and width. This means we can determine the width of this rectangle by dividing its area by its length. This means we need to divide a cubic polynomial by a linear polynomial. We can do this using polynomial long division.

We first divide their leading terms to get 2π‘₯2π‘₯=π‘₯. We add this to the quotient and then subtract π‘₯(2π‘₯+3) from the dividend. We note that π‘₯(2π‘₯+3)=2π‘₯+3π‘₯ and that ο€Ή2π‘₯+π‘₯βˆ’5π‘₯βˆ’3ο…βˆ’ο€Ή2π‘₯+3π‘₯=βˆ’2π‘₯βˆ’5π‘₯βˆ’3, as shown:

Since the degree of βˆ’2π‘₯βˆ’5π‘₯βˆ’3 is higher than or equal to the degree of the divisor, we need to apply this process again, with βˆ’2π‘₯βˆ’5π‘₯βˆ’3 as the new dividend.

We first divide their leading terms to get βˆ’2π‘₯2π‘₯=βˆ’π‘₯, which we then add to the quotient. We now need to subtract βˆ’π‘₯(2π‘₯+3)=βˆ’2π‘₯βˆ’3π‘₯ from the divided. This gives us the following:

We calculate that ο€Ήβˆ’2π‘₯βˆ’5π‘₯βˆ’3ο…βˆ’ο€Ήβˆ’2π‘₯βˆ’3π‘₯=βˆ’2π‘₯βˆ’3.

Since the degree of this polynomial is equal to the degree of the divisor, we need to apply this process one final time, with βˆ’2π‘₯βˆ’3 as the dividend.

We divide their leading terms to get βˆ’2π‘₯2=βˆ’1, which we add to the quotient. We then note that βˆ’1(2π‘₯+3)=βˆ’2π‘₯βˆ’3. This is equal to the dividend, so when we subtract, we will get 0. This gives us the following:

Hence, 2π‘₯+π‘₯βˆ’5π‘₯βˆ’32π‘₯+3=π‘₯βˆ’π‘₯βˆ’1, and the width of the rectangle is given by π‘₯βˆ’π‘₯βˆ’1.

In our final example, we will use polynomial long division and the given divisibility of two polynomials to determine the value of an unknown coefficient.

Example 5: Finding the Value of a Constant That Makes a Polynomial Divisible

Find the value of π‘˜ that makes the expression 30π‘₯+57π‘₯βˆ’48π‘₯βˆ’20π‘₯+π‘˜οŠ«οŠ¨οŠ©οŠͺ divisible by 5π‘₯βˆ’8.

Answer

We first recall that we say a polynomial is divisible by another polynomial if their division has a remainder of 0. This means we can apply polynomial long division to the given polynomials, and we know that the remainder must be the zero polynomial.

Before we apply polynomial long division, it is worth noting that the quintic polynomial is not given in descending powers of π‘₯. We should always reorder the dividend to have descending powers of π‘₯ since we always want to remove the leading terms of the dividends. So, we will use 30π‘₯βˆ’20π‘₯βˆ’48π‘₯+57π‘₯+π‘˜οŠ«οŠͺ as the dividend.

To apply polynomial long division, we first need to divide the leading terms. We get 30π‘₯5π‘₯=6π‘₯. We add this to the quotient and then subtract 6π‘₯ο€Ή5π‘₯βˆ’8ο…οŠ©οŠ¨ from the dividend. Since 6π‘₯ο€Ή5π‘₯βˆ’8=30π‘₯βˆ’48π‘₯, we will also include the term 0π‘₯οŠͺ to keep the columns with the same powers of π‘₯. We have

We have calculated that ο€Ή30π‘₯βˆ’20π‘₯βˆ’48π‘₯+57π‘₯+π‘˜ο…βˆ’ο€Ή30π‘₯βˆ’48π‘₯=βˆ’20π‘₯+57π‘₯+π‘˜.οŠͺοŠͺ

It is also worth noting that we add the term 0π‘₯ into our long division to keep the columns with the same powers of π‘₯.

Since this is not zero, we need to apply this process again, with βˆ’20π‘₯+57π‘₯+π‘˜οŠͺ as the dividend.

We first divide the leading terms to get βˆ’20π‘₯5π‘₯=βˆ’4π‘₯οŠͺ, and we add this to the quotient. Next, we need to subtract βˆ’4π‘₯ο€Ή5π‘₯βˆ’8ο…οŠ¨οŠ¨ from the dividend. We note that βˆ’4π‘₯ο€Ή5π‘₯βˆ’8=βˆ’20π‘₯+32π‘₯οŠͺ. We can then subtract this from the dividend as follows:

We can calculate that ο€Ήβˆ’20π‘₯+57π‘₯+π‘˜ο…βˆ’ο€Ήβˆ’20π‘₯+32π‘₯=25π‘₯+π‘˜.οŠͺοŠͺ

Since this is not zero, we need to apply this process again, with 25π‘₯+π‘˜οŠ¨ as the dividend.

We divide their leading terms to get 25π‘₯5π‘₯=5, and we add this to our quotient. We then subtract 5ο€Ή5π‘₯βˆ’8ο…οŠ¨ from the dividend as follows:

We calculated that ο€Ή25π‘₯+π‘˜ο…βˆ’5ο€Ή5π‘₯βˆ’8=ο€Ή25π‘₯+π‘˜ο…βˆ’25π‘₯+40=π‘˜+40.

Since we are told that the division is exact, we know that the remainder must be zero. Hence, π‘˜+40=0.

We can solve this to see that π‘˜=βˆ’40.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We define the quotient of two polynomials by finding the polynomial whose product with the divisor yields the dividend.
  • We can divide polynomials using long division.
  • We can check our answer by multiplying the divisor by the quotient.
  • If the remainder polynomial in the division of two polynomials is the zero polynomial, then we say that the divisor divides the dividend.
  • We should always reorder the dividend to have descending powers of π‘₯ since we always want to remove the leading terms of the dividends.

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