Lesson Video: Polynomial Long Division with Remainder Mathematics

Video Transcript

Polynomial Long Division with Remainder

In this video, we’re going to learn an algorithm to help us divide one polynomial by another polynomial. We’ll discuss how this is similar to regular long division and how to find the quotient and remainder polynomials. Finally, we’ll talk about the remainder theorem and how it relates to the factor theorem. Before we start talking about dividing two polynomials, let’s start by talking about dividing two numbers. Consider 231 divided by five. We know a lot of different ways of evaluating this expression. For example, if we were to just evaluate this directly, we might write the answer as 46.2. Another way of saying this would be 46 and one-fifth or 46 plus one over five.

But remember, the process we used to arrive at this answer is long division. There’s a lot of slightly different ways of doing long division. We’ll only go through one of these. To start, we call five our divisor. We want to see how many times five goes into 231. To start, we need to check how many times five goes into 200. We know five goes into 200 40 times. So we’ll write a four in the tens column. Next, we know that 40 times five is equal to 200, so we need to subtract this from 231. This, of course, leaves us with 31. We don’t need to do this process again. We need to see how many times five goes into 30.

Of course, we know five goes into 30 six times, so we need to add six to our answer of 40. Just as we did before, we need to subtract six times five. We know this is equal to 30. And this, of course, leaves us with one. If we were to try and do this process again, we would need to see how many times five goes into one. And, of course, five doesn’t go into one. So this tells us that our process is done and we were left with one. We call this term of one our remainder term. And we call 46 our quotient. And we can see this in what we had before. We have 231 divided by five is equal to our quotient 46 plus our remainder of one divided by our divisor of five.

Another common way of seeing this written is to multiply this equation through by five. This would give us 231 is equal to five times 46 plus one. And one thing worth pointing out is we can always guarantee that our remainder will be smaller than our divisor. This is because if it wasn’t, we could’ve just increased our quotient. So now, we’re going to ask the question, how does this relate to polynomial division? This time, instead of being given an integer divided by an integer, we’re going to be given a polynomial divided by a polynomial. And we’re going to find an algorithm which will help us see how many times d of 𝑥 goes into 𝑝 of 𝑥. We’ll call this polynomial long division. And this will be very similar to regular long division.

Just like with regular long division, we’ll find a quotient and remainder term. However, this time, because we’re dividing polynomials, our quotient and remainder will also be polynomials. We’ll call these 𝑞 of 𝑥 and 𝑟 of 𝑥. And our answer will be in exactly the same form we got for regular long division. We’ll find polynomials 𝑞 of 𝑥 and 𝑟 of 𝑥 such that 𝑝 of 𝑥 divided by d of 𝑥 is equal to 𝑞 of 𝑥 plus 𝑟 of 𝑥 divided by d of 𝑥.

Finally, remember, when we were doing regular long division, we could guarantee that our remainder would be smaller than our divisor. And we’ll have something similar with polynomial long division. We’ll be able to guarantee that the degree of our remainder is smaller than the degree of our divisor. And the reasoning for this is exactly the same as it is in regular long division. We keep removing multiples of d of 𝑥 until we can’t do this anymore. Let’s now go onto an example of using polynomial long division.

Use polynomial division to simplify three 𝑥 cubed plus two 𝑥 squared minus four 𝑥 plus one divided by 𝑥 plus one.

The question wants us to use polynomial division to simplify this expression. We’ll call the cubic polynomial in our numerator, that’s three 𝑥 cubed plus two 𝑥 squared minus four 𝑥 plus one, 𝑝 of 𝑥. And we’ll call the linear polynomial in our denominator, that’s 𝑥 plus one, d of 𝑥. This is our divisor. Using polynomial division is very similar to regular long division. We’ll set this up in exactly the same way. We’ll have our divisor 𝑥 plus one going into three 𝑥 cubed plus two 𝑥 squared minus four 𝑥 plus one.

When using regular long division, we want to see how many times our divisor goes into the highest term. We can do exactly the same thing here. We want to know how many times 𝑥 goes into the highest terminal polynomial. That’s three 𝑥 cubed. Of course, three 𝑥 cubed divided by 𝑥 is three 𝑥 squared. Or alternatively, we could write this as three 𝑥 squared times 𝑥 is equal to three 𝑥 cubed. Just like with regular long division, we want to write this in our quotient. Remember though, this is a term for 𝑥 squared, so we’ll write this in the column for 𝑥 squared.

In regular long division, the next step is to multiply this term we just added to our quotient by our divisor. We would then want to subtract this from our polynomial. We’ll do exactly the same thing here. First, we need to find three 𝑥 squared multiplied by our divisor 𝑥 plus one. If we evaluate this, we get three 𝑥 cubed plus three 𝑥 squared. We then want to subtract this from our polynomial 𝑝 of 𝑥. We’ll evaluate this term by term. First, three 𝑥 cubed minus three 𝑥 cubed is equal to zero, and two 𝑥 squared minus three 𝑥 squared is equal to negative 𝑥 squared.

It’s worth pointing out you’ll often see this term of zero omitted from working out. This is because we’ll always have a zero in this position. It doesn’t matter either way if you prefer leaving it in or taking it out. In this instance, we’ll be leaving it out. And then just like working a long division, we need to take down the rest of our terms. This gives us negative 𝑥 squared minus four 𝑥 plus one. Just like with regular long division, we now need to repeat our process. We need to see how many times 𝑥 goes into our highest term negative 𝑥 squared.

Well, we know negative 𝑥 squared divided by 𝑥 is negative 𝑥. We’ll write this in our quotient in the column we have for 𝑥. Now, just as we did before, we need to multiply our divisor of 𝑥 plus one by negative 𝑥. This gives us negative 𝑥 times 𝑥 plus one. And if we evaluate this, we get negative 𝑥 squared minus 𝑥. And remember, the next thing we need to do is subtract this from our polynomial negative 𝑥 squared minus four 𝑥 plus one. We’ll do this term by term. First, negative 𝑥 squared minus negative 𝑥 squared is negative 𝑥 squared plus 𝑥 squared. This is equal to zero.

Next, we have negative four 𝑥 minus negative 𝑥 is negative four 𝑥 plus 𝑥. This is equal to negative three 𝑥. Now, just as we did before, we need to take this constant of one down. So this gives us negative three 𝑥 plus one. Just like with regular long division, we now need to do this again. We keep going until we can’t do this anymore. Again, we need to see how many times 𝑥 goes into our highest-order term. That’s negative three 𝑥. This time, negative three 𝑥 divided by 𝑥 is equal to negative three. We’ll add this to our quotient. The next step is to multiply negative three by our divisor 𝑥 plus one and subtract this from negative three 𝑥 plus one.

So we want to evaluate negative three 𝑥 times 𝑥 plus one. If we distribute this over our parentheses, we get negative three 𝑥 minus three. And now, we subtract this from negative three 𝑥 plus one. We’ll do this term by term. The first term, we get negative three 𝑥 minus negative three 𝑥, which is negative three 𝑥 plus three 𝑥, which we know is equal to zero. Next, we want one minus negative three. Well, that’s one plus three, which we know is equal to four. But now, if we try and continue this process, we get a problem. We would want to see how many times 𝑥 goes into four. Well, this is just four divided by 𝑥.

This is not a polynomial. In other words, we can no longer divide this expression by our divisor. In fact, this will always happen when we’re left with an expression with a degree strictly less than our divisor. And just like with regular long division, we’ll call this our remainder term. We’ll call this 𝑟 of 𝑥 because often this will be a polynomial. And just the same as regular long division, we’ll call three 𝑥 squared minus 𝑥 minus three our quotient. We’ll call this 𝑞 of 𝑥.

And now, we can come to our answer in exactly the same way we do with regular long division. We can use our remainder and quotient to rewrite 𝑝 of 𝑥 divided by d of 𝑥 as 𝑞 of 𝑥 plus 𝑟 of 𝑥 divided by d of 𝑥, the quotient plus the remainder divided by the divisor. Now, all we have to do is substitute in our expressions for 𝑞 of 𝑥, 𝑟 of 𝑥, and d of 𝑥. And by doing this, we were able to rewrite the expression given to us in the question as three 𝑥 squared minus 𝑥 minus three plus four divided by 𝑥 plus one.

Let’s do another example to help us solidify what we’ve learnt.

Use polynomial long division to find the quotient 𝑞 of 𝑥 and the remainder 𝑟 of 𝑥 for 𝑝 of 𝑥 divided by d of 𝑥, where 𝑝 of 𝑥 is equal to 𝑥 of the seventh power plus 𝑥 to the sixth power plus 𝑥 to the fourth power plus 𝑥 squared plus 𝑥 plus one and d of 𝑥 is equal to 𝑥 cubed plus 𝑥 plus one.

The question wants us to use polynomial long division. We’re given our polynomial 𝑝 of 𝑥 and our divisor d of 𝑥. We need to find the quotient 𝑞 of 𝑥 and the remainder 𝑟 of 𝑥 when we divide 𝑝 of 𝑥 by d of 𝑥. Before we start answering this question, there’s a couple of things we should check. For example, we should check that both our polynomials 𝑝 of 𝑥 and d of 𝑥 are written in descending exponents of 𝑥. In this case, this is true, so we can just carry on with our long division. We’ll set up our long division. We have our divisor d of 𝑥 is dividing our polynomial 𝑝 of 𝑥.

Before we start doing our long division, there’s one more thing we can check. If we look at our polynomial 𝑝 of 𝑥, we can see there’s no term for 𝑥 to the fifth power and there’s no term for 𝑥 cubed. In regular long division, when this happens, we have a digit of zero in this place. However, because this is a polynomial, we just didn’t write these terms in. There’s a few different ways we could tackle this. We could just leave that as it is, or we could add terms zero 𝑥 to the fifth power and zero 𝑥 cubed. And both of these methods work, and you can use them if you prefer. However, in this video, we’re just going to leave a gap where these terms are to keep our columns aligned.

Now, let’s move on to our long division. The leading term in 𝑝 of 𝑥 is 𝑥 to the seventh power. We need to divide this by 𝑥 cubed. And of course, 𝑥 to the seventh power divided by 𝑥 cubed is 𝑥 to the fourth power. We’ll write this in our quotient, and we’ll write this in the column for 𝑥 to the fourth power terms. The next thing we need to do is multiply our divisor by the term in our quotient 𝑥 to the fourth power. Multiplying these together, we get 𝑥 to the fourth power times 𝑥 cubed plus 𝑥 plus one. And if we distribute this and simplify, we get 𝑥 to the seventh power plus 𝑥 to the fifth power plus 𝑥 to the fourth power.

We now want to subtract this from our polynomial 𝑝 of 𝑥. And remember, we want to keep each term in its respective column. We’ll start with 𝑥 to seventh power. Then we need to add 𝑥 to the fifth power. Finally, we add a term for 𝑥 to the fourth power. Now, we can just subtract this term by term. First, we get 𝑥 to the seventh power minus 𝑥 to the seventh power. This is equal to zero. You can write this term of zero in if you prefer. However, this term will always give us zero. So we’ll just leave this blank.

Next, we have 𝑥 to the sixth power minus zero. Of course, this is just equal to 𝑥 to the sixth power. Next, we have zero minus 𝑥 to the fifth power. This is negative 𝑥 to the fifth power. Then in our next column, we have 𝑥 to the fourth power minus 𝑥 to the fourth power. This is equal to zero. We’ll leave this bank. And remember, we need to bring down the rest of our terms. It’s worth pointing out some people prefer to leave these terms at the top until they’re needed. But we’re going to always bring these terms down.

We’re now ready to find the next term in our quotient. We need to divide 𝑥 to the sixth power by 𝑥 cubed. And 𝑥 to the sixth power divided by 𝑥 cubed is just equal to 𝑥 cubed. And remember, we write this in our column for 𝑥 cubed terms. The next step in our long division will be to multiply 𝑥 cubed by our divisor 𝑥 cubed plus 𝑥 plus one. This gives us 𝑥 cubed times 𝑥 cubed plus 𝑥 plus one. And if we distribute this and simplify, we get 𝑥 to the sixth power plus 𝑥 to the fourth power plus 𝑥 cubed.

We now need to subtract this from our polynomial. Remember, it’s important that we write each term in the correct column. We can then subtract this term by term. In our first column, we get 𝑥 to the sixth power minus 𝑥 to the sixth power, which is zero. In our second column, we get negative 𝑥 to the fifth power minus zero, which is just equal to negative 𝑥 to the fifth power. In our next column, we get zero minus 𝑥 to the fourth power, which is negative 𝑥 to the fourth power. We get a simpler story in our next column. We have zero minus 𝑥 cubed, which is negative 𝑥 cubed. Then once again, we bring down the remaining terms.

Once again, we need to find the next term in our quotient. We need to divide negative 𝑥 to the fifth power by 𝑥 cubed. Of course, if we do this, we get negative 𝑥 squared. Once again, we need to multiply the newly added term to our quotient by our divisor. Distributing this and simplifying, we get negative 𝑥 to the fifth power minus 𝑥 cubed minus 𝑥 squared. We then need to subtract this from our polynomial. Remember, we want to write each term in its correct column. We evaluate the subtraction term by term. This time, we get negative 𝑥 to the fourth power plus two 𝑥 squared. And then we bring the rest of our terms down.

And once again, we need to find the next term in 𝑞 of 𝑥. We need to divide negative 𝑥 to the fourth power by 𝑥 cubed. Doing this, we get negative 𝑥. Once again, we multiply negative 𝑥 by our divisor. And if we evaluate this, we get negative 𝑥 to the fourth power minus 𝑥 squared minus 𝑥. Now, we need to subtract this from our polynomial. Evaluating the subtraction and bringing down our term of one, we get three 𝑥 squared plus two 𝑥 plus one. And now, if we were to try and find the next term in our quotient, we would get three 𝑥 squared divided by 𝑥 cubed. This is three over 𝑥.

This is not a polynomial. This tells us we’re done. We can see that the polynomial we’re left with has a lower degree than our divisor. So we’ve found our quotient 𝑞 of 𝑥 and our remainder 𝑟 of 𝑥. And this gives us our final answer. We were able to show our quotient 𝑞 of 𝑥 is equal to 𝑥 to the fourth power plus 𝑥 cubed minus 𝑥 squared minus 𝑥 and our remainder 𝑟 of 𝑥 is equal to three 𝑥 squared plus two 𝑥 plus one.

Let’s now talk about a special case for our divisor d of 𝑥. Using polynomial long division on 𝑝 of 𝑥 divided by d of 𝑥, we know we can find polynomials 𝑞 of 𝑥 and 𝑟 of 𝑥 such that 𝑝 of 𝑥 is equal to d of 𝑥 times 𝑞 of 𝑥 plus 𝑟 of 𝑥. We want to talk about the case when we’re dividing by the linear polynomial 𝑥 minus 𝑎. This means we have 𝑝 of 𝑥 is equal to 𝑥 minus 𝑎 times 𝑞 of 𝑥 plus 𝑟 of 𝑥. But remember, this means the degree of our divisor is equal to one. And we know our remainder term must have a degree less than our divisor term.

So in the case when we’re dividing by a linear polynomial, we know our remainder term must have degree zero. We can just write this as the constant 𝑟. And this gives us a useful result. Let’s see what would happen if we were to substitute 𝑥 is equal to 𝑎. Substituting 𝑥 is equal to 𝑎, we get 𝑝 of 𝑎 is equal to 𝑎 minus 𝑎 times 𝑞 of 𝑎 plus 𝑟. Of course, 𝑎 minus 𝑎 is equal to zero. So we’re just left with 𝑝 of 𝑎 is equal to 𝑟. And this is a useful result to help us find the remainder when we’re dividing by a linear polynomial. We call this the remainder theorem. So let’s formalize what we mean by the remainder theorem.

The remainder theorem tells us if we divide a polynomial 𝑝 of 𝑥 by a linear polynomial 𝑥 minus 𝑎, the remainder must be the constant 𝑝 evaluated at 𝑎. This is a useful result to help us find our remainder term. Another thing worth pointing out is what happens when 𝑝 of 𝑎 is equal to zero. Well, when 𝑝 of 𝑎 is equal to zero, the remainder theorem tells us our remainder must be equal to zero. But what does our remainder being equal to zero mean? Well, if our remainder is equal to zero, then we must have that 𝑝 of 𝑥 is equal to 𝑥 minus 𝑎 times 𝑞 of 𝑥.

In other words, we must have 𝑥 minus 𝑎 is a factor of our polynomial 𝑝 of 𝑥. And this is a well-known theorem. We call this the factor theorem. If 𝑝 of 𝑎 is equal to zero, then 𝑥 minus 𝑎 is a factor of 𝑝 of 𝑥. Similarly, if 𝑥 minus 𝑎 is a factor of 𝑝 of 𝑥, then its remainder upon division is equal to zero. Let’s now take a look at an example where we use the remainder theorem.

Find the remainder when three 𝑥 cubed minus two 𝑥 squared plus four 𝑥 plus five is divided by three 𝑥 plus four.

The question is asking us to find the remainder term when a cubic polynomial is divided by a linear polynomial. One way of doing this is by using polynomial long division. However, we know this is a long process. Instead, let’s notice we’re dividing by a linear polynomial. And we’re only asked to find the remainder term. This should remind us of the remainder theorem. We recall the remainder theorems tells us when 𝑝 of 𝑥 is divided by a linear polynomial 𝑥 minus 𝑎, then the remainder is constant and equal to 𝑝 evaluated at 𝑎.

We do have to be a little careful with how we use the remainder theorem in this case. We are dividing by a linear polynomial three 𝑥 plus four, but it’s not in the form 𝑥 minus 𝑎. So instead of actually doing our long division, let’s call our quotient polynomial 𝑞 of 𝑥 and our remainder polynomial 𝑟 of 𝑥. This means we’ll have three 𝑥 cubed minus two 𝑥 squared plus four 𝑥 minus five is equal to three 𝑥 plus four times 𝑞 of 𝑥 plus 𝑟 of 𝑥 for some polynomials 𝑞 of 𝑥 and 𝑟 of 𝑥. We can see our divisor is a linear polynomial; it has degree one. Our remainder must have a lower degree than our divisor. This means it must have degree zero. In other words, it’s a constant. We’ll call this constant 𝑟.

And at this point, there’s two similar ways of solving this equation. If we solve our linear factor is equal to zero, this gives us 𝑥 is equal to negative four over three. One way to find the value of 𝑟 is to substitute this value directly into this expression. Doing this, we get our cubic polynomial evaluated at negative four over three is equal to zero times 𝑞 evaluated at negative four over three plus 𝑟. But this simplifies to just give us 𝑟. And this is a perfectly valid way of solving this equation. However, we’re going to do this by taking a factor of three outside of our divisor. Doing this, we can rewrite this as three times 𝑥 plus four over three.

And now, we’re starting to see something interesting. Let’s consider three as part of 𝑞 of 𝑥. So what do we now have? We have our cubic polynomial is equal to 𝑥 plus four over three times some polynomial plus a constant. In fact, what we’ve done here is found an expression for our quotient polynomial when we divide our cubic by 𝑥 plus four over three. In other words, the remainder term when we divide by 𝑥 plus four over three or when we divide by three 𝑥 plus four are the same. This means we can just use the remainder theorem to find our value of 𝑟. And we’ll do this by evaluating our cubic polynomial at 𝑥 is equal to negative four over three. This gives us the following expression. And then, by evaluating this expression, we were able to show that our remainder term must be equal to negative 11.

Let’s now go over the key points of this video. First, we were able to show, by using a similar method to long division, we can divide two polynomials. We know dividing a polynomial by its factor will give us remainder zero. We also know that our remainder polynomial will always have a lower degree than our divisor polynomial. Finally, we learned about the remainder theorem, which tells us when a polynomial 𝑝 of 𝑥 is divided by the linear polynomial 𝑥 minus 𝑎, our remainder polynomial will be constant. And it will be equal to 𝑝 evaluated at 𝑎.