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

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

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### Video Transcript

In this video, we’ll learn how to perform long division with polynomials.

Polynomials — that’s an expression that involves terms with positive whole-number powers — can sometimes be divided using simple methods like factoring. If those methods fail, however, we can use long division. This process looks a lot like the process for long division with numbers. So, let’s begin by recalling how we perform long division with numbers.

Calculate 3531 divided by 11.

In this question, the number 3531 is called the dividend and the number 11 is the divisor. The result we get when we do this division is called the quotient. We’re going to set this problem up using a sort of bus stop design. As we see, the dividend goes inside the bus stop and the number 11 goes on the outside. The first thing we do is divide the first digit of 3531 by 11. Three divided by 11 is zero. So, we place a zero above the three, and any remainders at this stage are ignored. We then take this number, zero, and we multiply it by the divisor. Zero multiplied by 11 is zero. So we put a zero underneath the three.

Then, we subtract the digits in this column. Three take away zero is three. Next, we bring down the number five. We can, if we require, bring down all of the digits. But the first one is sufficient. And now we divide 35 by 11. 35 divided by 11 is three remainder two. We ignore the remainder for now and add the whole-number answer above five. Now, we multiply this number three by 11. Three times 11 is 33. So, we put 33 underneath 35. And once again, we subtract. 35 minus 33 is two. Now, we bring down the three, and we divide 23 by 11. The whole-number solution is two. Of course, there’s a remainder, but we do ignore it.

Now, we multiply that two by 11 to give us 22. We write this 22 underneath the 23, and then we subtract again. 23 minus 22 is one. And so, we bring down the final digit. And the last thing we do is 11 divided by 11. 11 divided by 11 is one. So, we add a one above the one in our dividend. Finally, we multiply this one by the 11. Now, this step might not seem necessary, but we do perform it just to check whether we have a remainder. One multiplied by 11 is 11. And when we subtract 11 from 11, we get zero. If this answer here is zero, that tells us there is no remainder. 11 is a factor of 3531. And when we divide 3531 by 11, we get 321.

So, we’re now going to see what this might look like when we perform polynomial long division; that’s dividing algebraic expressions.

Find the quotient when two 𝑥 cubed plus seven 𝑥 squared minus eight 𝑥 minus 21 is divided by two 𝑥 plus three.

A quotient is the result we get by dividing one number by another. The same stands for algebraic expressions. So, we’re going to divide this cubic expression by this linear expression. The dividend — that’s the number we’re going to be dividing into, and here that’s the cubic — goes inside the bus stop. The divisor, that’s two 𝑥 plus three, goes outside. Now, this looks a lot like dividing with numbers. But instead of dividing digit by digit, we divide term by term.

We’re going to begin by dividing two 𝑥 cubed by the term with the highest power of 𝑥 in our divisor, so two 𝑥 cubed divided by two 𝑥. Let’s recap that to divide two 𝑥 cubed by two 𝑥, we simplify by dividing through by two. And then if we consider 𝑥 as being 𝑥 to the power of one, we know that when we divide algebraic terms with the same base, we simply subtract their powers. So, we get 𝑥 to the power of three divided by 𝑥 to the power of one as being 𝑥 squared. 𝑥 squared sits above the two 𝑥 cubed in our calculations.

Now, we take that 𝑥 squared, and we multiply it by each part of our divisor. Two 𝑥 times 𝑥 squared is two 𝑥 cubed, and three times 𝑥 squared is three 𝑥 squared. We’re now going to subtract each term from this expression two 𝑥 cubed plus three 𝑥 squared from two 𝑥 cubed plus seven 𝑥 squared. Two 𝑥 cubed minus two 𝑥 cubed is zero. And this is a good way to check that the earlier calculation we did is likely to be correct. Seven 𝑥 squared minus three 𝑥 squared is four 𝑥 squared. Now, we don’t really need this zero here, but what we do need to do is bring down the next term in our dividend. We’re going to bring down negative eight 𝑥.

Now, we’re going to divide four 𝑥 squared by two 𝑥. Well, we can simplify by dividing through by a factor of two. And then, once again, if we consider 𝑥 as being 𝑥 to the power of one, we subtract the powers and we get two 𝑥. So, two 𝑥 goes above seven 𝑥 squared. Now, we multiply two 𝑥 by each term in our divisor. Two 𝑥 times two 𝑥 gives us the four 𝑥 squared. Then, three times two 𝑥 gives us positive six 𝑥. Once again, we perform some subtractions. Four 𝑥 squared minus four 𝑥 squared is zero. Although, of course, we don’t really need to write that.

Then, we take negative eight 𝑥, and we subtract positive six 𝑥. Subtracting a positive is the same as just subtracting six 𝑥. So, we get negative 14𝑥. Finally, we bring down our very last term. That’s negative 21. Now, we divide negative 14𝑥 by two 𝑥. This time, the 𝑥’s cancel, and we get negative 14 divided by two, which is negative seven. And so, we put negative seven above the negative eight 𝑥. And we’re nearly there; we’re now going to multiply negative seven by each term in our divisor. Negative seven times two 𝑥 plus three is negative 14𝑥 minus 21.

Now, we should always perform this final subtraction. This will tell us if we have a remainder or not. Negative 14𝑥 minus negative 14𝑥 is the same as negative 14𝑥 plus 14𝑥, which is zero. Negative 21 minus negative 21 is also zero. And so, the remainder when dividing our cubic by two 𝑥 plus three is zero. The quotient when we perform this division is 𝑥 squared plus two 𝑥 minus seven.

In our next example, we’ll see how polynomial division, like the one we’ve just performed, can help us to simplify fractions.

Use polynomial division to simplify six 𝑥 cubed plus five 𝑥 squared minus 20𝑥 minus 21 over two 𝑥 plus three.

One method we do have for simplifying algebraic fractions is to factor where necessary. It’s not particularly straightforward to factor this cubic on our numerator. So instead, we’re going to recall that this line in a fraction actually just means divide. And we’re going to use polynomial long division. Our dividend, that’s the numerator of our fraction, goes inside the bus stop. The divisor, that’s the denominator, goes on the outside.

And then, we remember the first thing that we do is we take the first term in our dividend, that’s six 𝑥 cubed, and we divide it by the first term in our divisor, that’s two 𝑥. Six divided by two is three. Then, if we consider 𝑥 as being 𝑥 to the power of one, we know that we can subtract these exponents. And 𝑥 cubed divided by 𝑥 to the power of one is 𝑥 squared. This means that six 𝑥 cubed divided by two 𝑥 must be three 𝑥 squared.

Our next step is to multiply three 𝑥 squared by each term in our divisor. Three 𝑥 squared times two 𝑥 is six 𝑥 cubed. Notice that this is the same as the first term in our dividend, so we know we’ve probably started this correctly. We then calculate three times three 𝑥 squared. Well, that’s nine 𝑥 squared. Our next step is to subtract six 𝑥 cubed plus nine 𝑥 squared from six 𝑥 cubed plus five 𝑥 squared. Six 𝑥 cubed minus six 𝑥 cubed is zero. We don’t really need to write this zero. And then, five 𝑥 squared minus nine 𝑥 squared is negative four 𝑥 squared. We bring down the next term. Some people bring down all of the terms, but I prefer to keep things a little bit simpler.

And we’re now going to divide negative four 𝑥 squared by two 𝑥. Negative four divided by two is negative two. 𝑥 squared divided by 𝑥 to the power of one is 𝑥. So, negative four 𝑥 squared divided by two 𝑥 is negative two 𝑥. And we add negative two 𝑥 above five 𝑥 squared in our problem. Now, we multiply negative two 𝑥 by each term in our divisor. Two 𝑥 multiplied by negative two 𝑥 is negative four 𝑥 squared, and negative two 𝑥 times three is negative six 𝑥.

We then subtract each of these terms from negative four 𝑥 squared minus 20𝑥. Negative four 𝑥 squared minus negative four 𝑥 squared is negative four 𝑥 plus four 𝑥 squared. So that’s zero, and we don’t really need to write that. We then do negative 20𝑥 minus negative six 𝑥. That’s negative 20𝑥 plus six 𝑥, which is negative 14𝑥. We bring down negative 21. And we’re now going to divide negative 14𝑥 by two 𝑥. 𝑥 divided by 𝑥 is just one. So, we get negative 14 divided by two, which is negative seven. So, we add negative seven here. And once again, we divide this number by each term in our divisor. Negative seven times two 𝑥 is negative 14𝑥, and negative seven times three is negative 21.

We do one final subtraction, and this is a really important step to do because it tells us whether there’s a remainder or not. In fact, negative 14𝑥 minus 21 minus itself is just zero. And so, we’ve completed the division. When we simplify our algebraic fraction, we’re left with three 𝑥 squared minus two 𝑥 minus seven.

Now, at this stage, it’s really useful just to discuss briefly how we might check our solution. We perform an inverse operation. We take our quotient, here that’s the solution to the division, and multiply that by the divisor, remembering, of course, that the divisor is the algebraic expression here that we’re dividing by. We multiply three 𝑥 squared minus two 𝑥 minus seven by two 𝑥 plus three. And when we do, we should get the numerator, or the dividend.

In our next example, we’ll consider how to use polynomial long division to find missing terms.

Find the value of 𝑘 that makes the expression 30𝑥 to the fifth power plus 57𝑥 squared minus 48𝑥 cubed minus 20𝑥 to the fourth power plus 𝑘 divisible by five 𝑥 squared minus eight.

Now, for this expression to be divisible by five 𝑥 squared minus eight, that tells us that five 𝑥 squared minus eight is a factor of it. If this is the case, then when we perform the division, we should have no remainder, or a remainder of zero. So, let’s perform the long division. Now, if we look carefully at our expression, we see that the terms appear to be in a bit of a funny order. They are usually written in decreasing powers of 𝑥. And so, we might be tempted to rearrange them and write it as 30𝑥 to the fifth power minus 20𝑥 to the fourth power, and so on. Doing it this way, we’ll make the problem a little bit easier, but it is going to look a little bit strange.

Let’s begin as normal. We divide the term in our dividend with highest power of 𝑥 by the term in the divisor, also with the highest power of 𝑥. 30 divided by five is six. Then, when we’re dividing numbers whose bases are the same, we subtract their exponents. So, 𝑥 to the fifth power divided by 𝑥 squared is 𝑥 to the power of five minus two, which is 𝑥 cubed. This means that 30𝑥 to the fifth power divided by five 𝑥 squared is six 𝑥 cubed. And so, we write six 𝑥 cubed above this term. We now multiply this value by each term in our divisor. Six 𝑥 cubed times five 𝑥 squared gives us 30𝑥 to the fifth power. And then, when we multiply six 𝑥 cubed by negative eight, we get negative 48𝑥 cubed.

Now we’re going to line that up directly underneath the 𝑥 cubed terms. And we’re going to add in another term; we’re going to add in zero 𝑥 to the fourth power. This isn’t entirely necessary, but it can make it a little bit easier to follow what happens next. Our next step is to divide each of these three terms by the corresponding terms above. 30𝑥 to the fifth power minus 30𝑥 to the fifth power is zero. Negative 20𝑥 to the fourth power minus zero 𝑥 to the fourth power is negative 20𝑥 to the fourth power. And negative 48𝑥 cubed minus negative 48𝑥 cubed is zero.

Next, we bring down 57𝑥 squared. And we’re now going to divide negative 20𝑥 to the fourth power by five 𝑥 squared. Negative 20 divided by five is negative four, and 𝑥 to the fourth power divided by 𝑥 squared is just 𝑥 squared. And so, when we do this division, we get negative four 𝑥 squared. And this is the next term in our quotient. Remember, each time we find a term in our quotient, we multiply it by each part of the divisor. So, we’re going to work out negative four 𝑥 squared times five 𝑥 squared, which is negative 20𝑥 to the fourth power.

When we multiply negative eight by negative four 𝑥 squared, we get 32𝑥 squared. So, we can line this up directly under 57𝑥 squared. And then, we subtract each of these terms from the terms immediately above them. Negative 20𝑥 to the fourth power minus negative 20𝑥 to the fourth power is zero. Then, 57𝑥 squared minus 32𝑥 squared is 25𝑥 squared. And we’re nearly there. We bring down the final term; that’s the 𝑘.

Now, don’t worry too much that we don’t yet know what 𝑘 is. Remember, we’re looking for a remainder of zero. We’re going to divide 25𝑥 squared by five squared. Well, 25 divided by five is five, and 𝑥 squared divided by 𝑥 squared is one. So, the final term in our quotient is five. We take that five and we multiply it by five 𝑥 squared and negative eight. And that gives us 25𝑥 squared minus 40.

Now, we know that since we’re subtracting the final two terms, we’re going to get a remainder of zero. So, 25𝑥 squared plus 𝑘 minus 25𝑥 squared minus 40 has to give us zero. Well, 25𝑥 squared minus 25𝑥 squared is also zero. And so to ensure that our remainder is zero, and thus our expression is divisible by five 𝑥 squared minus eight, we can say that 𝑘 minus negative 40 itself must be equal to zero. 𝑘 minus negative 40 is, of course, 𝑘 plus 40.

And so let’s subtract 40 from both sides to solve for 𝑘. That gives us 𝑘 is equal to negative 40. And so, the value of 𝑘 that makes our expression divisible by five 𝑥 squared minus eight is 𝑘 equals negative 40. Note that at this point, we could go ahead and multiply our quotient. That’s six 𝑥 cubed minus four 𝑥 squared plus five by five 𝑥 squared minus eight. If we had performed the calculation correctly, we would end up with the expression 30𝑥 to the fifth power plus 57𝑥 squared minus 48𝑥 cubed minus 20𝑥 to the fourth power minus 40.

We’ll now recap some of the key points from this lesson. In this video, we saw that to perform polynomial long division, we perform a similar process to working with numbers. Instead of dividing each digit in turn, we actually divide each term in turn. We also saw that if we get a remainder of zero, in other words, our very final subtraction yields zero, that the divisor is a factor of the dividend. In other words, the dividend is exactly divisible by the divisor. And finally, we saw that we can check any division calculations by using inverse operations. In other words, we take the quotient, that’s our result, and multiply it by the divisor, that’s the algebraic expression we’re dividing by. If we get the dividend, then we know we’ve performed our calculation correctly.