LESSON 3:THE DIVISION OF POLYNOMIALS

A-SSE.A.2 MP.7

A-APR.C.4 MP.8

Opening Exercise

5 minutes, with explanation (2 slides)

a. Multiply these polynomials using the tabular method:

b. How can you use the expression in part (a) to quickly multiply ?

OPENING EXERCISE - ANSWER

a. Multiply these polynomials using the tabular method:

The product is

b. How can you use the expression in part (a) to quickly multiply ?

If we let , becomes and becomes .Since we know the product ,replace with and we can do:

DISCUSSION (QUESTIONS FOR CONSIDERATION)

Let’s look again at :

• How can a multiplication problem be written as a division problem?

• How can we rewrite the Opening Exercise as a division problem?

• Let . Substitute that value into each polynomial, and compare the results of multiplying and dividing the polynomials with the arithmetic problem. How do polynomial multiplication and polynomial division compare to multiplication and division of integers?

EXPLORATORY CHALLENGE

Using the tabular method, can we demonstrate division instead of multiplication?

1. Does ? Justify your answer.

STRUGGLE WITH THIS. WORK WITH IT UNTIL YOU CAN’T WORK ANY

MORE, AND THEN WORK WITH IT SOME

MORE.

Going ahead prematurely will only

hurt YOU.

PROMPTING QUESTIONS

The quotient will be the polynomial that would go along the top of the table. Remember we know it should be . (In the future, you wouldn’t know the answer ahead of time, but let’s use this fact that we DO know to get started). How can we get started to confirm this using only the dividend and divisor (the two remaining terms)?

Are there any cells in the table we can fill in based on the information we have? What about the top-left cell, for instance?

What must the first term in the line above the table be?

What goes in the rest of the cells in the first column? How can we continue using this pattern to fill in the remaining cells?

Compare your work on this problem with the Opening Exercise. How could you verify that really is the quotient? Explain.

SOLVING THE EXPLORATORY CHALLENGE

Remember, during multiplication ourFACTORS would go across the top and right sides.

In division, one of the factors will be our divisor andthe other one becomes our quotient.

Setting up the larger expression as the dividend is thetricky part! Just work backwards.

Remember that the results for what became ourproduct (now the dividend) were the sum of all thecells in that diagonal.

SOLVING THE EXPLORATORY CHALLENGE

Sample working through:

We know both the top left corner and the bottom right cornerbecause both of those boxes are the ONLY box along theirrespective diagonal.

MUST be pointing only to the top left box,and MUST be point only to the lower right.

SOLVING THE EXPLORATORY CHALLENGE

This next part is a little weirder. We know that diagonal has to add to, but how do we know that it’s a ? Why isn’t it a and ?

Well, that’s actually a result of the work we did before. Rememberwhen we filled in that ? For that cell to have worked, we neededthe column factor multiplied by the row factor to be .

The question now becomes “What times gives me ?”Our answer, must be the other factor. Put that on top.

Now we can fill out our lower left corner. We know that cell must be, or .

If that cell is and its diagonal is a sum of , the next box along thatdiagonal must be .

SOLVING THE EXPLORATORY CHALLENGE

Using that same logic, we can now figure out what goes above themiddle column.

We know that it must be a factor such that “Something times gives us .”

That something must be .

Now we can use the to fill out the middle, bottom cell.

Try this before you move on!

SOLVING THE EXPLORATORY CHALLENGE

Did you get the ?

What must happen along that diagonal ifthe whole thing sums to ?

Awesome! The whole process is almost done.Now we just need to figure out, “What times gives me ?”The answer, of course, is …

SOLVING THE EXPLORATORY CHALLENGE

One!

Now as our last step, we rewrite the terms with their appropriate signs(in this case everything was positive, but there are times where we mayhave negative results along the top; keep the signs together).

2𝑥3+15 𝑥2+27 𝑥+52 𝑥+5

=𝒙𝟐+𝟓 𝒙+𝟏

EXERCISE 3

Using the same method (Reverse Tabular Method of Multiplication) from before, find the quotient of .

Remember, our answer will be along the top (just like normal long division).

The divisor goes along the side.

The dividend is spread out over the diagonals.

Try to set this up before you move on!

EXERCISE 3 (SOLUTION)

Set up: Fill out the boxes you know: Use multiplication to discover the top:

ALRIGHT, THIS IS WEIRD…

Now we have a way to solve the table, but what happens when we don’t know how big the table is?

This is one of those things that you’re going to feel like you have to memorize; you don’t. With a little bit of experience, it’ll be obvious why this works out to what it does.

Let’s look at the following problem:

There are two MAJOR things you should notice about that problem (and what you’re looking for, moving forward):

1) We can tell the degree of both of those polynomials. The numerator is degree , and the denominator is degree .

2) There are gaps in the terms in the denominator. We see and an term, but no term. This doesn’t mean it isn’t there, it just means there are zero of them! It’ll help us during division to never lose those “missing” terms.

Let’s treat the problem like this:

SO HOW DO WE SET UP THE TABLE?

Here’s the weird part (and you thought we already covered the weird, didn’t you?):

The number of rows will be ONE MORE than the degree of the divisor.

The number of columns will be ONE MORE than the difference of the degrees of the dividend and the divisor.

That top one should make sense: This was the whole reason we added that thing. If we have a polynomial ofdegree , we expected terms (). Likewise, for any polynomial of degree , there are terms.

The bottom one is trickier when you read it like above, but again, it should make sense: Think of the VERY first thingwe should be able to know on the table – the top left cell. It’s always going to be the first term in the numerator. Thisalso causes you to always do the first term divided by the first term (in this case it’ll be ).Well, the result of that is , which means my quotient is degree and should have… terms. This means columns.

At this point you can attempt the example, or move on to see the table.

EXAMPLE/EXERCISE 4

IN YOUR NOTES

Please reflect, in your NOTE books, on the following questions. These will help drive our small group discussions:

• What strategies were helpful when you set up and solved these problems? What patterns did you notice as you solved these problems?

• What happens to the degree of the product when you multiply two polynomials?

• What happens to the degree of the quotient when you divide two polynomials?

• What happens to the leading coefficient when you multiply or divide polynomials? (We didn’t explicitly cover this one, but think about it in the context of your previous answers)