I don't understand polynomial division

[JohnnyIngrim]

At first he shows 2x+4 / 2 and you just divide both 2x and 4 by 2. But then in the next example he is dividing x^2+3x+6 by x+1 and he doesn't divide x^2 by x+1, 3x by x+1 and 6 by x+1. I do not understand how he does the problem.

 

[Mark44]

At first he shows 2x+4 / 2 and you just divide both 2x and 4 by 2. But then in the next example he is dividing x^2+3x+6 by x+1 and he doesn't divide x^2 by x+1, 3x by x+1 and 6 by x+1. I do not understand how he does the problem.

The first step is to divide x² by x, resulting in a partial quotient of x. To see how close this is, multiply that result by x + 1, to get x² + x.

Next, you subtract x² + x from x² + 3x, which results in 2x.

Bring down the next term, 6, and tack it onto the 2x, resulting in 2x + 6.

Divide 2x + 6 by x, which results in 2. See how close you got by multiplying 2 times x + 1 to get 2x + 2.

Subtract 2x + 2 from 2x + 6 to get 4, which is the final remainder. Since the degree of 4 is zero and the degree of x + 1 is 1, you stop.

Polynomial division is sort of like ordinary division of numbers. Take as an example 155 divided by 12.

To make this look a bit more like polynomial division, write 12 as 10 + 2, and write 155 as 100 + 50 + 5.
First, divide 100 by 10, which is 10. Write this 10 above the 50. Check how close this was by multiplying 10 times (10 + 2) which is 100 + 20.

Subtract 100 + 20 from the line above, which is 100 + 50. The difference is 30.

Bring down the final number, 5.

Divide 30 + 5 by 10, to get 3. To check how close we are, multiply 10 + 2 by 3 to get 30 + 6. Since this is larger than the number on the line above (30 + 5), we need to revise our partial answer downward. Trying again, we get a new partial answer of 2. Check by multiplying 10 + 2 by 2 to get 20 + 4. Subtract 20 + 4 from 30 + 5, which is 10 + 1 (or 11). Since 10 + 1 is smaller than our divisor, we're done.

We have shown that 155 divided by 12 is 12, with a remainder of 11.

 

[BvU]

Compare with the long division of 136 by 11: you don't start with 100/11, continue with 30/11 and end with 6/11, right ?

(So I just grabbed an easy value for x, namely 10).

What you do is you determine the coefficient (call it a) for the highest power of x in the quotient: (x²+3x+6 )/(x+1) = (ax + ...) and you subtract what you have so far from the dividend :

(x²+3x+6 )/(x+1) = ax + (x²+3x+6 ) - ax * (x+1) ) / (x+1)

Of course here a = 1, so you continue with (x²+3x+6 )/(x+1) = ax + (2x+6 ) / (x+1)Well well, two simultaneous responses!

 

[SteamKing]

At first he shows 2x+4 / 2 and you just divide both 2x and 4 by 2. But then in the next example he is dividing x^2+3x+6 by x+1 and he doesn't divide x^2 by x+1, 3x by x+1 and 6 by x+1. I do not understand how he does the problem.

In the first example, you can factor 2x + 4 into 2 * (x+2). If you then divide 2 * (x + 2) by 2, the twos cancel, leaving x+2 as the quotient. This is similar to dividing 75 by 5: you can write 75 as 15 * 5, which when divided by 5, leaves 15 as the quotient.

Now, dividing x² + 3x + 6 by x + 1 is not as simple, because the factors of x² + 3x + 6 are not immediately obvious. Here, one must fall back on the long division algorithm, where you find out how many times the divisor, x + 1, will go into the dividend, x² + 3x + 6. In this case, x + 1 goes into the dividend x times:

                      x
                    _______________
               x+1  ) x^2 + 3x + 6
                      x^2 + x
                    ---------------
                           2x + 6

The first part of the quotient is x with 2x+6 left over. Since this remainder is of at least equal degree as the divisor, a further partial division can be carried out:

                      x+2
                    _______________
               x+1  ) x^2 + 3x + 6
                      x^2 +  x
                    ---------------
                            2x + 6
                            2x + 2
                    ---------------
                                 4

So, x² + 3x + 6 divided by x+1 has a quotient of x+2 with remainder 4. You can always check the division by multiplying the quotient times the divisor and adding any remainder to the result. The answer should be the original dividend.

Perhaps you should practice first by doing long division using numbers, and then try it with polynomials.

 

[JohnnyIngrim]

The first step is to divide x² by x

Why? Isn't x+1 supposed to go in?

, resulting in a partial quotient of x. To see how close this is, multiply that result by x + 1, to get x² + x.

What do you mean "how close" it is? How close it is to what?

Compare with the long division of 136 by 11: you don't start with 100/11, continue with 30/11 and end with 6/11, right ?

No, you divide it into 13 first and carry 2 over onto the 6. I didn't get the rest of your post.

In the first example, you can factor 2x + 4 into 2 * (x+2). If you then divide 2 * (x + 2) by 2, the twos cancel, leaving x+2 as the quotient. This is similar to dividing 75 by 5: you can write 75 as 15 * 5, which when divided by 5, leaves 15 as the quotient.

Now, dividing x² + 3x + 6 by x + 1 is not as simple, because the factors of x² + 3x + 6 are not immediately obvious. Here, one must fall back on the long division algorithm, where you find out how many times the divisor, x + 1, will go into the dividend, x² + 3x + 6. In this case, x + 1 goes into the dividend x times:

What do you mean it goes into the dividen x times? In the previous example you never mentioned a dividend.

 

[Mark44]

Compare with the long division of 136 by 11: you don't start with 100/11, continue with 30/11 and end with 6/11, right ?

Assuming that you were replying to my post, actually I do start with 100/11, then 30/11, and finally 6/11, but only to draw a parallel to polynomial division.

 

[RUber]

You can't divide $x^2$ by $x+1$ evenly, so you are only comparing the highest order terms at anyone time.
You know that $(x+1)*x = x^2+x$, so you put that in as your best guess. In any case, you won't have any $x^2$ terms left over.
Then you move on to whatever is left over.
You can repeat this process as many times as you want, but most people finish with the $\frac 1{x+1}$ term, or the $x^{-1}$ place.
The goal at each step is to eliminate the highest order term in your numerator.

 

[JohnnyIngrim]

You can't divide $x^2$ by $x+1$ evenly, so you are only comparing the highest order terms at anyone time.

I don't get it... What do you mean comparing the highest order terms at anyone time?

You know that $(x+1)*x = x^2+x$, so you put that in as your best guess. In any case, you won't have any $x^2$ terms left over.

Best guess at what? How many times x+1 goes in? Like you said, it doesn't go in x^2+x times.

Then you move on to whatever is left over.

Whatever is left over of what where?

You can repeat this process as many times as you want, but most people finish with the $\frac 1{x+1}$ term, or the $x^{-1}$ place.
The goal at each step is to eliminate the highest order term in your numerator.

This is so confusing and I don't get it.

 

[SteamKing]

What do you mean it goes into the dividen x times? In the previous example you never mentioned a dividend.

I'm sorry. I thought these terms were clear, or could at least be looked up.

For an expression like a = b/c, b is called the dividend, c is called the divisor, and a is called the quotient. These terms are the same regardless of whether one is dividing numbers or polynomials.

Like I said. If you want to understand polynomial division, practice using the long division algorithm on numbers first.

 

[Mark44]

The first step is to divide x² by x

Why? Isn't x+1 supposed to go in?

It's much easier to figure out how many times x goes into x², isn't it? And x is reasonably close to x + 1.

, resulting in a partial quotient of x. To see how close this is, multiply that result by x + 1, to get x² + x.

What do you mean "how close" it is? How close it is to what?

We're actually dividing x² + 3x (the first two terms of the dividend, the thing that we're dividing) by x + 1 (the divisor), and we got a partial answer of x. To see how close this answer is to the correct answer, we multiply our partial quotient (x) by the divisor (x + 1), and get x² + x. This isn't quite the same as x² + 3x, so we subtract x² + x from x² + 3x to see how close our answer was. If it was spot on, the two quantities would have been the same and we would have gotten 0 instead of 2x that you see below.

                             x
                    _______________
               x+1  ) x^2 + 3x + 6
                      x^2 + x
                    ---------------
                           2x

Since our remainder, 2x, is the same degree as the divisor (x + 1) - both are degree 1, we're not done yet.

We continue by bringing down the next term from the dividend (+6)

                             x
                    _______________
               x+1  ) x^2 + 3x + 6
                      x^2 + x
                    ---------------
                           2x + 6

Now we divide 2x in the bottom line by x in the divisor, and get 2, and write that at the top (in the quotient).

                             x + 2
                    _______________
               x+1  ) x^2 + 3x + 6
                      x^2 + x
                    ---------------
                           2x + 6

Just as before, we see how close our partial answer is to the actual answer by multiplying the 2 in the quotient by the divisor (x + 1), to get 2x + 2.

                             x + 2
                    _______________
               x+1  ) x^2 + 3x + 6
                      x^2 + x
                    ---------------
                           2x + 6
                           2x + 2
                          ---------
                                4

How close did we come? We should have gotten exactly 2x + 6, but we instead got 2x + 2. If we subtract again (which is always how you check how close two things are) we get a final remainder of 4. Since the degree of 4 is 0 (there are no x terms) and the degree of the divisor is 1, we are done.

What this shows is that x² + 3x + 6 = (x + 1)(x + 2) + 4. Since there is a remainder of 4, this means that x + 1 does not divide x² + 3x + 6 evenly.
It also shows that
$$\frac{x^2 + 3x + 6}{x + 1} = x + 2 + \frac{4}{x + 1}$$

In the first example, you can factor 2x + 4 into 2 * (x+2). If you then divide 2 * (x + 2) by 2, the twos cancel, leaving x+2 as the quotient. This is similar to dividing 75 by 5: you can write 75 as 15 * 5, which when divided by 5, leaves 15 as the quotient.
Now, dividing x² + 3x + 6 by x + 1 is not as simple, because the factors of x² + 3x + 6 are not immediately obvious. Here, one must fall back on the long division algorithm, where you find out how many times the divisor, x + 1, will go into the dividend, x² + 3x + 6. In this case, x + 1 goes into the dividend x times:

What do you mean it goes into the dividen x times? In the previous example you never mentioned a dividend.

 

[Mark44]

Like I said. If you want to understand polynomial division, practice using the long division algorithm on numbers first.

Yes.

 

[RUber]

Think about when you learned how to FOIL or factor simple polynomials:
$(x+2)(x+4)=x^2+6x+8$
To go in reverse and factor, you know that the leading terms are probably x, since $x*x=x^2$.
(x+a) (x+b), then you look for a and b such that ab=8 and a+b=6.
If you were given one of the factors, say, (x+2), you might be able to just figure out what the other one was.
You could also do some long division.
$\frac{x^2+6x+8}{x+2} =\frac{x^2+6x}{x+2}+\frac{8}{x+2}$
$=x\frac{x+6}{x+2}+\frac{8}{x+2}=x\left(\frac{x+2}{x+2}+\frac{4}{x+2}\right)+\frac{8}{x+2}$
$=x\left(1+\frac{4}{x+2}\right)+\frac{8}{x+2}=x+\frac{4x+8}{x+2}=x+4$
The long division framework is a more efficient way to keep the algebra clear.

 

[JohnnyIngrim]

I'm sorry. I thought these terms were clear, or could at least be looked up.

For an expression like a = b/c, b is called the dividend, c is called the divisor, and a is called the quotient. These terms are the same regardless of whether one is dividing numbers or polynomials.

Like I said. If you want to understand polynomial division, practice using the long division algorithm on numbers first.

Yeah I still don't understand your explanation.

 

[JohnnyIngrim]

It's much easier to figure out how many times x goes into x², isn't it? And x is reasonably close to x + 1.

What do you mean reasonably close? If I divide 12 by 6 I don't divide 12 by 5 because 5 is reasonably close to 6.

We're actually dividing x² + 3x (the first two terms of the dividend, the thing that we're dividing) by x + 1 (the divisor), and we got a partial answer of x.

How is it a 'partial answer'? I don't get it.

 

[JohnnyIngrim]

Think about when you learned how to FOIL or factor simple polynomials:
$(x+2)(x+4)=x^2+6x+8$
To go in reverse and factor, you know that the leading terms are probably x, since $x*x=x^2$.

(x+a) (x+b), then you look for a and b such that ab=8 and a+b=6.
If you were given one of the factors, say, (x+2), you might be able to just figure out what the other one was.
You could also do some long division.
$\frac{x^2+6x+8}{x+2} =\frac{x^2+6x}{x+2}+\frac{8}{x+2}$
$=x\frac{x+6}{x+2}+\frac{8}{x+2}=x\left(\frac{x+2}{x+2}+\frac{4}{x+2}\right)+\frac{8}{x+2}$
$=x\left(1+\frac{4}{x+2}\right)+\frac{8}{x+2}=x+\frac{4x+8}{x+2}=x+4$
The long division framework is a more efficient way to keep the algebra clear.

I don't get how talking about factoring led to the long division way.

 

[Mark44]

It's much easier to figure out how many times x goes into x², isn't it? And x is reasonably close to x + 1.

What do you mean reasonably close? If I divide 12 by 6 I don't divide 12 by 5 because 5 is reasonably close to 6.

But you could write 6 as 5 + 1 and 12 as 10 + 2. Then this division problem would look more like the polynomial division we're trying to explain here.

                           2
                    ________
               5+1  ) 10 + 2
                      10 + 2
                    ---------
                           0

First, divide 10 by 5, getting 2, which I place at the top. At this point, 2 is the partial quotients.,
Next, multiply 2 times (5 + 1), which gives me 10 + 2. Write that below the dividend (the thing that's being divided.
Subtract: 10 + 2 - (10 - 2), which is 0.
We're done.

We're actually dividing x² + 3x (the first two terms of the dividend, the thing that we're dividing) by x + 1 (the divisor), and we got a partial answer of x.

How is it a 'partial answer'? I don't get it.

x is a partial answer because we're not done yet - we don't have the complete answer. Look at the sequence of steps that I laid out in the code blocks in post #10, and reread the explanations that I give with them. The first partial quotient was just x. We keep refining the process, getting more terms in the quotient until we're done, with a final quotient of x + 2, plus a remainder.

 

[JohnnyIngrim]

I was taught how to do long division in school but I don't understand it even after watching 5-10 videos and reading about it now.

Like in this video:


You put 8 in the 10s place. That implies it's 80 yes, but it's not 80 as we can see when it's finished. So why leave out the 6? It's 86, not 8, not 80. Also, in the middle of it, he multiplies 8 by 4 and gets 32 and brings it under the 34. Why?

 

[RUber]

You put 8 in the 10s place. That implies it's 80 yes, but it's not 80 as we can see when it's finished. So why leave out the 6? It's 86, not 8, not 80. Also, in the middle of it, he multiplies 8 by 4 and gets 32 and brings it under the 34. Why?

Yes, 8 in the 10s place is 80. 6 in the ones place is 6. 80+6 is 86.

Long division is all about sequential approximation.

for example if you have: 537/32
Starting with 10s, 20x32 is too much, 10x32 is a good starting point. (If you wanted to start with 12 or 15 it wouldn't hurt the process, but in long division you normally do one place column at a time. )
537 = 320 + 217.
So I know that 537/32 = 10 + 217/32.
I also know that 217/32 is less than 10.

I know that 3*7 = 21, but 32*7 = 224, too much, so I go with 6.
6* 32 = 192, 217-192 =25.
So 217 = 6*32 + 25.
That means that 537 = 10*32 + 6*32 + 25.
I can now write $\frac{537}{32}= \frac{10*32}{32}+\frac{6*32}{32}+\frac{25}{32} = 16 + \frac{25}{32}$

That is how long division works. If you don't enjoy the bookkeeping methods with the framework, then just break it up into fractions. Then you see all the pieces as you are putting them together.

 

[JohnnyIngrim]

Been away from this thread for a while because I gave up. I'm going to try again.

Yes, 8 in the 10s place is 80. 6 in the ones place is 6. 80+6 is 86.

Long division is all about sequential approximation.

All that came up when I googled that was some crazy stuff like this:

One of the general methods for the approximate solution of equations. In many cases the good convergence properties of the approximations constructed by this method allow one to apply it to practical computations.

Let

be some set and

an operator (not necessarily linear) on this set mapping it into itself. Suppose one has to find a fixed point of this mapping, i.e. a solution of the equation

I don't understand any of that and hence I don't understand sequential approximation.

for example if you have: 537/32
Starting with 10s, 20x32 is too much, 10x32 is a good starting point. (If you wanted to start with 12 or 15 it wouldn't hurt the process, but in long division you normally do one place column at a time. )
537 = 320 + 217.
So I know that 537/32 = 10 + 217/32.
I also know that 217/32 is less than 10.

I know that 3*7 = 21, but 32*7 = 224, too much, so I go with 6.
6* 32 = 192, 217-192 =25.
So 217 = 6*32 + 25.
That means that 537 = 10*32 + 6*32 + 25.
I can now write $\frac{537}{32}= \frac{10*32}{32}+\frac{6*32}{32}+\frac{25}{32} = 16 + \frac{25}{32}$

That is how long division works. If you don't enjoy the bookkeeping methods with the framework, then just break it up into fractions. Then you see all the pieces as you are putting them together.

I think this is similar to a set of videos I saw called the 'partial quotient method of division'. I am pretty sure I understand this but I don't understand how it explains the problem in the first post of the thread. For example, how would this work if it was 532 + 5 / 29 + 3?

 

[rcgldr]

sequential approximation

Polynomial division isn't the same as sequential approximation. Also using long hand division of decimal numbers as an example of polynomial division isn't very good either, because decimal long hand division involves borrows from higher order terms, and polynomial division does not involve borrows. It may help you to understand if you realize that only the most significant term of the divisor and the current divivdend is needed for each divide step, and that each quotient term is exact, not an approximation.

                       x + 2
            ________________
     x + 1  |   x^2 + 3x + 6

                x^2            (bring down the x^2)
                                      (x^2 / x = x)
             -( x^2 +  x )
               ----------
                     -1x
                   +  3x       (bring down the  3x)
                    ----
                      2x
                                       (2x / x = 2)
                   -( 2x + 2)
                    ---------
                         - 2
                       +   6   (bring down the   6)
                         ---
                           4

 

[JohnnyIngrim]

Polynomial division isn't the same as sequential approximation. Also using long hand division of decimal numbers as an example of polynomial division isn't very good either, because decimal long hand division involves borrows from higher order terms, and polynomial division does not involve borrows. It may help you to understand if you realize that only the most significant term of the divisor and the current divivdend is needed for each divide step, and that each quotient term is exact, not an approximation.

I don't understand what you mean by this sentence

                       x + 2
            ________________
     x + 1  |   x^2 + 3x + 6

                x^2            (bring down the x^2)
                                      (x^2 / x = x)
             -( x^2 +  x )
               ----------
                     -1x
                   +  3x       (bring down the  3x)
                    ----
                      2x
                                       (2x / x = 2)
                   -( 2x + 2)
                    ---------
                         - 2
                       +   6   (bring down the   6)
                         ---
                           4

That's what it said in the video but I don't understand why it works.

 

[rcgldr]

Assume that the most significant term of the dividend is: a x^n, and that the most signficant term of the divisor is b x^m, with m <= n, then the most significant term of the quotient will be: a/b x^(n-m). Then in the multiply and s++ubtract step, you have a x^n - ( a/b x^(n-m) ) (b x^m) = a x^n - a x^n = 0. This is the only quotient term that will clear out the most significant term of the dividend. After the multiply and subtract step, the next most signficant term will be some value times x^(n-1), depending on the result of the multiply and subtract step. You continue to do this until the most significant term in what's left of the dividend is some value times x^m.

Perhaps it would help to show a series of equations for (x^2 + 3x + 6) / ( x+1)

x^2 + 3x + 6 = (x) ( x+1) + (2x + 6)
x^2 + 3x + 6 = (x + 2) ( x+1) + (4)

 

[ehild]

You can write the divident x^2+3x+6 in the following way: (x^2+x)+(2x+2)+4, which is equal to x(x+1)+2(x+1)+4.
Divide by x+1: if you divide a sum, you have to divide all terms. So $\frac{x^2+3x+6}{x+1}=\frac{x(x+1)}{x+1}+\frac{2(x+1)}{x+1}+\frac{4}{x+1}=x+2+\frac{4}{x+1}$

 

[JohnnyIngrim]

Assume that the most significant term of the dividend is: a x^n, and that the most signficant term of the divisor is b x^m

What do you mean the most significant term and why does it matter here? Google returned nothing useful: https://www.google.ie/search?q=most significant term

with m <= n, then the most significant term of the quotient will be: a/b x^(n-m).

How do the previous things lead to this conclusion? Also, what do you mean a/b x^(n-m)? Do you mean a divided by the rest, or do you mean a divided by b multiplied by the rest, or what?

Then in the multiply and s++ubtract step, you have a x^n - ( a/b x^(n-m) ) (b x^m) = a x^n - a x^n = 0.

I don't know what this expression means.

This is the only quotient term that will clear out the most significant term of the dividend.

What do you mean 'clear out' and why is that the objective?

After the multiply and subtract step, the next most signficant term will be some value times x^(n-1), depending on the result of the multiply and subtract step. You continue to do this until the most significant term in what's left of the dividend is some value times x^m.

Don't get this for the same reasons as before.

Perhaps it would help to show a series of equations for (x^2 + 3x + 6) / ( x+1)

x^2 + 3x + 6 = (x) ( x+1) + (2x + 6)
x^2 + 3x + 6 = (x + 2) ( x+1) + (4)

I do not understand this either.

You can write the divident x^2+3x+6 in the following way: (x^2+x)+(2x+2)+4, which is equal to x(x+1)+2(x+1)+4.
Divide by x+1: if you divide a sum, you have to divide all terms. So $\frac{x^2+3x+6}{x+1}=\frac{x(x+1)}{x+1}+\frac{2(x+1)}{x+1}+\frac{4}{x+1}=x+2+\frac{4}{x+1}$

That seems to make sense but I don't understand the long division way.[/quote]

 

[vela]

What do you mean the most significant term and why does it matter here? Google returned nothing useful: https://www.google.ie/search?q=most significant term

The term with the highest exponent.

Also, what do you mean a/b x^(n-m)? Do you mean a divided by the rest, or do you mean a divided by b multiplied by the rest, or what?

Following the standard rules for expressions, it's (a/b) times x^(n-m). If rcgldr meant the latter, he would have written a/[b x^(n-m)].

x^2 + 3x + 6 = (x) ( x+1) + (2x + 6)
x^2 + 3x + 6 = (x + 2) ( x+1) + (4)

Here's the first step in gory detail:
\begin{align*}
x^2 + 3x + 6 &= (x^2 + 3x + 6) + x(x+1) - x(x+1) \\
&= x(x+1) + [(x^2+3x+6) - (x^2+x)] \\
&= x(x+1) + [2x+6]
\end{align*} In the first step, I added and subtracted the same thing, and then I regrouped the terms in the second step. Finally, I simplified.

Now compare this to dividing, say, 7 by 2. Two goes into seven three times, leaving a remainder of 1.
\begin{align*}
7 &= 7 + 3 \times 2 - 3 \times 2 \\
&= 3 \times 2 + [7 - 3 \times 2] \\
&= 3 \times 2 + 1
\end{align*} Same steps. I added and subtracted the same thing, regrouped terms, and simplified. The quotient is 3 and the remainder is 1.

In the polynomial case, after the first step, the quotient is $x$ and the remainder is $2x+6$. Note that highest power in the remainder is now $x$ whereas we started with a quadratic. The idea is to choose the quotient so that the highest-power terms cancel out so that the remainder is a lower-degree polynomial.

 

[JohnnyIngrim]

The term with the highest exponent.

And why does it matter? Why are we doing this?

Following the standard rules for expressions, it's (a/b) times x^(n-m). If rcgldr meant the latter, he would have written a/[b x^(n-m)].Here's the first step in gory detail:
\begin{align*}
x^2 + 3x + 6 &= (x^2 + 3x + 6) + x(x+1) - x(x+1) \\
&= x(x+1) + [(x^2+3x+6) - (x^2+x)] \\
&= x(x+1) + [2x+6]

He just did this in his head? And why?

\end{align*} In the first step, I added and subtracted the same thing, and then I regrouped the terms in the second step. Finally, I simplified.

Now compare this to dividing, say, 7 by 2. Two goes into seven three times, leaving a remainder of 1.
\begin{align*}
7/2 &= 7 + 3 \times 2 - 3 \times 2 \\
&= 3 \times 2 + [7 - 3 \times 2] \\
&= 3 \times 2 + 1
\end{align*} Same steps. I added and subtracted the same thing, regrouped terms, and simplified. The quotient is 3 and the remainder is 1.

I got lost at the first line. 3x2 - 3x2 = 0 right? How can 7/2 be equal to 7-0? Also, what has this to do with long division?

In the polynomial case, after the first step, the quotient is $x$ and the remainder is $2x+6$.

How is the remainder 2x+6 and the quotient x? I didn't see any division.

Note that highest power in the remainder is now $x$ whereas we started with a quadratic. The idea is to choose the quotient so that the highest-power terms cancel out so that the remainder is a lower-degree polynomial.

What do you mean they cancel out and why is this the idea?

 

[vela]

And why does it matter? Why are we doing this?

You asked what "most significant term" meant.

He just did this in his head? And why?

To attempt to explain to you how polynomial long division works.

I got lost at the first line. 3x2 - 3x2 = 0 right? How can 7/2 be equal to 7-0? Also, what has this to do with long division?

Sorry, the 7/2 on the LHS should have been just a 7. I'll fix it in the earlier post if I still can.

How is the remainder 2x+6 and the quotient x? I didn't see any division.

When you divide 7 by 2, for example, you get 3 for the quotient and a remainder of 1. One way to express these facts is to say $7 = 3x2 + 1$. More generally, if you divide $m$ by $n$, you have $m = qn + r$ where $0 \le r < n$.

Similarly, with polynomial division, if you divide $f(x)$ by $g(x)$, you can write $f(x) = q(x) g(x) + r(x)$ where $q(x)$ is the quotient, $r(x)$ is the remainder, and the degree of $r(x)$ is less than the degree of $g(x)$.

What do you mean they cancel out and why is this the idea?

What do you think I mean when I said the terms cancel out?

 

[rcgldr]

Lets try another example case, a x^5 + b x^4 + c x^3 + d x^2 + e x + f divided by m x^3 + n x^2 + o x + p. The first goal here is to get rid of the a x^5 in the dividend. The only way this is possible is with the first quotient term (a/m) x^2. Then look at what happens with the most significant terms of the dividend and the divisor, the first multiply and subtract step = a x^5 - ((a/m) x^2) (m x^3) = a x^5 - a x^5 = 0. If you look at the full multiply and subtact step, you have a x^5 + b x^4 + c x^3 + d x^2 + e x + f - ((a/m)x^2) (m x^3 + n x^2 + o x + p) = a x^5 + b x^4 + c x^3 + d x^2 + e x + f - (a x^5 + (a n / m) x^4 + (a o / m) x^3 + (a p / m) x^2 = (b - (a n / m)) x^4 + (c - (a o / m)) x^3 + (d - (a p / m) x^2) + e x + f. So the x^5 from the dividend has been elimiinated after the multiply and subtract step. To simplify the resulting term I'll define new coefficients for the current remainder, aa = (b - (a n / m)), bb = (c - (a 0 / m)), so rewriting this you have aa x^4 + bb x^3 + cc x^2 + dd x + ee divided by the same divisor as before, m x^3 + n x^2 + o x + p. For this step, the quotient term is (aa/m) x .

 

[ehild]

At first he shows 2x+4 / 2 and you just divide both 2x and 4 by 2. But then in the next example he is dividing x^2+3x+6 by x+1 and he doesn't divide x^2 by x+1, 3x by x+1 and 6 by x+1. I do not understand how he does the problem.

He does not divide x^{^2} by x+1, x by x+1 and 6 by x+1 as it does not lead anywhere. The expression will not become simpler, you get three fraction instead of one.

When doing long division, one rearranges the dividend, so as it becomes a sum, each term divisible by the divisor, except the constant term. x^2+3x+6= (x^2+x)+(2x+2)+4.
Now you can divide it by x+1, by terms. x^2+x divided by x+1 is x, 2x+2 divided by x+1 is 2, so you can write the result as $x+2+\frac{4}{x+1}$
When applying the long division method, you do the same, but mechanically, in a programmed way. You can program that method into a computer. .

Try to practice the divisions it the way above and you will notice that you do long division:)

 

[Joffan]

You can also think of this as doing division in an unknown number base (in which case it will help to able to use negative digits within the place notation).

                        1, 2 r 4
                    _______________
             1, 1  ) 1, 3, 6
                     1, 1
                    ------
                        2, 6
                        2, 2
                       ------
                           4

                        1,-2
                    _________
             1, 3  ) 1, 1,-6
                     1, 3
                    ------
                       -2,-6 
                       -2,-6
                       ======

Simply eliminating the highest place-value number each time, then dealing with any fallout in subsequent steps.