# Algebraic Long Division

#### Ebba Sen Pai

##### New member
My question is about the basics of Algebraic long Division. I was wondering how to tell where to put the initial quotient term when solving the a problem. Below I provide two examples that I cannot make out the reasoning of. As far as I can tell they are relatively similar in that they both have binomial divisors with degrees of 2, and divedends with degrees of 4. The first is from a Youtube channel called "The Chemistry Tutor", and the second is from my E-Book by Blitzer.
(Red text and underline is me)

What I cannot figure out is over which term inside the "house" (apologies, I don't recall correct term) my initial quotient term should go above. I generally understand the process well after that, but this initial step seems to vary and I cannot tell the reasoning behind it. Does my first terms in the quotient go above the 3rd term as demonstrated by "The Organic Chemistry Tutor", or does it go in the first above the first term as depicted by Blitzer?

(You don't need to read this part unless you want context). I am in Pre-Calc, and I only ever came across algebraic long division in my personal, self taught studies on Youtube when I did Algebra 1 (never in my school based textbook- we skipped that section). My Intermediate Algebra class had a "reviewing the basics section" where we touched on this very briefly but it never came back up in class after the 1st week. Suffice to say, I have no true backbone when it comes to this concept, and I truly appreciate anyone willing to share wisdom with me now on this subject I probably "should" know alot better, but simply haven't had any meaningful time with it.
As always, I thank you guys from the bottom of my heart.

#### Otis

##### Senior Member
I don't think it matters where you start writing the quotient. I do it the same way as shown in your book.

#### Harry_the_cat

##### Senior Member
I agree witn Otis. Doesn't really matter. I have always put it above the first term. I also would add in 0x into the divisor to avoid silly mistakes.

#### Dr.Peterson

##### Elite Member
My practice (not required, but a good idea in my opinion) is to arrange things so that there is a column for each power. That's why I do as shown in the YouTube example, putting each term of the quotient above the corresponding term (i.e. same degree) in the dividend. This helps you to notice (a) when you are finished, because you've used up all the spaces in the quotient, and (b) if you make a silly mistake somewhere writing a quotient term. It also matches how I do division of numbers. But it doesn't cause any real trouble if you do otherwise.

Putting 0 terms in where for missing terms (like the 0x^3 and 0x), into both the divisor and the dividend, is part of the same overall strategy.

#### Jomo

##### Elite Member
You really can decide for yourself. Once you write down some result for the quotient do you ever need to take into account where you put it? Try writing down the answer nowhere near where the dividend and divisor are and see if you get different answers.

#### JeffM

##### Elite Member
I think it is MUCH more important to understand why algebraic division works than the presentation.

$$\displaystyle \dfrac{6x^4 - 30x^2 + 24}{2x^2 - 8} =$$

$$\displaystyle \dfrac{3x^2(2x^2) + 3x^2(8 - 8) - 30x^2 + 24}{2x^2 - 8} =$$

$$\displaystyle \dfrac{3x^2(2x^2 - 8) + 24x^2 - 30x^2 + 24}{2x^2 - 8} =$$

$$\displaystyle 3x^2 - \dfrac{6x^2 - 24}{2x^2 - 8} =$$

$$\displaystyle 3x^2 - \dfrac{3(2x^2 - 8)}{2x^2 - 8} =$$

$$\displaystyle 3x^2 - 3 = 3(x^2 - 1).$$

In any case, you can see that dividing x^2 into x^4 is going to result in a term in x^2 so starting the quotient over the x^2 term of the dividend starts you off in an aesthetically pleasing way.

#### pka

##### Elite Member
I see absolutely no reason for anyone to waste time learning to do any of this.
Look at this website. The WolframAlpha website in available for almost all portable platforms.
For any teachers, instructors, collage professors who disagree please read opinion pieces by Keith Devlin< Prof. of Mathematics at Stanford & the MathGuy on NPR.

#### Ebba Sen Pai

##### New member
I don't think it matters where you start writing the quotient. I do it the same way as shown in your book.

Thanks! This makes so much more sense, and is even easier than I thought rather than harder.
I agree witn Otis. Doesn't really matter. I have always put it above the first term. I also would add in 0x into the divisor to avoid silly mistakes.
Thank you. I will be sure to account for 0x in the future.
My practice (not required, but a good idea in my opinion) is to arrange things so that there is a column for each power. That's why I do as shown in the YouTube example, putting each term of the quotient above the corresponding term (i.e. same degree) in the dividend. This helps you to notice (a) when you are finished, because you've used up all the spaces in the quotient, and (b) if you make a silly mistake somewhere writing a quotient term. It also matches how I do division of numbers. But it doesn't cause any real trouble if you do otherwise.

Putting 0 terms in where for missing terms (like the 0x^3 and 0x), into both the divisor and the dividend, is part of the same overall strategy.
Thank you. I will attempt to practice this. That does seem to make sense , and would help me keep things in sensible order.
You really can decide for yourself. Once you write down some result for the quotient do you ever need to take into account where you put it? Try writing down the answer nowhere near where the dividend and divisor are and see if you get different answers.
I did this and you were entirely correct. It feels good to have gone through the steps and seeing that this was true.
I think it is MUCH more important to understand why algebraic division works than the presentation.

$$\displaystyle \dfrac{6x^4 - 30x^2 + 24}{2x^2 - 8} =$$

$$\displaystyle \dfrac{3x^2(2x^2) + 3x^2(8 - 8) - 30x^2 + 24}{2x^2 - 8} =$$

$$\displaystyle \dfrac{3x^2(2x^2 - 8) + 24x^2 - 30x^2 + 24}{2x^2 - 8} =$$

$$\displaystyle 3x^2 - \dfrac{6x^2 - 24}{2x^2 - 8} =$$

$$\displaystyle 3x^2 - \dfrac{3(2x^2 - 8)}{2x^2 - 8} =$$

$$\displaystyle 3x^2 - 3 = 3(x^2 - 1).$$

In any case, you can see that dividing x^2 into x^4 is going to result in a term in x^2 so starting the quotient over the x^2 term of the dividend starts you off in an aesthetically pleasing way.
Thanks! I understand a bit of what went on there, but I am going to need to study it a little longer to ensure I understand all of it. It is helpful to remember that what I am doing in long division is really still "good old algebra" and just some nebulous formula that can only be performed in a singular fashion.
I see absolutely no reason for anyone to waste time learning to do any of this.
Look at this website. The WolframAlpha website in available for almost all portable platforms.
For any teachers, instructors, collage professors who disagree please read opinion pieces by Keith Devlin< Prof. of Mathematics at Stanford & the MathGuy on NPR.
Thank you for this information. I will begin practicing with Wolfram. I have used Symbolab in lower courses because it's accessibility to people unskilled in math, but I have been told by many others that Wolfram is the thing to become proficient in for higher math levels. It is my summer time, so I have no excuse but to capitalize on this wisdom and gain an edge now when it is relatively easy.

#### Harry_the_cat

##### Senior Member
I see absolutely no reason for anyone to waste time learning to do any of this.
Look at this website. The WolframAlpha website in available for almost all portable platforms.
For any teachers, instructors, collage professors who disagree please read opinion pieces by Keith Devlin< Prof. of Mathematics at Stanford & the MathGuy on NPR.
WolframAlpha takes the fun out of everything!!

#### pka

##### Elite Member
WolframAlpha takes the fun out of everything!!
Wtf does fun have to to with learning mathematics?

#### lookagain

##### Senior Member
My question is about the basics of Algebraic long Division. I was wondering how to tell where to put the initial quotient term when solving the a problem. Below I provide two examples that I cannot make out the reasoning of. As far as I can tell they are relatively similar in that they both have binomial divisors with degrees of 2, and divedends with degrees of 4. The first is from a Youtube channel called "The Chemistry Tutor", and the second is from my E-Book by Blitzer.
(Red text and underline is me)

View attachment 12800

The greatest common factor of the dividend and the divisor is 2, so it is useful to reduce the coefficients at the beginning:

$$\displaystyle \dfrac{6x^4 + 0x^3 - 30x^2 + 0x + 24}{2x^2 - 8} \ =$$

$$\displaystyle \dfrac{2(3x^4 + 0x^3 - 15x^2 + 0x + 12)}{2(x^2 - 4)} \ =$$

$$\displaystyle \dfrac{3x^4 + 0x^3 - 15x^2 + 0x + 12}{x^2 - 4}$$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Ebba Sen Pai, you are here in this thread to learn about the aspects of algebraic long division. Please ignore
those posts above of pka, because he is changing the subject away from the point of why you are here.

#### JeffM

##### Elite Member
I have to agree with pka about the low degree of utility in teaching algebraic long division when technology renders it obsolete, but seemingly you are being required to learn it so that is what you must do. Frankly, it is a lengthy mechanical process that appears to have no logic to it. In fact, the process mechanically replaces an obviously logical process that is usually hard to see. In the case of this problem, however, the logical approach is not too hard to see.

What monomial times $$\displaystyle 2x^2 - 8$$ results in an expression starting with $$\displaystyle 6x^4$$?

Obviously $$\displaystyle 3x^2(2x^2 - 8) = 6x^4 - 24x^2.$$

$$\displaystyle \dfrac{6x^4 - 30x^2 + 24}{2x^2 - 8} = \dfrac{6x^2 - 24x^2 - 6x^2 + 24}{2x^2 - 8} = \dfrac{3x^2(2x^2 - 8) - 3(2x^2 - 8)}{2x^2 - 8} =$$

$$\displaystyle \dfrac{(3x^2 - 3)(\cancel{2x^2 - 8)}}{\cancel{2x^2 - 8}} = 3x^2 - 3 = 3(x^2 - 1).$$

In this problem, what the mechanics of long division do is let you avoid seeing that you can factor the dividend by the divisor simply by replacing

$$\displaystyle -\ 30x^2$$ with $$\displaystyle -\ 24x^2 - 6x^2.$$

In most problems, the insight into how to factor the dividend by the divisor is very far from simple, and algebraic long division is a mechanical way to factor the dividend so as to permit cancelling the divisor (if that is possible).

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#### lookagain

##### Senior Member
Ebba Sen Pai,

if you are approaching this by factoring, then you are changing the problem, as well as you are supposing that the
quotient has a remainder of zero with it.

So, in this different problem (by factoring and not by algebraic long division), again do not ignore where I divided
out the greatest common factor of 2. Why? Because it would relatively overcomplicate the factoring to leave the
coefficients what they were elsewise.

$$\displaystyle 3x^4 - 15x^2 + 12 \ = \ 3(x^4 - 5x^2 + 4)$$

You're supposing that $$\displaystyle \ (x^2 - 4) \$$ divides that trinomial, so

$$\displaystyle 3(x^4 - 5x^2 + 4) \ = \ 3(x^2 - 4)( \ \ \ \ \ \ \ \ \ \ \ )$$

and that leaves the other factor to be $$\displaystyle \ (x^2 - 1) \$$ more readily. The "+ 4" in the trinomial gives the "- 1"
in the second factor.

Then, after moving around factors, $$\displaystyle \ 3(x^2 - 1) \ =$$

$$\displaystyle 3x^2 - 3 \ \ \ \ \ \ \ \ \ \ \$$ <-------- This is the form of the answer, not the factored form. You're giving the polynomial,
which is the quotient. That is also the result of dividing the original dividend by the original divisor.

#### Otis

##### Senior Member
… algebraic long division … appears to have no logic to it …
I think that might be true for a lot of students, in the beginning. Once we understand the process, the logic is the same as long division of one Whole number by another.

As for 'utility', it could be that some instructors cover the topic because it offers another opportunity to increase symbolic reasoning skills (i.e., working through steps of a familiar arithmetic process using symbolic numbers instead). In my case, I was taught polynomial long division in precalculus -- as a prelude to synthetic division. I still use synthetic division (occasionally).

$$\;$$

#### Otis

##### Senior Member
… Please ignore those posts above of pka, because he is changing the subject
I'm sure you've never done that.

$$\;$$

#### lookagain

##### Senior Member
I'm sure you've never done that.
Again, you have come on here with another non-argument, only to show how foolish you are. Stop deflecting
and pay attention to the point. When the student asks how to do it a certain way, the correct thing is not to change
the subject as pka has done and state to stop learning it and go elsewhere to some link, etc.
$$\;$$

What should be mmm4444bot's/Otis's reply said:
You're correct, lookagain.