[SOLVED]Analogy Between Long Division and Polynomial Long Division

Ackbach

Indicium Physicus
Staff member
Hopefully, anyone who has studied polynomial long division understands the link between it and regular long division. If you divide $58$ into $302985$, you could follow the usual long division procedure and obtain the answer. Alternatively, if you divide $x+4$ into $x^{3}+2x^{2}+6x+7$, you could use the exact same math to perform this division. However, polynomial long division offers one feature that I am not sure has its analogy in regular long division: minus signs. I can also divide $x-4$ into $x^{3}+2x^{2}+6x+7$ or $x+4$ into $x^{3}-2x^{2}+6x-7$. My question is this: is there an analogy in regular long division for minus signs in polynomial long division?

Klaas van Aarsen

MHB Seeker
Staff member
Hopefully, anyone who has studied polynomial long division understands the link between it and regular long division. If you divide $58$ into $302985$, you could follow the usual long division procedure and obtain the answer. Alternatively, if you divide $x+4$ into $x^{3}+2x^{2}+6x+7$, you could use the exact same math to perform this division. However, polynomial long division offers one feature that I am not sure has its analogy in regular long division: minus signs. I can also divide $x-4$ into $x^{3}+2x^{2}+6x+7$ or $x+4$ into $x^{3}-2x^{2}+6x-7$. My question is this: is there an analogy in regular long division for minus signs in polynomial long division?
When you do a long division with regular numbers, you are converting the result to decimal form: $\sum a_i 10^i$.
By convention all $a_i$ are between 0 and 9.

We don't have to do it that way though.
We can also use a different base than 10, say 16.
Or we can choose that all $a_i$ should be between -5 and +4.
That way you'll get a different representation of the result, similar to doing polynomial division.

Note that if you define x=10 in polynomial division, and also constrain the factors to be between 0 and 9, you get regular long division.

agentmulder

Active member
If you allow me to lose the place value in the answer then yes , i can make make it happen.

$\ \ \$ 6000 - 900 + 100 + 20 + 5 - 1
_________ ________ ________ ________ ________ _____
58 | 302985
-(348000)
________________ ___________ ________ ___
-45015
-(-52200)
______________ ______________ ______
7185
-(5800)
____ __________ ____________ ____
1385
-(1160)
_______ ____________ ___________ ______
225
-(290)
____ _______ _____________ ______ _____
-(65)
-(-58)
_________ _______________ ________
-7

You can stop now if you're bored or you can continue for as long as you like , what i have done is make a guess at each step until the subtraction produced a number that is obviously a remainder , -7. It doesn't matter if i go 'over' or 'under' at any step AND it doesn't matter by how much i am in error , the rules of arithmetic (and algebra indirectly) gaurantee the result is true.

$6000 - 900 + 100 + 20 + 5 - 1 - \frac{7}{58} = 5224 - \frac{7}{58} = 302985 ÷ 58$

agentmulder

Active member
An easier example that doesn't use too many numbers and operations would be to do 9 ÷ 7

$\ \ \$ 2 - 1
$\ \ \$ _______
7 | 9
-(14)
__________
-5
-(-7)
______
2

So the answer is $\ \ 2 - 1 + \frac{2}{7} = 1 + \frac{2}{7}$

Note that we can stop after the first 'guess' of 2 and we would get

$2 - \frac{5}{7}$

which is the correct answer. So at every step the algorithm produces a correct answer ... in my previous post as well ... it must be , otherwise we would not get the correct answer in the end because each subsequent calculation depends on the previous calculation , but some answers are more useful than others , depending on what you like or are looking for a particular representation of the same correct answer.

Ackbach

Indicium Physicus
Staff member
When you do a long division with regular numbers, you are converting the result to decimal form: $\sum a_i 10^i$.
By convention all $a_i$ are between 0 and 9.

We don't have to do it that way though.
We can also use a different base than 10, say 16.
Or we can choose that all $a_i$ should be between -5 and +4.
That way you'll get a different representation of the result, similar to doing polynomial division.

Note that if you define x=10 in polynomial division, and also constrain the factors to be between 0 and 9, you get regular long division.
If you allow me to lose the place value in the answer then yes , i can make make it happen.

$\ \ \$ 6000 - 900 + 100 + 20 + 5 - 1
_________ ________ ________ ________ ________ _____
58 | 302985
-(348000)
________________ ___________ ________ ___
-45015
-(-52200)
______________ ______________ ______
7185
-(5800)
____ __________ ____________ ____
1385
-(1160)
_______ ____________ ___________ ______
225
-(290)
____ _______ _____________ ______ _____
-(65)
-(-58)
_________ _______________ ________
-7

You can stop now if you're bored or you can continue for as long as you like , what i have done is make a guess at each step until the subtraction produced a number that is obviously a remainder , -7. It doesn't matter if i go 'over' or 'under' at any step AND it doesn't matter by how much i am in error , the rules of arithmetic (and algebra indirectly) gaurantee the result is true.

$6000 - 900 + 100 + 20 + 5 - 1 - \frac{7}{58} = 5224 - \frac{7}{58} = 302985 ÷ 58$
An easier example that doesn't use too many numbers and operations would be to do 9 ÷ 7

$\ \ \$ 2 - 1
$\ \ \$ _______
7 | 9
-(14)
__________
-5
-(-7)
______
2

So the answer is $\ \ 2 - 1 + \frac{2}{7} = 1 + \frac{2}{7}$

Note that we can stop after the first 'guess' of 2 and we would get

$2 - \frac{5}{7}$

which is the correct answer. So at every step the algorithm produces a correct answer ... in my previous post as well ... it must be , otherwise we would not get the correct answer in the end because each subsequent calculation depends on the previous calculation , but some answers are more useful than others , depending on what you like or are looking for a particular representation of the same correct answer.

Excellent! So the analogy was always there under my very nose; but because regular long division is hardly ever taught this way (ever taught this way?), it would not be obvious unless you go looking for it. This relieves the tension in my mind between the two; it seemed to me that the analogy had to be exact! You really are doing the same math.

agentmulder

Active member
You can also follow 'THEIR' rules about keeping place value and positive remainders until you decide to break the rules , Kobayashi Maru style.

1156432 ÷ 7

I'm going to follow the rules by putting 7 into 11 once then 7 into 45 6 times then 7 into 36 5 times , then break the rules and start 'guessing'

So far i've got this ,

$\ \ \ \ \ \$ 165' + 200 + 5
$\ \ \$ __________ ________
7 | 1156432
$\$ -(1155000)
$\ \$ ________ ________ ___
$\ \ \ \ \ \ \ \ \ \$ 1432
$\ \ \ \ \ \ \ \$ -(1400)
$\ \$ ________ ________ ________
$\ \ \ \ \ \ \ \ \ \ \ \ \ \$ 32
$\ \ \ \ \ \ \ \ \ \ \$ -(35)
$\ \$ ________ ________ ________
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ -3

That 165 represents 165000 because i'm 'honoring' place value , now start the guesswork and break the rule , any way you like ... i make a 'guess' that 7 will go into 1432 200 times then subtract 1400 and then 'guess' 7 will go into 32 5 times so that i can break the other rule about 'positive remainders'

Puting it all together gives $165000 + 200 + 5 - \frac{3}{7} = 165205 - \frac{3}{7}$

I could have made different 'guesses' , more wildly inaccurate 'guesses' , the procedure would have been longer but if do my arithmetic correctly , every step is a valid representation of the answer.

We could put a mark on 165 to let us know we're breaking the rule after that point and make a note

165' = 165000

as i have done above.