Polynomial Long Division - In depth Look on why it works!

In this video I show how you can easily divide polynomials through polynomial long division which is very similar to regular long division. I also explain WHY it works by expanding out the function and showing how you get to each step of the polynomial long division. This is a very important video to watch, even if you are well accustomed to polynomial long division because you may not have learned why this method works!

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[mes]

I don't always divide polynomials using long division but when I do I usually explain why the method works ;)

[farida.azarleon]

Really good idea to make a video of how it actually works, because we just learn that it works without learning why. The whys are always important :)

[torosushi]

Great video, I was never taught why it worked, I simply knew that it did. Thanks!

[chriskerley]

Thank you for explaining why it works. Very few people try to explain why, and a lot of students end up doing it by rote because of that. i.e They don't really understand what's going on, they just follow the rules.

[glennrickelton]

I was lost on your explanation of WHY algebraic long division works when I first watched your video but after viewing it again your approach makes sense. Well done! This is the only explanation I have found as to WHY this method works as opposed to simply learning by rote...Again, WELL DONE!

[沈博智-xy]

I watched this video ages ago to get my head wrapped around polynomial long division, and it worked.

Recently though, I thought of polynomial long division again and I prefer to see it a slightly different way

(x^2 - 9x - 10)/(x+1) = x^2/(x+1) + (-9x - 10)/(x+1)

If we look at x^2/(x+1) as x(x/(x+1)) [NB: the outside x is part of our quotient] and look at the inner part, we ask ourselves, what do we need to add or subtract to make that inner part equal 1 (numerator and denominator equal). Whatever this is will be precisely the 'subtraction part' of the polynomial long division (subtract what from -9x - 10)

In this case, it's 1, but don't forget the outside x, so this becomes -x.

x(x/(x+1)) = x((x+1-1)/(x+1)) = x((x+1)/(x+1) - 1/(x+1)) = x(1 - 1/(x+1)) = x - x/(x+1)

Going back to our original division:
(x^2 - 9x - 10)/(x+1) = x - x/(x+1) + (-9x - 10)/(x+1) = x + (-9x - 10 **- x**)/(x+1) = x + (-10x - 10)/(x+1)

Now let us focus on (-10x - 10)/(x+1)

I know that this is -10(x+1)/(x+1), but let's not do that to get a feel of why the long division works.

We will do the exact same thing, bring split it by the 'leading term' and the rest.

i.e. (-10x - 10)/(x+1) = -10x/(x+1) - 10/(x+1)

Focusing on -10x/(x+1) = -10(x/(x+1)) and looking at the inner part... [outside part -10 is the quotient part]
We see that I need to add and subtract 1, but there is a -10 out the front, so the 'subtraction step' becomes + 10 = -(-10)

-10x/(x+1) = -10(x/(x+1)) = -10((x + 1 - 1)/(x+1)) = -10((x+1)/(x+1) - 1/(x+1)) = -10(1 - 1/(x+1)) = -10 + 10/(x+1)

Going back to our original division

(x^2 - 9x - 10)/(x+1) = x - 10 *+ 10/(x+1)* - 10/(x+1) = x - 10 + (-10 **-(-10)**)/(x+1)) = x - 10


Summary, split by 'leading term and rest'
Factorise out the 'quotient part'.
Looking at inner part after factorising and ask what still needs to be added on to create an y/y situation. This will be the subtraction part.
Do Quotient + (rest - subtraction part times 'quotient part')/denominator

Repeat until remainder (rest - subtraction) has degree less than denominator.


In general:

For f(x) = a_(n+k)x^(n+k) + ... + a_0
and g(x) = b_nx^n + ... + b_0

and computing f(x)/g(x)

I will illustrate the first step:

f(x)/g(x) = (a_(n+k)x^(n+k) + a_(n+k-1)x^(n+k-1) ... + a_0)/(b_nx^n + b_(n-1)x^(n-1) + ... + b_0)

Split into leading and rest...

f(x)/g(x) = a_(n+k)x^(n+k)/(b_nx^n + b_(n-1)x^(n-1) + ... + b_0) + (a_(n+k-1)x^(n+k-1) ... + a_0)/(b_nx^n + b_(n-1)x^(n-1) + ... + b_0)

Focusing on a_(n+k)x^(n+k)/(b_nx^n + b_(n-1)x^(n-1) + ... + b_0)

The quotient will be a_(n+k)x^k/b_n since a_(n+k)x^(n+k) = (a_(n+k)x^k/b_n)(b_nx^n)

Focusing on the inner part (b_nx^n) with the denominator (b_nx^n + b_(n-1)x^(n-1) + ... + b_0), the 'inner subtraction part' is b_(n-1)x^(n-1) + ... + b_0
but dont forget the outer part, so outer times inner (outer is quotient part)

This simplifies to:
a_(n+k)x^(n+k)/(b_nx^n + b_(n-1)x^(n-1) + ... + b_0) = (a_(n+k)x^k/b_n) - + ... + b_0))/(b_nx^n + b_(n-1)x^(n-1) + ... + b_0)

Tying all this up into one would simplify be. 'Do Quotient + (rest - subtraction part times 'quotient part')/(denominator)


f(x)/g(x) = (a_(n+k)x^k/b_n) + (a_(n+k-1)x^(n+k-1) ... + a_0)/(b_nx^n + b_(n-1)x^(n-1) + ... + b_0) *- + ... + b_0))/(b_nx^n + ... + b_0)
= (a_(n+k)x^k/b_n) + ((a_(n+k-1)x^(n+k-1) ... + a_0) - + ... + b_0))**)/(b_nx^n + ... + b_0)

f(x)/g(x) = Quotient part with leading term + the other terms - **subtraction part**/denominator

Repeat the same algorithm till it becomes a remainder.


Note: subtraction part does not including the leading term, which justifies the leading term cancelling out in long division written way.