Regarding long division for rational algebraic expressions (1 Viewer)

[Slidey]

I suggest that you learn long division but don't use it for linear divisors.

Use synthetic division instead: http://www.purplemath.com/modules/synthdiv.htm

It's a great time saver, as it provides both the dividend and remainder, as with long division. It works for all divisors, however it's only a time saver for linear divisors and you won't find anybody to teach it for quadratic and higher divisors anyway. That is: use normal long division for quadratic divisors.

Cheers.

 

[Affinity]

the best way is to do it in your head... it's not as difficult as it seems because you write down your quotient as you go.

 

[YannY]

the best way is to do it in your head... it's not as difficult as it seems because you write down your quotient as you go.

And if i asked you to divide x^3-3x^2+2x+3 by x-2+i. How would you do that in your head - if you can i would have known you on TV already . But seriously synthetic division is pretty useful - i used it in my class test the other day and destroyed everybody else.

 

[Slidey]

And if i asked you to divide x^3-3x^2+2x+3 by x-2+i. How would you do that in your head - if you can i would have known you on TV already . But seriously synthetic division is pretty useful - i used it in my class test the other day and destroyed everybody else.

Yep! And it works for complex numbers, too! You'll rip up anybody doing it in their head or with long division.

But lol at it taking a year for anybody to reply to my thread.

 

[Affinity]

And if i asked you to divide x^3-3x^2+2x+3 by x-2+i. How would you do that in your head - if you can i would have known you on TV already . But seriously synthetic division is pretty useful - i used it in my class test the other day and destroyed everybody else.

you write x^2
think: -3x^2 - (-2 + i)x^2
then: -(1 + i) x ^2 + 2x
and continue to write: x^2 -(1+i) x
think 2x - (1+i)(2-i)x = 2x - (3 +i)x = -(1 + i)x [the only thing you really need to do here is the multuiplication, which is not that bad]
again bring down the next term, think -(1+i)x + 3
continue to write: x^2 - (1+i)x -(1+i)
and notice this is essentially the same as the last iteration, you get a remainder of -i

so (x^3-3x^2+2x+3)= (x-2+i)[x^2 - (1+i)x -(1+i)] - i

Edit: There are short cuts for if you know more about what's happening, for example, if you know that the remainder is 0 then you get the first and last term out easily, then figuring out the middle is easy.
for example, if you knew a priori that (x+1) divides (x^3 + 3x^2 + 3x + 1) then you know that the quotient must be in the form (x^2 + ax + 1)
working out a is not difficult then since the square term is x*(ax) + x^2 * 1 so a = 2


Ofcourse if you had funny coefficients like 0.12314 then it would probably be better to be a bit more methodical

 

[Affinity]

Yep! And it works for complex numbers, too! You'll rip up anybody doing it in their head or with long division.

But lol at it taking a year for anybody to reply to my thread.

most people don't reply to Sticky threads

 

[Slidey]

most people don't reply to Sticky threads

I think most people don't even read them. I know that I subconsciously just skip straight to the first non-stickied post when I view forums.

Still, it's good to have discussion in them, as with your explanation of doing it in your head, or YannY's endorsement of the efficacy of this method.

 

[hurikai]

Cool, this'll come in handy. Thanks.

 

[nayyarv]

Yeah, synthetic division is really fast for linear factors

i dont bother with long division anymore, it doesnt compare with synthetic

apparently its in the IB (International Bacclaureate) and since my maths teacher teaches the IB as well, he showed us synthetic division

However its useless for dividing by quadratics, so dont forget about it, though most of the division u need to do, will be for linear factors

 

[duy.le]

2 part question
i) can u use synthetic division for dividing p(x) with ax+b?
ii) are you allowed to use it in the hsc?

 

[namburger]

2 part question
i) can u use synthetic division for dividing p(x) with ax+b?
ii) are you allowed to use it in the hsc?

i) yes
ii) in my school exam, i just go from step ax^3 + bx^2 + cx + d to (x+e)( .... ) for example, didnt get any marks deducted

 

[azureus88]

How would you use synthetic divison for linear divisons that have non-monic coefficients? eg. (4x^3 - 4x^2 + 7x +14)/(2x+1)

 

[Timothy.Siu]

How would you use synthetic divison for linear divisons that have non-monic coefficients? eg. (4x^3 - 4x^2 + 7x +14)/(2x+1)

u wud divide the divisor by 2 and the question by 2 and then multiply it later, i know this doesn't really make sense but it does when u do it so for this one,

u go (2x^3-2x^2+7/2x+7)/(x+1/2), this would keep the same ratio so the quotient will be the same,
so in the end u get (x+1/2)(2x^2-3x+5)+4.5=(2x^3-2x^2+7/2x+7)
multiply 2 to both sides
(2x+1)(2x^2-3x+5)+9= (4x^3 - 4x^2 + 7x +14)

 

[Slidey]

2 part question
i) can u use synthetic division for dividing p(x) with ax+b?

Yes. In fact this is the main time you should use synthetic division - linear factors (i.e. of form ax+b). For division by quadratic and higher it's wiser to use long division.

ii) are you allowed to use it in the hsc?

Yes.

 

[INFINITO]

CLICK ON THIS ATTACHMENT FOR GENERAL SYNTHETIC DIVISION THEOREM, View attachment ADAM LARKIN.pdf

 

[sadpwner]

Does it only work when dividing quadratics? Also, I've heard that you don't get marks if the method isn't in the syllabus.

 

[InteGrand]

Does it only work when dividing quadratics? Also, I've heard that you don't get marks if the method isn't in the syllabus.

Horner's method (aka synthetic division) works when higher degree polynomials too. And if the question specifically was "divide P(x) by D(x)" or something, then you should probably use long division. However, if you just need to divide a polynomial in order to do a bigger question (e.g. to find an integral), you can just use Horner's method.