Bazza's 3/4 Mathematics questions

[WhoTookMyUsername]

What's the best way to factorise a quadratic where a =/=1 ?
Yeah i should know this but i don't really know how to best approach this (in year 11 i just get by with some dodgy grouping method)

a guide anyone xD?

EDIT: on further contemplation i think i almost do it right. i find a factor pair of ac that adds to b, but how do i "skip" the grouping stage? In year 10 we were taught to group, then factorise, but i know the grouping stage is not necessary, what thought process do i have to use to skip it?


Rohitpi edit: changed title to not confuse with GMA

[enwiabe]

For , you always know that



This will give you the factors and .

But we don't want fractional factors, which will give you a scalar multiple of the quadratic you wish to find. So, if one of the factors is a fraction, e.g. , multiply out to get as a factor.

This is much better explained with an example.

So you can therefore write,

Example, suppose

or

Thus factors are and

The first factor we need to multiply through by 3 so we get

So we can therefore write

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


Edited: to fix up the maaaaaaaaaaaajor fail.

[monkeywantsabanana]

I've been using the method of multiplying the a with the c and finding numbers that multiply that give ac and add to give b... I don't know what it's called...

You can complete the square? I find it really handy and easy although many many would disagree...

There's this other method called I think... "cross-swords" or "chopsticks" where you write the factors then multiply them diagonally to get the factors...

You can also... I guess use the quadratic formula and work backwards from your solutions but I don't think you can use that because I think you might lose a in the process... maybe you can for when a = 1 but correct me if I'm wrong

That's all I know. (:

[monkeywantsabanana]

So you can therefore write,

What happens if the answer was ?

[enwiabe]

So you can therefore write,

What happens if the answer was ?

Argh you're very right. This is what happens when I math on no sleep -_-.

Yeah it'll be a multiple of it

I was just discovering that I was writing out an example for Bazza

[WhoTookMyUsername]

thanks enwiabe, though it's very useful for more complex problems it's probably excessive work for easier questions (i was trying to ask about these moreso)
Edit: actually i think i know how to do this now in my head for easier problems as well (ty guys)


2) Is long division of polynomials ever actually necessary (apart from if it's specified)? Inspection has worked well for me so far (does it fail with any particular type of question?)

[enwiabe]

For , you always know that



Giving you the factors and .

So you can therefore write,

thanks for that explanation enwiabe. Though it's very good for more complicated problems i don't think that the level of working out is probably not necessary for a question such as

atm to solve this i go

factor pair of ac (4)  that adds to b is 4 and 1

______________________________________
so

_____________________________________________I want to remove these two lines and to be able to skip in my head... Exactly how do i do this?


=[/s]


Edit: actually i think i know how to do this know xD (props to gossamer)


2) Is long division of polynomials ever actually necessary (apart from if it's specified)? Inspection has worked well for me so far (does it fail with any particular type of question?)

Except that my explanation was sadly wrong I'm editing it now to make it right -_-

[enwiabe]

Bazza, that method works if and only if there are clear factors. This method works universally, even when you get surds and other crap. The other method that works best is completing the square.

Re: your 2nd question, I'm a fan of 'by inspection', too. I never use long division. By inspection has never failed me yet, and I can't conceive of a scenario where it would. Perhaps kamil or Ahmad can?

[WhoTookMyUsername]

when attempting to find the domain of a hybrid function, let's say one equation holds true when
and this simplifies to is there a way you can tell for questions like these that the domain is (-inf,0] U [3, in) ?
like with doesn't this become ? how do you tell which one is what should be "flipped" (x<0)

thanks
(again, i should definitely know this but ... bio... )

[pi]

Yes there is (link below), but personally, I find a "quick-sketch" far easier

Link: All you need to know about inequalities!

It's best explained in those attachments

What's the best way to factorise a quadratic where a =/=1 ?
Yeah i should know this but i don't really know how to best approach this (in year 11 i just get by with some dodgy grouping method)

a guide anyone xD?

Have you tried looking into the "half-quadratic formula"?

[WhoTookMyUsername]

when calculating whether or not a certain point is a local minimum or maximum (using differentiation without a graph for cubic, quartic etc.)
let's say i have the turning  points for and they are (1,2) (-1,2) It's much faster to just do f(.5), f(-.5) etc. If these values are resultant values are higher it's a local min, if it's lower it's a max. But the book deliberatly avoids this and calculates around 6 points for the gradient forming a \_/ \_/ picture.

My question is is this necessary? Does the method i feel is faster have any pitfalls? Do the examiners accept it? etc.

thanks

[pi]

You should be doing 6 points, that is how you produce a gradient table that is a valid proof. I know it may be tedious, but its VCAA...

Otherwise you could look into second derivatives for the second type of valid proof.

[kamil9876]

Good question, there are pitfalls. How do you know that say one if them is not an inflection point and hence not a local minimum? Likewise I think the gradient table generalizes better to more than 2 stationary points (but that would be evil).

edit: I actually may have misinterpreted your question, not sure on the exact alternative method you are proposing.

[WhoTookMyUsername]

well basically the way i was doing it is very similar but just saves a little bit of time

i'm doing f(1.5) and f(.5) and f(-1.5) and f(.5) But the only difference is don't need to draw a little gradient table. Though after contemplation i think drawing the gradient table would take about 5 seconds and isn't worth worrying about.

thanks rohitpi and kamil

[WhoTookMyUsername]

Have you tried looking into the "half-quadratic formula"?

M... what is this? i just googled it and nothing came up :S