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Factorising Cubic Equations - Maths Help

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I have a practise C1 exam tomorrow for my AS Level Maths & I am revising at the moment & cannot work out how to factorise a cubic equation which I have from a different past paper. I could probably… Read More
8y, 3m agoPosted 8 years, 3 months ago
I have a practise C1 exam tomorrow for my AS Level Maths & I am revising at the moment & cannot work out how to factorise a cubic equation which I have from a different past paper.

I could probably get it after a while but if anyone knows a way to generally factorise cubics I would be greatful if you could explain as my notes dont seem to make sence to me.

The said equation is x ^ 3 - 3x ^ 2 - 6x + 8

I have tried getting a linear & quadratic from it but I cant seem to get it right. In the end I need 3 linear factors of it.

If anyone could explain I would be very greatful. Its a non calculator paper so shouldnt be too complex.

Thanks alot

Cheers
James
8y, 3m agoPosted 8 years, 3 months ago
Options

(10)
#1
You can use synthetic division to factorise cubics: http://www.purplemath.com/modules/synthdiv.htm

You'll need to guess a factor, such as 1 or -2 etc. It's a factor if the remainder is 0.

I learnt this at Higher level (A level) Maths, and it is very useful.
#2
You need to remove a factor of x from the equation to make it a quadratic, then simplify from there. Or so I think... 3 years since I've done A2 maths.
#3
Managed it after about an hour of messing around on various sites & trial & error.

It was...

(x-1)(x^2 - 2x -8)

(x-1)(x+2)(x-4)

Need to go over them before tomorrow though haha!

Cheers for the replies!
#4
The factor was 1. Managed to work it out synthetic division.

At least you got it sorted.
#5
How did you get the factor mate?

May come in handy tomorrow.

Cheers
James
#6
One way to get started is to look at the unit (the element without x). In this case its 8 - which, assuming all the factors use integers, can only be factored with 1,2 & 4 or 1,1 & 8. As there are a couple of negative elements in the equation but the '8' is positive, you know that 2 of the factors must be (x - ..) instead of (x + ..). So just from these simple deductions you've reduced the number of possibilities to only 5, which are easy to multiply out and find the correct one...:
(x - 1)(x - 2)(x + 4)
(x - 1)(x + 2)(x - 4)
(x + 1)(x- 2)(x - 4)
(x - 1)(x - 1)(x + 8)
(x - 1)(x + 1)(x + 8)
#7
use the F of (x) thing. I did C1 in January (epic fail)... C2 and Mechs in june :(
#8
James...
How did you get the factor mate?

May come in handy tomorrow.

Cheers
James

The factor was just guessed. If if is non calculator, the factor will be a small number that you can do in your head. If the remainder is 0 (using synthetic division), then its a factor.
1 Like #9
Here is a sketchy method I made in Paint:
#10
Thanks again for the replies, been a great help & made it sound alot more simple, hopefully the exam goes alright, got a C in the last one but got back from Australia the day before & missed 3 weeks of revision & didnt know I had the test, should be better prepared this time hopefully!

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