# How do you factorise this...Watch

#1

I kinda remember there being a general formula for cubics, is it something like:

?

And if it is, how would I go about solving this?

Thanks

EDIT: is it just comparing coefficients?
0
8 years ago
#2
Hey 2710 - see if I can get it right this time!

There is a standard expression for x^3 - y^3
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#3
Lol

1) What is that expression?
2) What about a standard expression for x^3 + or - constant
3) How about a normal cubic? Does my expression work for normal cubics with x and x^2 coefficients in the equation?

Thanks
0
8 years ago
#4
Try find a root of the equation first.
0
8 years ago
#5
OK I can tell you - or you can divide (x^3 - y^3) by (x-y) obvious factor - long division or inspection (where does the x^3 come from etc)
0
#6
Try find a root of the equation first.
Well, isn't that what I'm asking help for?
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#7
(Original post by ian.slater)
OK I can tell you - or you can divide (x^3 - y^3) by (x-y) obvious factor - long division or inspection (where does the x^3 come from etc)
I don't think I have learnt division with two different variables :P

And isn't it just what I wrote in the first post?
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8 years ago
#8
Synthetic Division

or just

x=3
0
#9
(Original post by Fraser.)
Synthetic Division

or just

x=3
I do not know what Synthetic division is. I know the answer is 3 >__>. The question says to factorise.

And @ Lou Reed, if that is the correct formula, thank you. I guess thats one more formula to learn
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8 years ago
#10
(Original post by 2710)
Lol

1) What is that expression?
2) What about a standard expression for x^3 + or - constant
3) How about a normal cubic? Does my expression work for normal cubics with x and x^2 coefficients in the equation?

Thanks
This is what you're looking for. See if you can compare the expression there to the one you have to solve.

HINT:
Spoiler:
Show
27=3^3, so let a=x and b=3
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8 years ago
#11
Obviously (x^3 - 27) = (x-3)(x^2 + some x's + some constant)
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8 years ago
#12
(Original post by 2710)
I don't think I have learnt division with two different variables :P

And isn't it just what I wrote in the first post?
Welllll, if you add 27 to both sides, and then find x, you can get it in that nice fancy form that yourself and the above posters are mentioning. For this particular question that's the easiest way of doing it.
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#13
Welllll, if you add 27 to both sides, and then find x, you can get it in that nice fancy form that yourself and the above posters are mentioning. For this particular question that's the easiest way of doing it.
Will examiner's like that? It's kinda like me taking a shortcut, and then plugging the values back in. Meh.

And thanks people.

@The Nack, with those formulas, can I just put all of them as +ve and then when I compare coefficients, it will sort the signs out? Because Its kinda hard remembering those :P
0
8 years ago
#14

Find the factors of -27 (1,-1,3,-3,9,-9,27,-27) then employ the factor theorem

Decide then where you want to go with it (more factor finding, division etc)
0
8 years ago
#15
(Original post by 2710)
Will examiner's like that? It's kinda like me taking a shortcut, and then plugging the values back in. Meh.

And thanks people.

@The Nack, with those formulas, can I just put all of them as +ve and then when I compare coefficients, it will sort the signs out? Because Its kinda hard remembering those :P
Well you need to find a factor of the equation first before you can factorise a cubic. So yes examiners will like that. Lots.
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8 years ago
#16
(Original post by 2710)
@The Nack, with those formulas, can I just put all of them as +ve and then when I compare coefficients, it will sort the signs out? Because Its kinda hard remembering those :P
No that wouldn't work, you'd end up with an answer that had the wrong sign. The best thing to do is just to remember those two forumlae, they're not that complicated to be honest. Maybe if you study them you can find a way to remember which one to use in the exam?

EDIT: it's worth noting that if you expand out the RHS of each of those formulae I showed you, then you will simply get the expression on the LHS, which would let you check in the exam whether you have chosen the correct one. You just have to remember that the signs of one goes +-+ and the other goes -++ .
0
#17
Well you need to find a factor of the equation first before you can factorise a cubic. So yes examiners will like that. Lots.
Look, stop being a smartass. I do not need to find a root first, as most of these posters have said. I can use a general equation and then compare coefficients. I did not have to find the 'root' for that, I get the root at the end. If you are gonna be sarcastic like that, id prefer if you did not post
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#18
(Original post by TheNack)
No that wouldn't work, you'd end up with an answer that had the wrong sign. The best thing to do is just to remember those two forumlae, they're not that complicated to be honest. Maybe if you study them you can find a way to remember which one to use in the exam?

EDIT: it's worth noting that if you expand out the RHS of each of those formulae I showed you, then you will simply get the expression on the LHS, which would let you check in the exam whether you have chosen the correct one. You just have to remember that the signs of one goes +-+ and the other goes -++ .
Ok thanks, I guess its not that hard after all lol
0
8 years ago
#19
(Original post by 2710)

I kinda remember there being a general formula for cubics, is it something like:

?

And if it is, how would I go about solving this?

Thanks

EDIT: is it just comparing coefficients?
It seems like you're being led on a wild goose chase with some of these methods. Seriously, just look up factor theorem.
0
8 years ago
#20
(Original post by 2710)
Look, stop being a smartass. I do not need to find a root first, as most of these posters have said. I can use a general equation and then compare coefficients. I did not have to find the 'root' for that, I get the root at the end. If you are gonna be sarcastic like that, id prefer if you did not post
You don't really have to do that with the formulae I showed you. In the formulae I suggested, you have a linear expression to solve (a+-b) and the quadratic expression to solve. That should be pretty straightforward, at least moreso than trying to solve a cubic directly.
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