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# How to factorise a cubic?

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1. I'm guessing this is pretty simple, but I've never really got the hang of factorising polynomials.

I know that -1 is a root and the cubic equation is , what methods are there of now factorising and which is the quickest?

thanks!
2. I would use polynomial long division as you know one factor.
3. Find something that you can divide it by using the factor theorem, then do long division, it will become a quadratic (or better) then it's simple.
4. Use algebraic division to find the quadratic. Factorise the quadratic. remember though you are dividing it by x+1 not x-1
5. (Original post by Brit_Miller)
I would use polynomial long division as you know one factor.
How do you do that though, I'm not great at this sort of thing, thanks
6. (Original post by Benniboi1)
How do you do that though, I'm not great at this sort of thing, thanks
https://www.dlsweb.rmit.edu.au/lsu/c...20division.pdf

Check out something like that, it goes through the steps.
7. If you know -1 is a root, this means (x+1) is a factor.
Therefore, using long division divide the cubic equation by (x+1), you will be left with a quadratic you can hopefully factorise and you will get three brackets e.g. f(x)=(x+1)(x+p)(x+q).

If you don't like long division, as I know a few people don't, you can always find the other factors how you found (x+1), you subbed in -1 to get 0, you could try other numbers until you also get 0, for example IF you found out that subbing in 2 gave you 0, then (x-2) is also a factor. (That was just a random number, I don't know if that is actually an answer!)

Hope I helped.
8. (Original post by Benniboi1)
I'm guessing this is pretty simple, but I've never really got the hang of factorising polynomials.

I know that -1 is a root and the cubic equation is , what methods are there of now factorising and which is the quickest?

thanks!
If -1 is a root, then (x+1) is a factor.

So, (x+1)f(x) = x^3 + 6x^2 - 9x - 14, for some polynomial f.

deg LHS = deg RHS, so deg f(x) = 2. (Or in other terms, f(x) = Ax^2 + Bx + C, for constants A, B and C).

(x+1)(Ax^2 + Bx + C) = x^3 + 6x^2 - 9x - 14

Now it's just a case of expanding out the LHS and comparing co-efficients with the RHS. ie. Ax^3 = x^3, so A=1 etc.
9. (Original post by Zuzuzu)
If -1 is a root, then (x+1) is a factor.

So, (x+1)f(x) = x^3 + 6x^2 - 9x - 14, for some polynomial f.

deg LHS = deg RHS, so deg f(x) = 2. (Or in other terms, f(x) = Ax^2 + Bx + C, for constants A, B and C).

(x+1)(Ax^2 + Bx + C) = x^3 + 6x^2 - 9x - 14

Now it's just a case of expanding out the LHS and comparing co-efficients with the RHS. ie. Ax^3 = x^3, so A=1 etc.
I actually really like this method, it feels like it's familiar but I've never done it before.

thanks!
10. thanks very much everyone, only 18 minutes later and I've got 3 ways of doing it!
11. (Original post by Benniboi1)
I'm guessing this is pretty simple, but I've never really got the hang of factorising polynomials.

I know that -1 is a root and the cubic equation is , what methods are there of now factorising and which is the quickest?

thanks!
Always found this to be the easiest way and it's super quick as well, I've linked an image of me solving it with steps, hope it helps!
Attached Thumbnails

12. (Original post by Benniboi1)
thanks very much everyone, only 18 minutes later and I've got 3 ways of doing it!
Try this one out, it's quick and easy!
13. (Original post by Ali_Ludley)
Always found this to be the easiest way and it's super quick as well, I've linked an image of me solving it with steps, hope it helps!
thanks very much!
14. (x+1)f(x) = x^3 + 6x^2 - 9x - 14, for some polynomial f.
-> f(x) = x^3 + 6x^2 - 9x - 14/(x+1)

Use synthetic division (essentially putting the denominator in the numerator, with the aim of cancellations):

x^3 + 6x^2 - 9x - 14/(x+1) = x^2(x+1) + 5x(x+1) -14(x+1) / (x+1)

(x+1)'s cancel, leaving....
15. The method I would use is:

x^3 + 6x^2 - 9x - 14

= (x+1)(ax^2 + bx + c)
Since you know that -1 is a root, so (x+1) must be a factor.

Now to work out the values of a, b and c, multiply it out again:
= ax^3 + (a+b)x^2 + (b+c)x + c

Match up these unknown coefficients with the coefficients of the original cubic, and you get:
a = 1, b = 5, c = -14

So you can now write the expression as:
(x+1)(x^2 + 5x - 14)

And now you can factorise the quadratic, which is easy enough.
= (x+1)(x+7)(x-2)

I personally find this easier than polynomial division.
16. (Original post by tazarooni89)
The method I would use is:

x^3 + 6x^2 - 9x - 14

= (x+1)(ax^2 + bx + c)
Since you know that -1 is a root, so (x+1) must be a factor.

Now to work out the values of a, b and c, multiply it out again:
= ax^3 + (a+b)x^2 + (b+c)x + c

Match up these unknown coefficients with the coefficients of the original cubic, and you get:
a = 1, b = 5, c = -14

So you can now write the expression as:
(x+1)(x^2 + 5x - 14)

And now you can factorise the quadratic, which is easy enough.
= (x+1)(x+5)(x-2)

I personally find this easier than polynomial division.
Urm, only just looked at this... But your answer is wrong, the factors are (x+1)(x+7)(x-2)
Just pointing that out, didn't really look at your method.
x^3 + 4x^2 - 7x - 10

EDIT: Ah i see now, you just factorised the last bit wrong, nothing wrong with the method, might want to change the +5 to a +7
17. (Original post by Ali_Ludley)
Urm, only just looked at this... But your answer is wrong, the factors are (x+1)(x+7)(x-2)
Just pointing that out, didn't really look at your method.
x^3 + 4x^2 - 7x - 10

EDIT: Ah i see now, you just factorised the last bit wrong, nothing wrong with the method, might want to change the +5 to a +7
Yeah sorry, typo - fixed

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