The Student Room Group

Factorising cubics

Is it possible to factorise a cubic without knowing any roots or do you have to guess the first root?
there is a handy formula for finding the roots:

cardano.png

:rofl:
Reply 2
Original post by the bear
there is a handy formula for finding the roots:

cardano.png

:rofl:

Thanks
As well as the bears helpful reply, heres a bit more ...
https://www.youtube.com/watch?v=N-KXStupwsc
but you should be able to spot one fairly trivial root for a cubic at a level. So either try -2,-1,0,1,2, ... and/or use a bit of insight about coefficients to spot the roots sign / value. So for instance
x^3 + x^2 + x + 1 = 0
must a negative root as if x>0, then the cubic > 0. Also x=-1 must be a root as the odd power terms and the even power terms will sum to zero. So factor out (x+1) and factorize the remaining quadratic. You could also think about d being the product of the roots so if theyre integers, you can look for the factors of d.

Quite a few guides out there
https://www.youtube.com/watch?v=MUSETrY1r0M
http://www.mash.dept.shef.ac.uk/Resources/web-cubicequations-john.pdf
But for 1/2 hr you could try sketching some of the cubics in the preivous pdf (without looking)
* the asymptotic behavaiour is determined by the sign of "a". This give you the sign of the cubic for large |x|
* at x=0, the tangent to the cubic is given by cx + d. This alone can often give a good idea of where simple (small) roots are (assuming the coefficients are not too large). The previous examples tangent is x+1, so the tangents root is x=-1 which is spot on in this case.
* sub a few values in if all else fails and use factor theorem to get roots. Use sign changes in the cubic to hunt down the interval that contains a root.
* the stationary points are given by solving the quadratic derivative = 0. These will be related to the roots. Takes a bit of time though, so its a fall back.
(edited 1 year ago)
Reply 4
Thank you
Original post by Bigflakes
Thank you

Once you have one factor then it's straightforward to get the quadratic without needing to divide.
Reply 6
Yep thanks

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