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###### Factorising cubics

Original post by the bear

there is a handy formula for finding the roots:

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.

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)

Original post by Bigflakes

Thank you

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

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