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Finding xintercept of cubic functions? / Factoring cubic functions 
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#1
Mar2310, 08:37 PM

P: 11

1. The problem statement, all variables and given/known data
My problem is that whenever I have to use cubic functions  whether it's finding the roots (when dy/dx kis a cubic function) or finding the xintercepts... With quadratic functions I usually use the quadratic formula. I try to figure out how to factor these equations in my head but I can rarely get it right. Right now my problem is with finding the xintercepts of y = 2x^{3}  3x^{2}  12x 2. Relevant equations Finding the local maximum/minimum when dy/dx = 4x^{3} + 12x^{2}  4x  12 3. The attempt at a solution I've tried using newton's method on the relevant equation question, but that failed. x_{1} equalled 3 and x_{2} equalled 109, using an initial guess of 1. I won't put working because I think this really isn't what I'm supposed to do. Don't really need answers to these questions, just really need some help on how to deal with cubic functions. Thanks a lot if anyone can help. 


#2
Mar2310, 08:54 PM

P: 150

Hi liquidwater,
To the first problem, you can circumvent dealing with a cubic altogether by factoring. For the second one, try factoring by grouping. I hope this helps! 


#3
Mar2310, 09:17 PM

P: 11

That's what I'm really having trouble with... "With quadratic functions I usually use the quadratic formula. I try to figure out how to factor these equations in my head but I can rarely get it right." I can't think of the right combination of numbers to use. 


#4
Mar2310, 10:28 PM

Mentor
P: 21,258

Finding xintercept of cubic functions? / Factoring cubic functions



#5
Mar2310, 11:38 PM

P: 11

Should I be able to this kind of thing in my head, just by looking at it? Or is there some sort of formulae/tricks that I can use? I can factorize quadratics fine (but I often just turn to quadratic formula). 


#6
Mar2310, 11:58 PM

Mentor
P: 21,258

The first one is pretty easy, since each term has a factor of x.
2x^{3}  3x^{2}  12x = x(2x^{2}  3x  12). Now it's just a matter of factoring 2x^{2}  3x  12. Pretty obviously, the factorization is going to look like (2x + ?)(x + ?). Now it's a matter of trial and error of picking factors of 12 that combine to give you a middle term of 3x. In factoring 12, one factor will have to be negative and one positive. For the other polynomial, your derivative, that's going to be a bit harder, but you can start by noticing that each coefficient is a multiple of 4. 4x^{3} + 12x^{2}  4^{x}  12 = 4(x^{3} + 3x^{2}  x  3 ). One approach that works in special cases is to notice x^{3} + 3x^{2} and  x  3 can be factored by a method called factoring by grouping, already mentioned by tjackson3. Another more general approach is to use the Rational Root Theorem to see if the cubic has any roots that are rational numbers. If p/q is a root of the cubic, then one factor is (x  p/q). This theorem says that if p/q is a root of a_{n}x^{n} + a_{n1}x^{n1} + ... + a_{1}x + a_{0}, then p has to divide a_{0}, and q has to divide a_{n}. For this cubic, x^{3} + 3x^{2}  ^{x}  3, if p/q is a root, then p has to divide 3 and q has to divide 1. This means that the possible roots are +/3 or +/1, or equivalently, that x + 3, x 3, x + 1, or x  1 are the possible linear factors. Each of these can be tested using long polynomial division or (easier) synthetic division. 


#7
Mar2410, 12:50 AM

P: 11




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