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Finding x-intercept of cubic functions? / Factoring cubic functions

  • #1

Homework Statement


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 x-intercepts... 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 x-intercepts of
y = 2x3 - 3x2 - 12x

Homework Equations



Finding the local maximum/minimum when dy/dx = 4x3 + 12x2 - 4x - 12

The Attempt at a Solution


I've tried using newton's method on the relevant equation question, but that failed. x1 equalled -3 and x2 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.
 

Answers and Replies

  • #2
150
0
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
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!
Thanks for replying,

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.
 
Last edited:
  • #4
33,183
4,867
Thanks for replying,

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.
Then you should spend some extra time going over factorization problems. Not being able to do simple factorization problems correctly will be a deterrent to getting the nuts and bolts of more complex problems.
 
  • #5
Then you should spend some extra time going over factorization problems. Not being able to do simple factorization problems correctly will be a deterrent to getting the nuts and bolts of more complex problems.
Thanks... I can't factorize cubic functions, I should have explicitly asked for help with that.

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).
 
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  • #6
33,183
4,867
The first one is pretty easy, since each term has a factor of x.
2x3 - 3x2 - 12x = x(2x2 - 3x - 12).
Now it's just a matter of factoring 2x2 - 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.
4x3 + 12x2 - 4x - 12 = 4(x3 + 3x2 - x - 3 ).

One approach that works in special cases is to notice x3 + 3x2 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 anxn + an-1xn-1 + ... + a1x + a0, then p has to divide a0, and q has to divide an.

For this cubic, x3 + 3x2 - 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
The first one is pretty easy, since each term has a factor of x.
2x3 - 3x2 - 12x = x(2x2 - 3x - 12).
Now it's just a matter of factoring 2x2 - 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.
4x3 + 12x2 - 4x - 12 = 4(x3 + 3x2 - x - 3 ).

One approach that works in special cases is to notice x3 + 3x2 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 anxn + an-1xn-1 + ... + a1x + a0, then p has to divide a0, and q has to divide an.

For this cubic, x3 + 3x2 - 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.
:approve: Thanks a lot! I'll start working on this stuff.
 

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