Factoring

• Oct 8th 2012, 12:46 PM
Nervous
Factoring
How do you factor
$18x^3-57x^2-85x+100$ ?

The first thing I tried was grouping:
$3x(6x^2-19)-5(17x-20)$

That's not going to work, so I tried factoring out an x and then factoring the trinomial. However, I never got past factoring the trinomial...

I've gotten as far as
$x(18x^2-57x-85)+100$

However, I cannot factor the trinomial. I made a list of every possible combination of the factors of 18 and the factors of 85, and their results. None of them added up to -57.

Factors of 18:
$(1,18)(2,9)(3,6)$
Factors of 85:
$(1,-85)(-1,85)(5,-17)(-5,17)$

After that, I made the long list of these sums:
$(1*5)+(18*-17)=+-301$
(I put in the plus/ minus to save time, since switching the negative would change nothing.)
$(1*17)+(18*5)=+-73$
$(1*1)+(18*1)=+-...$
You get it. I went through every possible combination, but never got to a +-57.

So how do I get the answer:
$(x-4) (3x+5) (6x-5)$
• Oct 8th 2012, 01:37 PM
johnsomeone
Re: Factoring
Quote:

Originally Posted by Nervous
How do you factor
$18x^3-57x^2-85x+100$ ?

You first reserve hours for lots of fun being frustrated at computationally intensive trial and error. Seriously, it's hard to factor something like that, especially with that leading 18 coefficient. There are various tricks (there's even a formula, like the quadratic formula, for cubics (Cardano's Formula) - but it's rather messy), but none will make you happy unless you get lucky and one of them works prefectly. In general, it sucks.

If it's going to factor over the integers (meaning it has a rational solution) then it will have a root which is a ratio of a divisor of 100 divided by a divisor of 18.
Divisors of 100: {1, 2, 4, 5, 10, 20, 25, 50, 100}
Divisors of 18: {1, 2, 3, 6, 9, 18}
Thus plug each of these:
{1, 2, 4, 5, 10, 20, 25, 50, 100}
and {1/2, 5/2, 25/2}
and {1/3, 2/3, 4/3, 5/3, 10/3, 20/3, 25/3, 50/3, 100/3}
and {1/6, 5/6, 25/6} and {1/9, 2/9, 4/9, 5/9, 10/9, 20/9, 25/9, 50/9, 100/9}
and {1/18, 5/18, 25/18}
into this $18x^3-57x^2-85x+100$ and check if any of them gives you 0.
Then you're HALF done, because you have to check all the same numbers, but with a minus sign in front.
And when you're done plugging those 72 different fractions into that cubic, if none of them give you 0, then all you can say is that "either I made a computational mistake evaluating at least one of them - which is a virtual certainty - or there are no rational roots to that cubic, and so it can't factor nicely over the integers".
Have fun.

Quote:

Originally Posted by Nervous
The first thing I tried was grouping:
$3x(6x^2-19)-5(17x-20)$

That's not going to work, so I tried factoring out an x and then factoring the trinomial. However, I never got past factoring the trinomial...

I've gotten as far as
$x(18x^2-57x-85)+100$

Such approaches can work, but virtually never do. Unless you're fantastic at factoring, I'd suggest not wasting your time even trying to get it to work like this.
I'll give you an example where it works, but again, in my experience this rarely seems to be a beneficial thing to spend too much time looking for (unless you're some kind of polynomial multiplying savant):

$5z^3-3z^2-15z+9$

$= (5z^3 - 15z) +( - 3z^2 + 9)$

$= 5z(z^2 - 3) -3 (z^2 - 3)$

$= (5z - 3)(z^2 - 3)$.
• Oct 8th 2012, 03:53 PM
Nervous
Re: Factoring
I want to summarize your post in order to confirm that I understand you...

So, in short: Don't bother with quadratic formula. (The only reason I didn't to begin with was that this problem came from a lesson prior to the lesson on quadratics, so I assume there was another way to solve it.)

Second, I could try plugging in those 72 different values, but since I'll likely get something wrong, it would be indeterminately more difficult than a mere 72 values.

And in the end, you said that my other methods probably won't work either.

If I understand you correctly, then I can only commend you for this post. Thanks for the long, detailed post. You truly did say a lot about my problem, I even learned a little from it. Is there anything else you could tell me, anything you might have left out? If you're having trouble thinking of a possible something else, might I suggest something along the lines of how to reasonably factor that expression.
• Oct 8th 2012, 05:54 PM
johnsomeone
Re: Factoring
Quote:

Originally Posted by Nervous
So, in short: Don't bother with quadratic formula.

Correct. The quadratic formula is no help at all with cubic equations - at least until you've found one root. (Well, the quadratic formula is cleverly used in Cardano's Formula as the solution to a cubic, but that's another story. Google it if you're curious, but it's a mess.)

Quote:

Originally Posted by Nervous
Second, I could try plugging in those 72 different values, but since I'll likely get something wrong, it would be indeterminately more difficult than a mere 72 values.

I think doing that would be insane.

Quote:

Originally Posted by Nervous
And in the end, you said that my other methods probably won't work either.

That's not quite what I meant. What I meant was the ONE method that it seemed you were trying: pulling apart terms of the polynomial, factoring them individually, and then hoping that you can find some common factor that would be a factor of the entire cubic, is usually not successful. It can work in special cases, but in general I wouldn't waste too much time looking for such a lucky break. Give it a shot if you like, but understand that it will usually not work.
There are all kinds of things that might let you "see" how to solve a given cubic equation, but they're specific to the particulars of the equations. If you see them, awesome.
One factoring trick worth knowing is Eisenstein's Criterion - but it's of no help here. Eisenstein's Criterion, when it applies, will tell you when you don't even need to bother with those 72 rational root possiblities. Google it if you're curious.

Quote:

Originally Posted by Nervous
Thanks for the long, detailed post.

Quote:

Originally Posted by Nervous
Is there anything else you could tell me, anything you might have left out? If you're having trouble thinking of a possible something else, might I suggest something along the lines of how to reasonably factor that expression.

If I were given this problem, I would try 1, -1, 2, -2, 4, -4, 5, -5, 10, and -10. And then I would go do the laundry and never again waste anymore time with it. It turns out that, according to you, 4 is a root, so that would get it. (And once you get one root, you can polynomial division to be left with a quadratic, which can then be solved.)
If a human life hung in the balance and I absolutely had to solve it w/o access to a computer, and none of {1, -1, 2, -2, 4, -4, 5, -5, 10, -10} worked, then I'd use calculus to get a sketch of the graph, and plug in a few points, and then maybe decide on a few of those 72 values that seem like reasonable tries. If that still got me nowhere, then I'd apply Cardano's formula.
• Oct 8th 2012, 07:59 PM
Nervous
Re: Factoring
I really like that idea of using calculus, that would be an instance of math being an art instead of a science. But how would I do that, once I had the graph or the derivative? Use critical points, inflection points, what?
• Oct 8th 2012, 09:29 PM
johnsomeone
Re: Factoring
I'd find the critical points. The inflection points don't matter as much, though I might find them too, since they're very easy to find. Solving for the critical points requires solving a quadratic, so is doable. That and the general shape of a cubic would give me some idea of where the roots are. I'd evaluate the cubic at various integers to get a better idea of where the roots are. I'd be able to tell whether it had 1 or 3 real roots (or two, since a double root would show up as a critical point). Then I'd test the rational root possiblities (those 72 values), but only for the ones that seemd like that'd have a chance. Whether it has 1 or 3 real roots tells me something about the discriminant of its (still unknown) quadratic factor, which perhaps could be of help.