# Factoring problems help

1. Feb 27, 2008

### Andy111

For some of these we can not factor exactly, so we must be as exact as we can, I kind of understand how to do it, but these are some problems that confused me a bit. We're not allowed to use graphing calculators for these by the way.

1. The problem statement, all variables and given/known data
1)1-3x$$^{2}$$=2x$$^{3}$$
2)x$$^{4}$$-11x$$^{2}$$+28x=0
3)4x$$^{3}$$=13x-6
4)2x$$^{3}$$-8x=0

2. Relevant equations
See above

3. The attempt at a solution

2. Feb 27, 2008

### rocomath

1) Let's start by solving for 0 and with respects to a POSITIVE leading term.

$$2x^3+3x^2-1=0$$

I'm assuming you're currently learning synthetic division? What would your next step be?

3. Feb 27, 2008

### Tedjn

In particular, there is no surefire way to factor a cubic equation (or any polynomial more complicated than a quadratic) besides educated guessing and checking. For what do you need to check? How should you make your guesses? Ask those questions, then apply rocophysics's suggestion.

4. Feb 27, 2008

### Andy111

What we do is plugin numbers to find between which intervals the sign changes (because that's where a zero will be), then we keep bisecting the interval until we get a value that when plugged in is close to zero. I did this and found two of the zeros (.5 and -1). So I got (2x-1)(x+1) but there should be one more answer, shouldn't there?

5. Feb 27, 2008

### rocomath

One thing that sucks about that method is that, you don't have to have a sign change for values before and after your roots. If $$(x-a)^n$$ and n is even, then it touches the x-axis but bounces back ... doesn't cross.

Use the Rational Roots Theorem to find which roots to test, and then once you find your first 0 using Synthetic Division. Your equation breaks down to a quadratic and then you can easily use the quadratic equation if it still doesn't factor nicely.

6. Feb 27, 2008

### Andy111

No, I don't know the rational roots theorem, or if I have learned it, I never equated a name to it.

7. Feb 27, 2008

### Andy111

And my teacher said not to use division because it would take too long on the test and we wouldn't have time to finish.

8. Feb 27, 2008

### rocomath

So exactly what are you learning? B/c plug and chug isn't going to cut it for harder problems.

9. Feb 27, 2008

### Tedjn

The Rational Root Theorem should be your first idea for a question such as this. What it does is test for easy roots that could help you reduce your higher degree polynomial into quadratics (at least in homework problems that's what usually happens).

Let's take a cubic equation as a case in point. What follows applies to all polynomials, but an example with a cubic is easier. You will find that we can factor our cubic as follows:

$$a_3x^3 + a_2x^2 + a_1x + a_0 = (qx - p)(b_2x^2 + b_1x + b_0)$$

where constants b0 through b2 are unknown.

Now, it is clear that if qx - p = 0, then the cubic equals 0. Thus, x = p/q is a root of the cubic--if we plug it into the cubic, we get 0. It is also clear from the factorization that a3x3 = qb2x3, thus q is a factor of a3. We also notice that a0 = -pb0, so p is a factor of a0.

Now, it is not guaranteed that q and p be factors of the leading and constant terms, because they may be irrational, in which case b2 and b0 are also irrational. However, if the cubic equation has a rational zero p/q, then p and q must be factors of a0 and a3. In general, for any polynomial which has a rational root p/q, p must be a factor of the constant term, and q must be a factor of the leading term.

This is the Rational Root Test. You try to find a easy, rational root of the cubic. Then you can use synthetic or long division to reduce the cubic to a quadratic.

Now, this means if you have the cubic equation ax3 + bx2 + cx + d, any rational root p/q will have $\pm p$ divide d and $\pm q$ divide a. Thus, you should try various $\pm p/q$ into the cubic to see if you get a zero. If so, you have found a rational root, and you know that qx - p is a factor. You can use division to find the quadratic that is left.

Last edited: Feb 27, 2008
10. Feb 29, 2008

### Gib Z

Although there is a general method for cubics (and quartics for that matter), both are quite lengthy. This limitation comes from only allowing things like cube roots and other rational operations in the solution. Eg if we allowed the inverse of f(x) = x^3 + x, we could have a formula for the 5th degree polynomial. Best chance here is rational roots theorem.

11. Feb 29, 2008

### HallsofIvy

Staff Emeritus
When in doubt, assume your teacher is not an evil person! While "real world" problems can be horrendous, class problems typically are not. What I would do with 1 is try simple values. Put x= 0, 1, -1, 2, -2, etc. and hope that you find a root. After you have one, you can reduce the equation to a quadratic.

(2) already is a quadratic- in x2- so that should be easy.

(3) Same advice as for (1).

(4) is trivial to factor.

12. Mar 2, 2008

### Andy111

Thanks for the help. Knowing the rational root theorem really helps.