Factor problems help | Not allowed to use graphing calculators

[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.

Homework Statement

1)1-3x$$^{2}$$=2x$$^{3}$$
2)x$$^{4}$$-11x$$^{2}$$+28x=0
3)4x$$^{3}$$=13x-6
4)2x$$^{3}$$-8x=0

Homework Equations

See above

The Attempt at a Solution

 

[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?

 

[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.

 

[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?

 

[rocomath]

What we do is plugin numbers to find between which intervals the sign changes (because that's where a zero will be). 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?

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.

 

[Andy111]

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

 

[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.

 

[rocomath]

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

 

[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.

 

[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.

 

[HallsofIvy]

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.

 

[Andy111]

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