# Help in factorization of a third degree polynomial

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## Homework Statement

Looking to factor $-2x^3-3$ and having an issue. To my understanding, the Fundamental Theorem of Algebra tells us that it is at least theoretically possible to factor any polynomial of degree n.

## The Attempt at a Solution

So my first step to factor this was to use the Rational Zeros Theorem, and synthetically divide $-2x^3-3$ by these possible rational zeros. After multiple attempts, no possible rational zeros were found to correspond to any factors of $-2x^3-3$. So this tells me that we either have real but irrational zeros, or non-real zeros.
Where can I go from here to factor this thing out in knowing this?

## Answers and Replies

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SammyS
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## Homework Statement

Looking to factor $-2x^3-3$ and having an issue. To my understanding, the Fundamental Theorem of Algebra tells us that it is at least theoretically possible to factor any polynomial of degree n.

## The Attempt at a Solution

So my first step to factor this was to use the Rational Zeros Theorem, and synthetically divide $-2x^3-3$ by these possible rational zeros. After multiple attempts, no possible rational zeros were found to correspond to any factors of $-2x^3-3$. So this tells me that we either have real but irrational zeros, or non-real zeros.
Where can I go from here to factor this thing out in knowing this?
Solve $\ -2x^3-3=0\$ and use the Factor Theorem.

Mark44
Mentor

## Homework Statement

Looking to factor $-2x^3-3$ and having an issue. To my understanding, the Fundamental Theorem of Algebra tells us that it is at least theoretically possible to factor any polynomial of degree n.
That's not what it says.
To quote the wiki article, https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra,
The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.
In fact, Galois provided that there is no general factorization formula for fifth-degree and higher polynomials. That's not to say that you can't factor some fifth-degree or higher polynomials.
opus said:

## The Attempt at a Solution

So my first step to factor this was to use the Rational Zeros Theorem, and synthetically divide $-2x^3-3$ by these possible rational zeros. After multiple attempts, no possible rational zeros were found to correspond to any factors of $-2x^3-3$. So this tells me that we either have real but irrational zeros, or non-real zeros.
Where can I go from here to factor this thing out in knowing this?
Write it as $-2(x^3 + \frac 3 2)$. Do you have any formulas for factoring cubics of the form $x^3 + a^3$ or $x^3 - a^3$?

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Are you referring to the Zero Factor Theorem? I thought that was used for quadratics, and the reason for using things like the Rational Zeros Theorem, Descartes Rule of Signs, Intermediate Value Theorem, Upper and Lower Bounds, etc, was to drop a polynomial down to a quadratic so that we can use the Zero Factor Property?

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That's not what it says.
To quote the wiki article, https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra,

Write it as $-2(x^3 + \frac 3 2)$. Do you have any formulas for factoring cubics of the form $x^3 + a^3$ or $x^3 - a^3$?
But isn't that article quote saying the same thing? If it does have at least one complex root, then it can be factored, can't it?

Ok I do know the sum and difference of cubes formulas. Are you suggesting that I go with solving something like $-2(x^3 + 1.44714243)$ with the second term in the parentheses being equal to $3/2$ when cubed?

Mark44
Mentor
But isn't that article quote saying the same thing? If it does have at least one complex root, then it can be factored, can't it?
No, they're not saying the same thing. The Fund. Thm. of Algebra says that at least one complex root exists, but it doesn't tell you how to find it. You can't factor a polynomial if you don't know the roots.
opus said:
Ok I do know the sum and difference of cubes formulas. Are you suggesting that I go with solving something like $-2(x^3 + 1.44714243)$ with the second term in the parentheses being equal to $3/2$ when cubed?
No, it would be $-2(x^3 + 1.5)$, so you should be able to use one of your formulas to get a linear factor and a quadratic (which will have complex roots only).

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Could you please explain the 1.5? My train of thought here is that, by your recommendation, I use the sum of cubes formulas which is $a^3+b^3=(a+b)(a^2-ab+b^2)$.
But to use this, I need $-2(x^3+\frac{3}{2})$ to be in the form $-2(a^3+b^3)$. To do so, I would take the cube root of the 1.5, which will give me 1.44714243 which, when cubed, is equivalent to 1.5

Mark44
Mentor
Could you please explain the 1.5? My train of thought here is that, by your recommendation, I use the sum of cubes formulas which is $a^3+b^3=(a+b)(a^2-ab+b^2)$.
But to use this, I need $-2(x^3+\frac{3}{2})$ to be in the form $-2(a^3+b^3)$. To do so, I would take the cube root of the 1.5, which will give me 1.44714243 which, when cubed, is equivalent to 1.5
$-2(x^3 + (\sqrt[3]{1.5})^3)$ -- Now the part in parentheses fits your sum of cubes formula. It's best to leave the terms exact for now, not as decimal approximations.

In what you wrote in post #5, you had $-2(x^3 + 1.44714243)$, which is incorrect. The decimal approximation should have been to the power 3.

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Ok that makes sense.
Now as a more general question, which is why I was asking this question in the first place:

Consider a rational function $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$
If I want to graph this, there are a number of things I need to find. In all cases, I'll need to find the x-intercepts (if any) and the vertical asymptotes (if any).
Now in a rational function $f(x)=\frac{P(x)}{q(x)}$, our zeros, and x-intercepts are determined by $P(x)$. This is because we have a fraction, and for a fraction to equal zero, the numerator has to equal zero. Now for the vertical asymptotes, these are determined by $q(x)$, because any values that make the denominator equal to zero and are thus undefined and the vertical asymptote is at the zero x=c.

Now to find the zeros and vertical asymptotes, I need to be able to factor both the numerator and the denominator.

Given this problem, $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$, my text states that the fraction is already simplified. Which I take it to mean, that the numerator and the denominator have no common factors other than 1. Now to determine if this fraction is simplified, we would need to factor the numerator and denominator. But the text did no such factoring.

So, is there a way to notice that the numerator and denominator here have no common factors without factoring them? I'm finding a little bit of difficulty in this step. The prior section went over finding all of the zeros for higher degree polynomials, and that was easy with the use of the Rational Zeros Theorem, and accompanying theorems. However, in these graphing problems, they aren't nearly as neat or straightforward and I'm having a hard time factoring these. I'm starting to think that the last section had cherry-picked, very easy polynomials to factor. Whereas in reality, they aren't that straight forward.

StoneTemplePython
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Given this problem, $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$, my text states that the fraction is already simplified. Which I take it to mean, that the numerator and the denominator have no common factors other than 1. Now to determine if this fraction is simplified, we would need to factor the numerator and denominator. But the text did no such factoring.

So, is there a way to notice that the numerator and denominator here have no common factors without factoring them? I'm finding a little bit of difficulty in this step.
What I've written below is more of a point of interest -- something to think about for the future-- as I don't think it will directly help you now. Again emphasizing monic polynomials, you have

$f(x) = 2\frac{p(x)}{q(x)} = 2 \frac{x^3 + x^2 -2x + 1}{x^3-27}$

if hypothetically I told you that $p(x)$ has $k$ distinct roots (over $\mathbb C$) and $q(x)$ had $r$ distinct roots, then there are no common roots if and only if the 'combined' polynomial given by $g(x) = p(x)q(x)$ has $(k + r)$ distinct roots. Why?

There's another, more direct approach that is harder to describe that uses resultants.

In both cases, the underlying approaches come from Sylvester and require linear algebra to interpret and solve, I'm afraid.

I'm starting to think that the last section had cherry-picked, very easy polynomials to factor. Whereas in reality, they aren't that straight forward.
I think a lot of introductory textbook problems are this way. Solve easy stuff first... hopefully you can return to these questions and ideas you have at some future point once you have more powerful tools at your disposal.

Last edited:
epenguin
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Given this problem, $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$, my text states that the fraction is already simplified. Which I take it to mean, that the numerator and the denominator have no common factors other than 1. Now to determine if this fraction is simplified, we would need to factor the numerator and denominator. But the text did no such factoring.

So, is there a way to notice that the numerator and denominator here have no common factors without factoring them? I'm finding a little bit of difficulty in this step. The prior section went over finding all of the zeros for higher degree polynomials, and that was easy with the use of the Rational Zeros Theorem, and accompanying theorems. However, in these graphing problems, they aren't nearly as neat or straightforward and I'm having a hard time factoring these. I'm starting to think that the last section had cherry-picked, very easy polynomials to factor. Whereas in reality, they aren't that straight forward.
Perhaps the most obvious and elementary way to test whether or not you have a common factor in this, as you say, made easy case, is to factorise the denominator. It fairly obviously has the factor (x - 3). So you could divide the numerator by that and see whether there is any remainder, or easier just substitute 3 for x in it and see whether the result is 0 - if it is then (x - 3) is a factor of it.
To get the other factor of the denominator you could divide it by (x - 3). You get a real quadratic (one with nonreal roots). You can divide the numerator by that and see whether you get a nonzero remainder.

The Sylvester method is not very far from these things in spirit. I think what you have done has well prepared and given you an advantage for what you need to do beyond now, i.e. read at the next chapters in your algebra book which might be about complex numbers (in connection with polynomial equations in particular) or cube roots of unity, or cubic equations etc.

scottdave
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Great point @epenguin just divide the polynomials and see if there is a remainder, without having to factor.

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Thanks everyone. It appears these things aren't quite as simple as I originally thought! But I think I've got a better idea now. So can I expect to see these again down the road, and be introduced to new methods of solving these?

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Can I summarize by saying that if have a similar rational, with say $P(x)$ is of degree 6 and $q(x)$ is of degree 5, and neither has obvious factors, and I divide $\frac{P(x)}{q(x)}$ and a remainder results, then $q(x)$ doesn't divide evenly into $P(x)$, and there are no common factors?

StoneTemplePython
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Can I summarize by saying that if have a similar rational, with say $P(x)$ is of degree 6 and $q(x)$ is of degree 5, and neither has obvious factors, and I divide $\frac{P(x)}{q(x)}$ and a remainder results, then $q(x)$ doesn't divide evenly into $P(x)$, and there are no common factors?
Have you tried testing this idea over a simple example using what you know about synthetic division? For example, consider

$\frac{p(x)}{q(x)} = \frac{(x-3)(x-5)(x-6)}{(x-3)(x+3)} = \frac{x^3 - 14x^2 + 63x - 90}{x^2 - 9}$

so doing synthetic division, you can split the division into two steps. What you want is:

$(x^2-9)g(x) = p(x)$

but do the easy part first, i.e. first synthetically divide out $(x-3)$. What do you get? Is their a remainder? Then from that result synthetically divide out $(x+3)$. Now what happens?
- - - -
Coming up with simple examples to test your own ideas on... is an important habit to develop.