How do I factor this cubic polynomial?

[shreddinglicks]

Homework Statement

-x^3 - 6x^2 -12x -8

Homework Equations

The Attempt at a Solution

I don't know, I just know the roots are -2 with multiplicity 3.

 

[Mark44]

Homework Statement

-x^3 - 6x^2 -12x -8

Homework Equations

The Attempt at a Solution

I don't know, I just know the roots are -2 with multiplicity 3.

If -2 is a root, then x - (-2) = x + 2 is a factor.
Do you know how to do polynomial long division? If you don't, do an internet search using that as the search phrase.

 

[shreddinglicks]

If -2 is a root, then x - (-2) = x + 2 is a factor.
Do you know how to do polynomial long division? If you don't, do an internet search using that as the search phrase.

How does that help me factor and find the roots? Let's assume I have this cubic equation and need to find the roots, what is the process of obtaining them?

 

[hilbert2]

Multiply $(x+2)^2$ with $-(x+c)$ and deduce from the result what $c$ must be to produce the polynomial in the question.

 

[shreddinglicks]

Multiply $(x+2)^2$ with $-(x+c)$ and deduce from the result what $c$ must be to produce the polynomial in the question.

But what if I don't know the roots?

 

[hilbert2]

Edit: Incorrect answer removed, sorry for any confusion. -hilbert2

 

[RPinPA]

It can be tedious if there are a lot of possible roots to try, but the Rational Root Theorem will tell you all possible roots.

Here's your polynomial:

-x^3 - 6x^2 -12x -8

The RRT says that any rational root must be of the form +-p/q where p is a root of the constant term (8, ignoring the sign) and q is a root of the leading coefficient.

That means the only possible rational roots, are +-1, +-2, +-4, +-8. So you evaluate the polynomial at each of those values, one by one.

 

[shreddinglicks]

It can be tedious if there are a lot of possible roots to try, but the Rational Root Theorem will tell you all possible roots.

Here's your polynomial:The RRT says that any rational root must be of the form +-p/q where p is a root of the constant term (8, ignoring the sign) and q is a root of the leading coefficient.

That means the only possible rational roots, are +-1, +-2, +-4, +-8. So you evaluate the polynomial at each of those values, one by one.

I will look into this.

 

[FactChecker]

1) If you don't know the roots, it takes a lot of luck (or a rigged example) for the roots to be rational. The general process for real roots is a formula that is MUCH more complicated than the quadratic equation. (I can not do it by memory, and I don't know anyone who can.) See https://en.wikipedia.org/wiki/Cubic_function#General_formula .
2) If you know that the roots are rational, the process described above will work. Once you find one root, you can do a polynomial division and solve the simpler second order polnomial.

 

[shreddinglicks]

Thanks FactChecker, I was just curious. I'm studying for a test that will cover systems of linear ODE's. The book I was reviewing had a problem with that characteristic polynomial. I wanted to see if I could get the eigenvalues without a calculator.

 

[FactChecker]

Thanks FactChecker, I was just curious. I'm studying for a test that will cover systems of linear ODE's. The book I was reviewing had a problem with that characteristic polynomial. I wanted to see if I could get the eigenvalues without a calculator.

When you see a final answer like a third order zero at -2, you can assume that someone started with a simple final answer of integer roots and worked backward to get the problem statement. That kind of final answer is very unusual otherwise. Other techniques are usually needed to find the roots.

 

[shreddinglicks]

When you see a final answer like a third order zero at -2, you can assume that someone started with a simple final answer of integer roots and worked backward to get the problem statement. That kind of final answer is very unusual otherwise. Other techniques are usually needed to find the roots.

OK, thanks. I guess it would be safe to assume I won't see something like this on my exam.

 

[FactChecker]

People who make up in-classroom tests usually make sure that the answer is "rigged" to be solvable using techniques available to students. If it's a take-home test, anything goes and it may be a lot harder.

 

[shreddinglicks]

People who make up in-classroom tests usually make sure that the answer is "rigged" to be solvable using techniques available to students. If it's a take-home test, anything goes and it may be a lot harder.

It's in class and the teacher specified no calculators allowed.

 

[symbolipoint]

The example is a very typical one which occurs as material for College Algebra or Pre-Calculus, in which students will learn about Rational Roots Theorem and the Zeros of Polynomials.

You have the theorem which tells you the possible roots to examine are -1, -2, -3, -4, -8, 1, 2, 3, 4, 8.

First, you can pull out a factor of -1, so you have
-1(x^3+6x^2+12x+8).

You will find that the first possible root which gives 0 remainder in Synthetic Division is -2.
Keep on checking possible roots using synthetic division.
You should find that the ONLY roots which will work are -2;
and you will find that the multiplicity for this root is 3.

I will assume that this is a very clear solution to your request for help on this factorization.

Spoiler: only if you still need

In case you want to see the full result:
-1(x+2)^3

 

[shreddinglicks]

The example is a very typical one which occurs as material for College Algebra or Pre-Calculus, in which students will learn about Rational Roots Theorem and the Zeros of Polynomials.

You have the theorem which tells you the possible roots to examine are -1, -2, -3, -4, -8, 1, 2, 3, 4, 8.

First, you can pull out a factor of -1, so you have
-1(x^3+6x^2+12x+8).

You will find that the first possible root which gives 0 remainder in Synthetic Division is -2.
Keep on checking possible roots using synthetic division.
You should find that the ONLY roots which will work are -2;
and you will find that the multiplicity for this root is 3.

I will assume that this is a very clear solution to your request for help on this factorization.

Spoiler: only if you still need

In case you want to see the full result:
-1(x+2)^3

Thanks, I'll will certainly look into this when I have a little more time after my exam.

 

[symbolipoint]

Thanks, I'll will certainly look into this when I have a little more time after my exam.

Do not wait. Look into BEFORE your examination so you have more chance to learn what you need to know how to do!

 

[Ray Vickson]

If you don't know the roots and have to solve this kind of problem, you have to try dividing the polynomial with $x-c$, where $c$ is a number that the constant factor $-8$ is divisible with (the possibilities are $\pm 1, \pm 2,\pm 4$ and $\pm 8$). For a cubic equation with integer coefficients and three real roots, a solution can always be found that way.

You don't need three real roots. Furthermore, real roots cannot always be found that way: only integer roots (if any) can be found using the rational root theorem. A a cubic equation can have integer coefficients and one or three real roots, but with all roots irrational, needing either a complicated solution formula or a numerical solving method.

 

[hilbert2]

You don't need three real roots. Furthermore, real roots cannot always be found that way: only integer roots (if any) can be found using the rational root theorem. A a cubic equation can have integer coefficients and one or three real roots, but with all roots irrational, needing either a complicated solution formula or a numerical solving method.

Ok, sorry for being careless.

 

[LCKurtz]

You don't need three real roots. Furthermore, real roots cannot always be found that way: only integer roots (if any) can be found using the rational root theorem. A a cubic equation can have integer coefficients and one or three real roots, but with all roots irrational, needing either a complicated solution formula or a numerical solving method.

So the OP doesn't misunderstand, you mean for this equation only integer roots can be found because the leading coefficient is $-1$. More generally, the rational root can indeed find the rational roots, if any, of polynomials with integer coefficients.

 

[symbolipoint]

So the OP doesn't misunderstand, you mean for this equation only integer roots can be found because the leading coefficient is $-1$. More generally, the rational root can indeed find the rational roots, if any, of polynomials with integer coefficients.

The original poster for the question is assumed to be a college algebra student trying to get some help on dealing with polynomial functions and the Rational Roots Theorem. Usually all one needs is the discussion in the textbook. Just this topic, some students find hard enough.

In the real world, someone might analyze data and try to find a polynomial function to fit. Rational Roots theorem could be insufficient in such cases.

 

[RPinPA]

I will look into this.

I made a very serious careless error from typing too fast. I said this

the Rational Root Theorem will tell you all possible roots.

and that is dead wrong. It's called the Rational Root Theorem because it tells you all possible rational roots. If there are any roots that are rational, it has to be one of those. It gives no information about other kinds of roots, complex or irrational. Except that you know any that aren't rational have to be irrational or complex.

I said it more correctly later in my post, but the damage was done.

 

[neilparker62]

I don't know, I just know the roots are -2 with multiplicity 3.

Seems like they are pretty much giving you the answer - what factor results from a root at -2 ? Repeat it 3 times!

 

[Mark44]

Seems like they are pretty much giving you the answer - what factor results from a root at -2 ? Repeat it 3 times!

I think you are misunderstanding what the OP meant in the same way I did in an earlier post. My new interpretation is that the OP knows what the roots are from the posted answer, but is asking how one would find these roots. Several posters mentioned the rational root test as a starting point for doing this.

 

[neilparker62]

I think you are misunderstanding what the OP meant in the same way I did in an earlier post. My new interpretation is that the OP knows what the roots are from the posted answer, but is asking how one would find these roots. Several posters mentioned the rational root test as a starting point for doing this.

Oh - ok sorry for that. Well it depends on what one can assume. If one root of the cubic is rational and can be obtained from the question ( -2 in this case) - or by trial and error - you could write something like:

f(x) = -(x+2)(x^2 + kx + 4) where the first and last terms in the quadratic factor are obtained by inspection and k by simple equation (2kx + 4x = 12x). Again , assuming one rational root, the nature of the other two can be determined from the coefficients in the quadratic factor. If it is a 'pre-cooked' example, the quadratic will most likely have rational roots.

 

[symbolipoint]

Oh - ok sorry for that. Well it depends on what one can assume. If one root of the cubic is rational and can be obtained from the question ( -2 in this case) - or by trial and error - you could write something like:

f(x) = -(x+2)(x^2 + kx + 4) where the first and last terms in the quadratic factor are obtained by inspection and k by simple equation (2kx + 4x = 12x). Again , assuming one rational root, the nature of the other two can be determined from the coefficients in the quadratic factor. If it is a 'pre-cooked' example, the quadratic will most likely have rational roots.

Yes (your f(x) example). Some other information may be needed, but you could use general solution formula of quadratic expression to find an expression for the roots - which may be irrational (or possibly, complex), depending on what is given. I am not sure about your second expression of 2kx+4x=12x.

 

[neilparker62]

Yes (your f(x) example). ... I am not sure about your second expression of 2kx+4x=12x.

Comes from obtaining terms of order x from the product. Alternatively terms of order x^2 in which case 2x^2 + kx^2 = 6x^2 using the OP's given cubic.