Factoring x^3 + x^2 -1: Solution & Explanation

[Miike012]

Homework Statement

Does x^3 + x^2 -1 factor? and if yes... how?

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

Yes it has a factor however it is not rational. So you will need to change your methods for finding that root.

 

[Miike012]

I plugged it into my calc and it says it has an x-int.

 

[Miike012]

Which method would you use?

 

[chiro]

Which method would you use?

There is actually an algorithmic way to find the roots for polynomials with degree less than five.

The other way to do it is to guess a root for the polynomial (note that it could be complex) and then divide the polynomial by (x - c) where c is the root.

Since your equation is a cubic, if you guess one root, you can convert it to the product of (x-c) with a quadratic which you can solve using the standard quadratic formula.

I noticed that your polynomial is monic as well which means you have further properties that I can not recall.

 

[Miike012]

Interesting... I need to find and learn all these properties and this algorithm you are talking about...
If you or anyone else knows where I can find this info, please post the site and I'll be happy to check it out, thank you.

 

[Miike012]

And as far as picking a root... that could take all day couldn't it? I am sure its between 1-2 but that could be any decimal between those points.

 

[rock.freak667]

And as far as picking a root... that could take all day couldn't it? I am sure its between 1-2 but that could be any decimal between those points.

Your root is not rational, so you will need to use an iterative method.

 

[Miike012]

Never learned that method. Should I know that?... I've only taken algebra...
And how do you know that it is irrational?

 

[Miike012]

The only rules that I know are the diff of two sqares, and sum and dif. of two cubes...
Are there any others that I should know of?

 

[rock.freak667]

The only rules that I know are the diff of two sqares, and sum and dif. of two cubes...
Are there any others that I should know of?

How exactly do you wish to factor it? Like (x-a)(x-b)(x-c), if so then one of your roots are irrational and you can't factor it with algebraic manipulation.

 

[Miike012]

I wasnt referring to using diff of two sqares, and sum and dif. of two cubes on this problem... I just want to know if there are other methods that I should know for future reference.

How exactly do you wish to factor it? Like (x-a)(x-b)(x-c), if so then one of your roots are irrational and you can't factor it with algebraic manipulation.

 

[rock.freak667]

I wasnt referring to using diff of two sqares, and sum and dif. of two cubes on this problem... I just want to know if there are other methods that I should know for future reference.

You should know the remainder and factor theorem.

If f(x) is your polynomial and (x-a) is a factor of f(x) then f(a)=0.

So you check the last number of your polynomial (the term without 'x' in it), in this case it is -1.

The only way to get is to multiply 1 by -1, so you'd try f(1) and f(-1), if any give zero, then it is a root.

 

[eumyang]

And as far as picking a root... that could take all day couldn't it? I am sure its between 1-2 but that could be any decimal between those points.

No, it's not between 1 and 2.

Your root is not rational, so you will need to use an iterative method.

Never learned that method. Should I know that?... I've only taken algebra...
And how do you know that it is irrational?

I didn't learn the iterative method until I was in college, but I don't know if that's typical.

As for rational vs. irrational roots, read up on the Rational Zeroes Theorem.

 

[Miike012]

I know the rat zero therm.

Is it hard to learn the iterative method?

 

[HallsofIvy]

There are several different methods for numerically solving a polynomial equation that involve "iteration". Probably the fastest is "Newton's method" that works by approximating the graph of the polynomial by its tangent line at each point.

To find an (approximate) solution to f(x)= 0, near x= a, write the equation of the tangent line as y= f'(a)(x- a)+ f(a) where f'(a) is the derivative at x= a. That will be equal to 0 when f'(a)(x- a)+ f(a)= 0 which, solving for x, leads to x= a- f(a)/f'(a). Take that value of x to be the new "a" and iterate (repeat). For $f(x)= x^3+ x^2- 1= 0$, $f'(x)= 3x^2+ 2x$ so at, say a= 1, we would have $x= a- f(a)/f'(a)= 1- 1/5= 0.8$. Now, taking a= 0.8, we would have $x= 0.8- 0.152/3.77= .760$ to three significant figures. Keep doing that until you find that whatever number of decimal places you want keep repeating the same digits.

If you haven't taken Calculus and don't know how to find the derivative, a simpler but slower method is "bisection". Taking x= 0, $f(0)= 0^3+ 0^2- 1= -1$ while taking x= 1, $f(1)= 1^3+ 1^2- 1= 1$. Since any polynomial is "continuous" the only way the value can go from negative to positive is to pass through 0- there is some x between 0 and 1 such that f(x)= 0.

We don't know exactly where that x is so let's just "bisect" the interval and try x= 1/2.
$f(0.5)= (0.5)^3+ (0.5)^2- 1= -0.625$. Since that is negative, there must be a root between 0.5 and 1. Again, try half-way between (just because it is easy to calculate): try x= 0.75. $f(0.75)= (0.75)^3+ (0.75)^2- 1= -0.015625$. That is almost 0 so we are close to a root! It is also still negative so we know that root must be between 0.75 and 1. Halfway between is (0.75+ 1)/2= 0.875 so calculate f(0.875), determine whether it is positive or negative and continue.

By the way, in "elementary" algebra "factoring" normally means "factoring with integer coefficients". Your example, $x^3+ x^2- 1$ cannot be factored with integer coefficeints. One way to see that is to use the "rational root theorem". If a polynomial, $ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z$, with integer coefficients, has a rational root $x= \alpha/\beta$, with $\alpha$ and $\beta$ integers, then it has a factor $\beta x- \alpha$ and so $\beta$ must be an integer factor of the leading coefficient, a, and $\alpha$ must be an integer factor of the constant term, z.

In $x^3+ x^2- 1= 0$ the leading coeffient is 1 and the constant term is -1 which has, as integer factors, only 1 and -1 so the only "possible" rational roots are 1 and -1 and it is easy to see that they do not satisfy the equiation. Therefore, $x^3+ x^2- 1$ cannot be factored with integer or rational coefficients.

 

[eumyang]

I know the rat zero therm.

Apparently not well enough, because you asked earlier:

And how do you know that it is irrational?

I'll just quote a portion of HallsofIvy's excellent post:

In $x^3+ x^2- 1= 0$ the leading coeffient is 1 and the constant term is -1 which has, as integer factors, only 1 and -1 so the only "possible" rational roots are 1 and -1 and it is easy to see that they do not satisfy the equiation. Therefore, $x^3+ x^2- 1$ cannot be factored with integer or rational coefficients.

 

[Miike012]

I must have read over that part from Hallso. Thanks Eum.