- #1

- 1,011

- 0

## Homework Statement

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Miike012
- Start date

- #1

- 1,011

- 0

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

- #2

rock.freak667

Homework Helper

- 6,230

- 31

- #3

- 1,011

- 0

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

- #4

- 1,011

- 0

Which method would you use?

- #5

chiro

Science Advisor

- 4,790

- 132

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.

- #6

- 1,011

- 0

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.

- #7

- 1,011

- 0

- #8

rock.freak667

Homework Helper

- 6,230

- 31

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

- #9

- 1,011

- 0

And how do you know that it is irrational?

- #10

- 1,011

- 0

Are there any others that I should know of?

- #11

rock.freak667

Homework Helper

- 6,230

- 31

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.

- #12

- 1,011

- 0

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.

- #13

rock.freak667

Homework Helper

- 6,230

- 31

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.

- #14

eumyang

Homework Helper

- 1,347

- 10

No, it's

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

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.

- #15

- 1,011

- 0

I know the rat zero therm.

Is it hard to learn the iterative method?

Is it hard to learn the iterative method?

- #16

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

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 [itex]f(x)= x^3+ x^2- 1= 0[/itex], [itex]f'(x)= 3x^2+ 2x[/itex] so at, say a= 1, we would have [itex]x= a- f(a)/f'(a)= 1- 1/5= 0.8[/itex]. Now, taking a= 0.8, we would have [itex]x= 0.8- 0.152/3.77= .760[/itex] 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, [itex]f(0)= 0^3+ 0^2- 1= -1[/itex] while taking x= 1, [itex]f(1)= 1^3+ 1^2- 1= 1[/itex]. 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 lets just "bisect" the interval and try x= 1/2.

[itex]f(0.5)= (0.5)^3+ (0.5)^2- 1= -0.625[/itex]. 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. [itex]f(0.75)= (0.75)^3+ (0.75)^2- 1= -0.015625[/itex]. 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, [itex]x^3+ x^2- 1[/itex] cannot be factored with integer coefficeints. One way to see that is to use the "rational root theorem". If a polynomial, [itex]ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z[/itex], with integer coefficients, has a rational root [itex]x= \alpha/\beta[/itex], with [itex]\alpha[/itex] and [itex]\beta[/itex] integers, then it has a factor [itex]\beta x- \alpha[/itex] and so [itex]\beta[/itex] must be an integer factor of the leading coefficient, a, and [itex]\alpha[/itex] must be an integer factor of the constant term, z.

In [itex]x^3+ x^2- 1= 0[/itex] 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, [itex]x^3+ x^2- 1[/itex] cannot be factored with integer or rational coefficients.

- #17

eumyang

Homework Helper

- 1,347

- 10

Apparently not well enough, because you asked earlier:I know the rat zero therm.

I'll just quote a portion of HallsofIvy's excellent post:And how do you know that it is irrational?

In [itex]x^3+ x^2- 1= 0[/itex] 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, [itex]x^3+ x^2- 1[/itex] cannot be factored with integer or rational coefficients.

- #18

- 1,011

- 0

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

Share: