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### Factoring and the Factor Theorem

```
Date: 09/28/98 at 17:04:26
From: Josh Katus
Subject: Factoring 3rd Degree Polynomials

I am currently taking a 12th grade calculus course my teacher
recommended me to you to get help in factoring the equation:

x^3 - x^2 + x - 2 = 4

We know the answer is x = 2 but we are unsure how to get it. We tried

x^3 - x^2 + x = 6

Then we factored out an x:

x(x^2 - x + 1) = 6

That is as far as we managed to get.

Concerned in Calc Class
```

```
Date: 09/29/98 at 07:08:22
From: Doctor Anthony
Subject: Re: Factoring 3rd Degree Polynomials

First write the equation with everything on the left and zero on the
right.

f(x) =  x^3 - x^2 + x - 6 = 0

Then use the Factor Theorem, which states:

If f(a) = 0, then (x-a) is a factor of f(x).

This is easily proved, for if we divide f(x) by (x-a) and the quotient
is q(x) with remainder r, then:

f(x) = (x-a)q(x) + r

and   f(a) =  0 + r.

So if    r = 0  (x-a) must be a factor of f(x).

We therefore evaluate f(1), f(-1), f(2), f(-2), ... and if any of
these proves to be zero we immediately have a factor of f(x).

Since f(x) = x^3 - x^2 + x - 6 = 0:

f(1) = 1-1+1-6  which is not = 0, and similarly f(-1) NOT = 0

f(2) = 8 - 4 + 2 - 6 = 0

so by Factor Theorem x-2 is a factor.

Divide by x-2 using long division.

x^2 + x + 3
____________________
x-2)x^3 -  x^2 + x - 6
x^3 - 2x^2
------------
x^2 + x - 6
x^2 -2x
----------
3x - 6
3x - 6
-------
0   0

So f(x) = (x-2)(x^2 + x + 3)

The quadratic factor does not factorize further, so that completes the
factorization.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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