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### Factoring (x - 1)(x^2 - x - 2)

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Date: 11/04/97 at 23:22:03
From: Michelle
Subject: Algebra 2

I just have a question about how to factor. I know I did it in Algebra
but I'm not remembering how. The question is (X-1)(Xsquared-x-2). I
have gotten to a point, but then I don't know how to finish.  I have

xcubed _ xsquared - 2 x - xsquared - x + 2 = 0

xcubed + 3x + 2 = 0

Thanks, Michelle
```

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Date: 11/05/97 at 16:16:01
From: Doctor Keith
Subject: Re: Algebra 2

Great question.

I will go through the general ideas and path to solve the problem.
If something does not seem clear write back and I will clarify it.
For notational simplicity I will use x^2 for xsquared and sqrt(x) for
the square root of x.

First let's look at what we are trying to do. We are given a
polynomial, which in your case is (x - 1)(x^2 - x - 2), and we want to
find the factors.

The factors of a polynomial are terms of the form (x - a) such that

(x-a)(x-b)(x-c) = (x-1)(x^2 -x -2)

i.e. the product of the factors equals the polynomial. Many people
call this finding the zeros of a function because if we take x = a or
x = b or x = c that term becomes zero, and thus the whole polynomial
is zero at that point.

Example:

let x = a in the above factorization:
f(x) = (x-a)(x-b)(x-c)
f(a) = (a-a)(a-b)(a-c)
= (0)(a-b)(a-c) = 0

Okay, what did that tell us?  First we can see that we are given one
of the factors of your problem, namely (x-1)!  We thus only need to
factor (x^2 - x - 2).

As a general guideline I would not multiply through unless I had no
other alternative. For instance (x^2 - x - 2)(x^2 - 2x - 4) is easier
to factor by finding the factors of (x^2 - x - 2) and the factors of
(x^2 - 2x - 4) than to find the factors of the product of the two.
Note that if the factors of (x^2 - x - 2) = (x-a)(x-b) and the factors
of (x^2 - 2x - 4) = (x-c)(x-d) then the factors of
(x^2 - x - 2)(x^2 - 2x - 4) = (x-a)(x-b)(x-c)(x-d).

The reason I mention zeros of a function is that this is often how
factoring is referred to in later math courses and it is good to know
when those come by that you are just doing something you already know.

Now we need to find the factors of a quadratic, namely (x^2 -x -2).
There are several ways to do this, and I will show you my favorites:

for ax^2 + bx + c the factors are (x-d)(x-e) with d, e given by

-b + sqrt(b^2 -4ac)
d = -------------------
2a

-b - sqrt(b^2 -4ac)
e = -------------------
2a

This method works for all equations nicely and is really good for
have a = 1, b = -1, c = -2 thus

1 + sqrt(1 - 4(-2))     1 + sqrt(1 +8)         1+3
d = -------------------- = ------------------- =  ----- = 2
2                      2                2

1 - sqrt(1 - 4(-2))    1-3
e = -------------------- = ---- = -1
2                2
so:

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

2) The second method is a structured guessing method.

It could be solved explicitly (no guessing) but that would take
as much work as the original problem. The advantage is that this
can be good on tests if you are a quick guesser.

Note that (x-a)(x-b)= x^2 - (a+b)x + ab = x^2 - x - 2

Thus we can use this by requiring:

-(a+b) = -1
ab = -2

Now here is how we do the educated guessing. Since the order of the
factors doesn't matter, let a >= b in general. From the second
requirement we see that the product of a and b is negative; thus
the two terms have opposite signs (assuming we are using real
numbers, which is probably valid until college level math).

Since a >= b we must have a > 0 and b < 0 and thus a > b.
From the first requirement we see that a + b = 1, and since
a > b and b = -|b| (|x| is the absolute magnitude, and just
means dropping the sign for real numbers) we have

a = 1 - b = 1 + |b| > |b|

which tells us that the magnitude of a is bigger than the magnitude
of b. Moreover, since b < 0 and a > |b| then
-2 = -a|b| > -|b|^2  so
-sqrt(2) < b < 0.  This means that since a>|b| then a>sqrt(2).

Now we have narrowed down our area of guessing.  I usually try
letting one of the terms be +/-1 and the other be whatever will
satisfy the second requirement. We then check and see if the
first requirement is satisfied. Since we know a can't be 1 but
b could be -1, I will pick b = -1 and thus a(-1) = -2 so a = 2.
Now we try this in our first requirement and get -(2-1) = -(1) = -1
and both conditions are satisfied on the first guess.

(x - a)(x - b) = (x - 2)(x - -1) = (x - 2)(x + 1)

In most classroom problems this will work within a couple guesses.
What if it hadn't worked on the first try?  Well, if the sum of our
guesses were too big we would make b slightly more negative (know -
sqrt(2)<b<-1 ) and find the a that works for this new guess in
ab = -2 and try again.  If the sum were too small, then we'd want
to make b less negative (know -1<b<0) and again find the a that
goes.

As you can see, it converges quickly for simple problems, but for
ab = 2.1345643433 this method is a lost cause.

3)  Completing the square

This is a particularly pretty way of solving the problem.  What we
do is note that (x-a)(x-a) = x^2 -2ax + a^2. Now we can write any
quadratic as (x-a)^2 +b = x^2 -2ax +a^2 + b. So we require that in

-2a=-1  or  a = 1/2

then we have that

b + a^2 = b + 1/4 = -2    or b = -2.25 = -9/4

Now the factors are given by:

(x-a+sqrt(-b))(x-a-sqrt(-b))

so for us, since sqrt(-b) = 3/2 we get

(x-1/2+3/2)(x-1/2-3/2) = (x+1)(x-2)

Okay, we have derived the same answer by three methods. Which is best?
That depends on what is easiest for you. I will usually try a couple
of guesses using method 2, because with a little practice you can get
really good with it (even though it took more writing to explain
because it is more intuitive), especially since I now know the range
of expected values. If I don't get it quickly I will go to method 1
(quadratic equation), as this always works. When do I use method 3?
When I want to prove something in my work, as the beauty and neatness
of the expression really are impressive, and that is a big part of
writing research papers.

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

and we are done.

I hope this has helped.  If anything is unclear write back and I will
explain it more.

-Doctor Keith,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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