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Factoring a Degree 2 Polynomial


Date: 3/21/96 at 0:40:13
From: Brandon
Subject: Math

I don't get this.  I have tried, but not succeeded!
I am in grade nine.

Questions like x exponent2 - 12x - 85  I don't get these factor 
things !

Please help me a.s.a.p
Thanks, Brandon


Date: 3/21/96 at 13:45:12
From: Doctor Ethan
Subject: Re: Math

Hey Brandon,

Well, factoring can be tricky.  There are a few tricks. 
Let's see if we can use them.

I will use x^2 to mean x exponent2, okay?

So we have 

x^2 - 12x - 85

Here is the idea.  We want to write this as the product of two 
binomials that look like 

(x + a)(x + b)

where a and b are any number. Right?

So we want to figure out what a and b have to be.

Well, let's work backwards for a second.

(x + a)(x + b) = x^2 + ax + bx + a*b  = x^2 + (a+b)x + a*b

Right?  

Well then, if we have the equation that you gave us, we know 
two things.

1. (a+b) = -12

2. a*b  = -85

Let's work with number 2 first.

If a times b is -85, what can a and b equal?

Let's make a chart.

  a  |   b
-----------
  1  |  -85
  5  |  -17
  17 |   -5
  85 |   -1

Now, which one of these possibilities gives us a + b = -12?

Well, let's try them.

1 + (-85) = -84  Nope.
5 + (-17) = -12  Yep.
17 + (-5) =  12  Nope.
82 + (-1) =  84  Nope.

Only one of them works, so 

(x + 5)(x + (-17))  =  (x + 5)(x - 17) 

is the answer.

Hope that this helps.

-Doctor Ethan,  The Math Forum


Date: 3/21/96 at 14:31:12
From: Doctor Terrill
Subject: Re: Math

Dear Brandon,

Factoring isn't an easy thing to figure out.  Usually, the answer 
isn't obvious, and you have to think a lot about the problem 
before you can figure it out.  I'll try to explain how I think 
about factoring.

First of all, instead of x exponent 2 I'll write x^2: the 
caret symbol means that the 2 should be up above the x, like an 
exponent.

When you factor a number, like 12, you write it as a product of 
prime numbers: for instance,

                  12 = 2 * 2 * 3    (* means times).

To factor x^2 - 12x - 85, we want to write it as a product of 
other algebraic expressions. In other words, when we factor
x^2 - 12x - 85, we will be able to write it as the product of two 
or more algebraic expressions.  

So, we want to figure out what a, b, c, and d will have to be to 
make the equation

            (a + b) (c + d) = x^2 - 12x - 85

true.

First, let's multiply out (a+b) (c + d).  A good way to remember 
how to do this is to think of (a + b) as one number at first.  

For instance,

                  7 * (c + d) =  7 * c   +  7 * d.

So, if we think of (a + b) as one number, we get

 (a + b) (c + d) = (a + b) * c + (a + b) * d.

Now, we can multiply out (a + b) * c and (a + b) * d.
This gives us 

ac + bc + ad + bd.

So, we want to figure out what a, b, c, and d have to be
to make the equation

x^2 - 12x - 85 = ac + bc + ad + bd.

Since we have an x^2 on the left-hand side, we know that
something on the right will probably equal x^2. 
Let's suppose that ac = x^2.

We know that ac = x^2 when a = x and c = x.  We'll try putting
these values into our equation, and see what happens.
(We could also have a = -x and c = -x, or a = x^2 and c = 1, 
but it's most likely that a = x and c = x.)

Substituting a = x and c = x into our equation gives us

x^2 - 12x - 85 = x ^2 + bx + ax + bd. 

We can factor the expression on the right by taking the x 
out of bx + ax.  This gives us

x^2 - 12x - 85 = x^2 + (b + d)x + bd.

Since we have a -12x on the left, we know that (b + d)x will have
to equal -12x. 

Also, since -85 is the only part of the equation on the left that
doesn't have an x in it, bd will have to equal -85.  ( x^2 has an
x in it, as does (b+d)x, so those can't equal -85.)

So, we know that b + d = -12, and bd = -85.
We need to find an a and b that work in both these equations.  
One way to do this is by making a list of all the b's and d's that
work in the equation bd = -85:

-85 = -1 * 85

= -85 * 1

= -17 * 5

= -5 * 17.

If all we wanted was b and d such that bd = -85, then any of these
numbers would work: for instance, we could have b = -1 and d = 85.

But we have other restrictions on b and d: they have to add up to 
-12. That only works for one set of numbers, -17 and 5, because
  
-17 * 5 = -85, AND -17 + 5 = -12.

So, b is -17 and d is 5.

We can substitute these into x^2 - 12x - 85 = (a + b) (c + d). 
This gives us:

x^2 - 12x - 85  = (x - 17) (x + 5).

We can check if this is the correct answer by multiplying it out:

(x - 17) (x + 5)  = (x - 17) * x + (x - 17) * 5 

= x^2 - 17x + 5x - 17 * 5

= x^2 - 12x - 85,

which is what we wanted. 

So the factorization of x^2 - 12x - 85 is  (x - 17) (x + 5).

It took a lot of steps to find the answer, but it gets
easier the more you do it.  

Here's one for you to try:  

            Try factoring x^2 + x - 6.

            Remember, we want to be able to write it in the form 

                    (a + b) (c + d),

and since we have an x^2 in the equation, a and c are probably both x.

So you want to find b and d in the equation

(x + b) (x + d) = x^2 - 12x - 85.

Try taking it from there.

I hope this helped,
	
-Doctor Terrill,  The Math Forum

    
Associated Topics:
High School Polynomials

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