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Factoring 5x^2 + 2x -1

Date: 01/22/2001 at 20:56:29
From: Doug
Subject: Factoring

I can't figure out how to factor this:
5x^2 + 2x -1

The best I can do is:
(5x + 1)^2 + 2(x-1)

...but I'm not even sure that that is factoring.

Date: 01/23/2001 at 13:42:37
From: Doctor Keith
Subject: Re: Factoring

Hi Doug,

There are many ways to factor an expression.  One good example
from the Dr. Math archives is 

  What is Factoring?   

It explains the FOIL method.  

Another method I like to use is the quadratic formula. The steps for 
using the quadratic formula are:

1) Write down the formula:

        -b  sqrt(b^2 -4ac)
    r= ---------------------

2) Now we need to write what a, b, and c are.  To do this we recall 
that a, b, and c are defined by:

    a*x^2 + b*x + c

so for your problem:

    5*x^2 + 2*x - 1

you should then write what a, b, and c are.

3) Plug your a, b, and c from part 2 into the formula in part 1. This 
will give you the two roots. Note that one root is defined by the plus 
sign and the other by the minus:

         -b + sqrt(b^2 -4ac)
    r1= ---------------------

         -b - sqrt(b^2 -4ac)
    r2= ---------------------

4) Now you are almost done. All you have to do is put the results from 
3 into the factored form:

    a*x^2 + b*x + c = a*(x - r1)(x - r2)

And that is it, you have factored the quadratic!


On a final note, the method you were trying is called completing the 
square. You want to be careful when you do this that you don't 
accidentally change the formula. You had:

(5x + 1)^2 + 2(x-1) = (5x + 1)(5x + 1) + 2x - 2
                    = 25x^2 + 10x + 1 + 2x - 2
                    = 25x^2 + 12x - 1

Note that this is not your original problem. When completing the 
square you do the following:

1)  write your formula out:

    a*x^2 + b*x + c

2) We want to make the first two terms match a perfect square. An 
example of a perfect square is

    (x+d)^2 = x^2 + 2dx + d^2

To achieve this we factor the 'a' out of the first two terms to get:

    a*x^2 + b*x + c = a(x^2 + (b/a)x) +c

3) Now we want to make the terms in the parenthesis (x^2 + (b/a)x) 
look like the right-hand side of the perfect square example we gave 
above (x^2 + 2dx + d^2).  To do this we say that:

     b/a = 2d
       d = b/2a

4) We rewrite our equation using this:

    a*x^2 + b*x + c = a(x^2 + (b/a)x) +c
                    = a(x^2 + 2(b/2a)x) +c

5) We still need the d^2 term (or (b/2a)^2 ) so we just add and 
subtract it (a net 0).

    a*x^2 + b*x + c = a(x^2 + (b/a)x) +c
                    = a(x^2 + 2(b/2a)x) +c
                    = a(x^2 + 2(b/2a)x + (b/2a)^2 - (b/2a)^2) +c

6) We now have that the first three terms in parenthesis 
(x^2 + 2(b/2a)x + (b/2a)^2) are actually a perfect square, 
so we can rewrite it as:

    a*x^2 + b*x + c = a(x^2 + (b/a)x) +c
                    = a(x^2 + 2(b/2a)x) +c
                    = a(x^2 + 2(b/2a)x + (b/2a)^2 - (b/2a)^2) +c
                    = a( (x + (b/2a))^2 - (b/2a)^2 ) +c

7) All we have to do now is to multiply through by a:

    a*x^2 + b*x + c = a(x^2 + (b/a)x) +c
                    = a(x^2 + 2(b/2a)x) +c
                    = a(x^2 + 2(b/2a)x + (b/2a)^2 - (b/2a)^2) +c
                    = a( (x + (b/2a))^2 - (b/2a)^2) +c
                    = a(x + (b/2a))^2 - a(b/2a)^2 +c

And we have completed the square. Now you should note that completing 
the square and factoring are related but not the same.  You can use 
completing the square to factor by setting  the last line equal to 

  0 = a(x + (b/2a))^2 - a(b/2a)^2 +c

and solving for x.  The answer will be the quadratic formula with an x 
where I put an r. I think it is easier to just use the quadratic 
fromula myself.

I suggest trying each of the methods. Some are easier for some people, 
and some work better on certain problems. In the end this is where 
your  experience comes in.

Hope this helps. If you have any questions write back.

Best wishes,

- Doctor Keith, The Math Forum   
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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