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Factor for X; Consecutive Odd Integers


Date: 01/19/98 at 14:30:29
From: Sarah 
Subject: Algebra factoring

(x-2)(x+3) = 6  Solve by factoring for x.

Also...

Label your variables, write an equation, solve, label answers.

Find three consecutive odd integers such that the sum of the first, 
twice the second, and three times the third, is 106.

      x = first odd integer

  x + 1 = second odd integer

  x + 3 = third odd integer

    x+2(x+1) + 3(x+3) = 106


Date: 01/23/98 at 10:23:44
From: Doctor Loni
Subject: Re: Algebra factoring

Good questions, Sarah.  Let's see if I can help you.

For the first one:

Solving a factoring problem is much easier if you make sure 
everything is equal to 0. That way we can set each part of 
the equation equal to zero. For instance, if:

      (x + 2) (x - 4) = 0 

then either (x + 2) = 0 or (x - 4) = 0 will make the equation true, 
because anything times zero is zero.

  so if     x + 2 = 0     x = -2

     if     x - 4 = 0     x = 4

So the solution would be that x would equal -2 or 4. If you put -2 
back in for x you would get:

     (-2 +2)(-2 -4) = 0(-6) = 0   Substituting 4 for x would give you 
                                  zero also.

In your equation, we first need to get everything equal to zero

      (x -2)(x + 3) = 6

Subtracting 6 from both sides gives you:

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

Since we can't really do much with the equation when it looks like 
this, let's multiply out the stuff in parentheses.

       x^2 + 3x - 2x -6      -6     = 0

Now we can combine terms:

      x^2 + x - 12 = 0

Now we can factor.  First write your parentheses:

   (         )(        )

(I am assuming here that you have some experience with factoring and 
know the basics of how to do it. If this is not true, write back and I 
will explain the basics of factoring to you.)

Since the first terms in the parenthesis multiply to give you the x^2 
term in the trinomial, your first factoring step would look like this:

      (x        )(x       )

Now you should be able to finish the factoring on your own.

For the second problem, you had the right idea, but you made an 
incorrect assumption. You assumed that because you are finding 
consecutive odd integers, the second integer would be x + 1 and the 
third odd integer would be x + 3. There is actually a difference of 
two between odd integers.

For instance, if you solve this equation and find that your first 
odd integer is 15, what will the next odd integer be? It will be 17. 
That is 2 more than 15, not one. Don't let the fact that they are odd 
integers fool you into thinking that the difference between odd 
numbers is an odd number. The distance from one odd number to the next 
is 2, just the same as it is for consecutive even numbers.

If       x = the first consecutive odd integer

     x + 2 = the second consecutive odd integer 
and
     x + 4 = the third consecutive odd integer

x + 2(x +2) + 3(x + 3) = 106   (you knew how to set up this part)

Now solve for x and that will be equal to your first odd integer. 
The second odd integer will be found by adding 2 to x, the third by 
adding 4 to x.

If this didn't quite answer your question, or you need more help, let 
me know!

-Doctor Loni,  The Math Forum
 Check out our web site!  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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