Algebra 1 Equation Division
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The Great Divide
Equation Division

 
 So how do you factor something 
 if you don't have a perfect square or cube?
 
 How would we find the factors?
 
 Would we take a guess?
 
 Yep! That's one way to do it ...
 
 Let's say we have the expression:
 

 X3 + 4X2 + X - 6

 
 Does this factor?
 
 Who knows!
 
 Let's guess that (X - 1) might be a factor.
 
 Swell, how can you tell for sure?
 
 Well, how would you check to see if 3 was a factor of 114?
 You would divide 114 by 3 and see if it came out evenly.
 That is, see if it divides with no remainder.
 If it does, then 3 is a factor of 114.
 Let's see ...
 

 
 Great 114 = 3 x 38.
 How does that help us?

 

 We can use the same procedure to see if 
 X - 1 is a factor of X 3 + 4X 2 + X - 6.
 We just divide the expression by X - 1.
 
 SAY WHAT???
 
 It looks a little goofy, but trust me, it works.
 
 First set up the problem ...
 

 
 The highest power of X in the whole expression is X 3.
 The highest power of X in the thing we're dividing by is X.
 X 3 divided by X is X 2.
 That means, if this thing is going to work,
 the first term we get from the division will be X 2.
 Write it in ...

 

 
 See that we line it up over the 4X 2
 because it's the same power of X.
 
 Now we multiply the term on top (the X 2)
 by the thing we're dividing by (the X-1)
 write the product under the expression
 and subtract it from the expression.
 This is exactly what we do in any division problem.
 

 

 
 So what's left starts with 5X 2.
 If (X - 1) divides into this expression,
 it will go 5X times, since the X from the (X - 1)
 times 5X equals 5X 2.
 THIS IS THE HEART OF THIS WHOLE 
 EQUATION DIVISION THING 
 Before you're done with this page, you need to understand
 why the next term we write on top is 5X.
 If you do, the rest is details.
 If you don't, you don't know this concept.
 
 Here's the next step ...

 
 Hey, this looks like it might work.
 What's left starts with 6X.
 If (X - 1) is going to divide into this, it will go 6 times
 since 6 x X = 6X.
 Let's see what we're left with.
 

 

 
 Since we have a remainder of zero,
 we know that (X - 1) times (X 2 + 5X + 6) equals X 3 + 4X 2 + X - 6.
 

 (X - 1) (X2 + 5X + 6) = X3 + 4X2 + X - 6

 
 and if we look close, we might even see that
 (X 2 + 5X + 6)  factors to (X + 3)(X + 2).
 That means we can say ...
 

(X - 1) (X + 3) (X + 2) = X3 + 4X2 + X - 6

 
 So swell, fine, wonderful, this actually works.
 But it has one big problem.
 We had to guess the factor correctly to make this work.
 If we had guessed, say, (X + 1) which is NOT a factor,
 the division would not have come out evenly.
 We would have to keep on guessing at factors
 until we got lucky and guessed one right.
 
 TRUE!
 
 The thing is, that before computers came along
 this was about the only way we had short of calculus
 to find these factors.
 
 Now a fancy calculator can check loads of factor guesses per second.
 Unfortunately, you still need to learn this method.
 
 Why?
 Well, partly because lots of the time 
 you aren't looking for all of the factors of some big expression
 you just want to know if some special thing is a factor
 and that's easy to check.
 Just divide and see if the remainder is zero.
 
 Example:
 
 Is 2X + 3 a factor of 10X 5 + 9X 4 - X 3 + 14X 2 - 17X - 12
 
 Another way to ask this is: Does 2X + 3 divide evenly into
 10X 5 + 9X 4 - X 3 + 14X 2 - 17X - 12 (without a remainder)?
 
 Let's see ...

 

 
 2X + 3 can't be divided evenly into 18, so we're stopped.
 This one didn't work out evenly.
 We have a remainder of 18.
 That means 2X + 3 is not a factor.
 
 If we want to get really technical,
 we could use this remainder (18)
 and the thing that we were dividing by (2X + 3)
 to create a funny last term of our division
 and have it work out evenly.
 We could call the last term:
 

 
 Then when we multiply this by (2X + 3) we would get the 18
 that would make things work out evenly.
 
 That would give is:

 

 
 While this little trick let us finish the problem,
 most of the time we only want to call things factors
 if they work out cleanly without the tricky fraction term at the end.
 

   copyright 2005 Bruce Kirkpatrick

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