



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: 





X^{3}
+ 4X^{2} + 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 X1) 


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) (X^{2}
+ 5X + 6) = X^{3} + 4X^{2} + 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) = X^{3} + 4X^{2} + 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 
