



OK,
this has all been great up to now,



but
where this stuff really gets used 


is
when we have something like: 





X^{2} +
12X + 32 





and
we try to figure out what two "factors" 


someone
might have multiplied together to get this. 





When
we start working on one of these 


we
don't know if we can find two factors 


but
let's try. 





To
wind up with an X
^{2}
term, 


we
need to multiply X times X. 





That
means that if we are going to be able to factor this thing, 


the
factors will start out like this: 





(X
)(X ) 





The
next thing we do is see 


that
all of the terms are positive. 





That
means if we CAN factor this 


all
of the factors will be positive numbers. 


That
means we can put in a bit more info ... 





(X
+ )(X
+ ) 





The
last (and hardest) part is to figure out 


what
number we need to finish off the factors. 





If
any numbers will work, 


they
will be numbers that multiplied together 


give
us the last number in our original problem. 


That
number was 32. 





So
we line up all of the pairs of numbers 


that
multiplied together give us 32. 


They
are: 





1 × 32 2
× 16 4 × 8 





For
the numbers to work, 


they
also have to add up to the number part 


of
the middle term (the 12). 





So
let's check to see if any of them do ... 





1
+ 32 = 33 2 + 16 =
18 4 + 8 = 12






So
4 and 12 are our winners. 


The
two factors we have built so far are both the same 


so
it doesn't matter which number goes where. 





So: 


X^{2} +
12X + 32 = (X + 4)(X + 8) 





Example: 





Let's
try to factor 


X^{2}
 4X  32






The
first term is still X ^{2}, so to get it 


the
factors still need to start out with: 





(X
)(X ) 





Since
we have some negative numbers 


in
the thing we are trying to factor 


we
know that there has to be a negative 


in
here somewhere. 





Since
the 32 is negative, 


we
know that only one of the two numbers is negative. 





If
both of the numbers had been negative, 


the
32 would have wound up being positive. 


Since
both of the factors are the same so far 


the
negative can go on either one. 


So we
have ... 


(X
+ )(X
 )






Now
we need to figure out what the numbers are. 


The
term on the end is 32 again, 


so
the numbers that might work 


are
the same as before. 


(Don't
worry about the minus sign yet) 





1 x
32 2 x 16 4 x 8 





NOW,
since we have one negative and one positive number 


we
are looking for numbers where the difference 


is
4. 


That's
pretty easy to spot. 





4  8
= 4






So
the numbers we need are 4 and 8. 


That
makes the factors: 





(X + 4)(X  8) 





That
makes the whole thing ... 





X^{2} 
4X  32 = (X + 4)(X  8) 





Remember,
there's no guarantee 


that
we will be able to factor these things. 


If
we can't find a pair of numbers 


that
multiply together to give us the last term 


and
the sum or difference gives us the number part 


of
the middle term 


we
can't factor the expression. 


At
least with the tricks we have so far ... 





Great,
Let's try one more. 





Example: 





Find
the factors of: 





2X^{2} 
15X + 18 





OK,
this one is a little trickier. 


We
have 2X ^{2} now instead of X
^{2}. 


To
get 2X ^{2}, we need to multiply 2X times X. 


That
means our factors start out as: 





(X
)(2X ) 





The
sign on the middle term is negative 


so
we know that there has to be at least one 


negative
number in the factors. 





The
third term is a positive number. 


We
know that since at least one of the number factors must be negative 


and
the third term is positive 


that
BOTH of the number factors must be negative. 


That
means our factors so far are: 





(X
 )(2X
 ) 





Now
we need to know what pairs of numbers 


can
be multiplied together to get 18 ... 





1 x
18 2 x 9 3 x 6 





Now
we're ready for the middle term (addition) part. 


Up
to now we've just tried adding or subtracting 


each
of the multiplication pairs to find the middle term number. 


Here,
the X factors are X and 2X. 


that
means in this problem one of the numbers 


gets
included TWICE when the middle term number gets built. 





OK,
the number we are trying to get to is 15. 


One
of the numbers will count twice. 





 1  1 
18 =  20




 1 
18  18 =  37 




2  2  9 =  13 




2  9  9 =  20 




3  3  6 =  12 




3  6  6 = 18 
We have a
winner! 





So
we need to use the 3 and the 6. 


We
also need to arrange them 


so
that the 6 gets multiplied times the 2. 


That
means the 6 needs to be in the other factor 


than
the one the 2X is in. 





So
the factors are: 





(X  6)(2X  3) 





Which
makes the whole thing ... 





2X^{2}
 15X + 18 = (X  6)(2X  3)






copyright 2005 Bruce Kirkpatrick 
