



There is
a shortcut that we can use 


to see if
something like (X  2) is a factor 


of some
big polynomial like 2X ^{6}  4X
^{5} + 3X
^{3}
 11X ^{2} + 17X  14 


To use
this shortcut, the factor that we're checking 


must
start with X, not something like X
^{2} or 4X. 





Example: 





Check to
see if (X  2) is a factor of 





2X^{6}
 4X^{5} + 3X^{3}  11X^{2} + 17X  14






Here's
what we do ... 





Step
1) 


Make sure that there is a term 


for every
power of X less than the biggest one. 


If any
are missing (like X4 here) write a zero in it's place ... 





2X^{6}
 4X^{5} + 0 + 3X^{3}  11X^{2} + 17X  14 





Step
2) 


All powers of X are now present? Good! 


Now write
down just the coefficients 


(the
numbers without the X's or exponents). 





2  4 + 0 + 3 
11 + 17  14 





Step
3) 


Now look at the term that we want to check. 


Look at
it as: 








we need
whatever goes in the box. 


Since
our term is (X  2), we have a 2 in the box.



BUT if we
were checking (X + 5) 


we would
need to write it as (X  (5)). 


That
means that 5 would be in the box. 





Step
4) 


Take whatever goes in the box AND 


the
coefficients from step 2 and write them all down. 


Then draw
a line under the numbers ... 











Step
5) 


Take the first coefficient (here it's a 2) 


and write
it below the line ... 











Step
6) 


Multiply the number that you wrote under the line 


times the
number in the box. 


Write the
answer under the next number to the right (2 x 2 = 4) ... 











Step
7) 


Add the number you just wrote down 


and the
number above it. 


Write the
answer below the line (4 + 4 = 0) ... 











Step
8) 


Repeat steps 6 and 7 until you run out of numbers
on the
right ... 











Step 9) 


IF the last number that you write down under the line is a zero, 


the thing
that you were checking IS a factor of the big polynomial. 


If the
last number is NOT a zero, 


the thing
that you were checking is not a factor. 





Step 10) 


If the thing that you were checking 


is a
factor of the big polynomial, 


the other
factor is built from the numbers under the line. 


We write
powers of X next to the numbers. 


Start on
the left with an exponent one smaller 


than the
biggest exponent in the big polynomial. 


As we go
to the right, the exponents get one smaller each number. 


That is
... 


2X^{5} +
0X^{4} + 0X^{3} + 3X^{2}  5X + 7 





We can
leave out any term with a zero coefficient. 





2X^{5} +
3X^{2}  5X + 7 





That
means ... 





2X^{6}
 4X^{5} + 3X^{3}  11X^{2} + 17X  14 = (X
 2)(2X^{5} + 3X^{2}  5X + 7)






That's
just swell. 


Now we've
got good news, bad news, and good news. 





Good
News: Once you get the hang of it, 


this is
much easier to do than equation long division. 





Bad
News: This trick only works with X ± (a number) 


and we
still need to guess at what the factors might be. 





Good
News: This little trick has another tricky use! 





Say that
we wanted to find the value of: 





2X^{6}
 6X^{5}  8X^{4}  X^{2} + 6X 
10 when X = 4






Up to
now, what we would have to do is calculate 4
^{6}, 


then
multiply that by 2, 


then
calculate 4 ^{5}, 


then
multiply that by 6, 


add that
to the first calculation, and so on. 


It's a
big mess and takes forever! 


BUT NOW,
we have a new way to figure this out. 


Here's
what we do ... 





Set up
the polynomial coefficients 


just like
we did in the example. 


Fill in
any missing powers of X with zeros ... 





2
6 8 0
1 6 10 





The X
value that we want to solve for 


goes in
the box on the left ... 











Draw the
line and do the trick ... 











Here's
what makes this so great. 


The last
number that you wrote down (the 2) 


is the
value of 2X ^{6}
 6X ^{5}  8X
^{4}  X
^{2} + 6X  10
when X = 4. 


It's
true! It works! 


Don't
believe me? 


Check it
out the old way ... 





copyright 2005 Bruce Kirkpatrick 
