~SYNTHETIC DIVISION & FACTORING~

twinkle

~These techniques are used at all levels, including calculus & beyond.
For this course, you'll probably will use these for finding certain types of limits & finding factors or zeroes of poly functions. (If they are rational numbers)

~What's behind the technique?

~If f(x) is a poly fcn and you divide it by a linear expression, of the form
(x-b), then there will be a quotient & remainder.
The remainder, R=f(b). [i.e.,  f(x)/(x-b)=q(x)+ R/(x-b), where q(x) is the quotient (one degree less than f(x)].
Note that the remainder R is expressed over the divisor.

~If the remainder R=0, then f(x)=q(x)(x-b), which means that (x-b) is a factor of f(x).
Also, If R=f(b)=0, then x=b would be one root of the equation f(x)=0 (also called a zero of the fcn).
The other solutions would come from the quotient, q(x)=0.
So, to find them, you would have to work with the quotient.

~Synthetic division gives us an easy way to find remainders & quotients. So, when we get a R=0, we also find factors & thus roots of f(x)=0.

~When you use Synthetic division, you are dividing f(x) repeatedly by expressions of the form x-b. It will let you know when a R=0 pops up. When that does, you have found a linear factor of f(x) & a root of f(x)=0. Then working on the quotient, q(x), the same way, you can find the others.

~Note: If f(x) doesn't contain linear factors, forget it. Use some other method.

~How does Synthetic division work? Follow these steps.

1) write the poly fcn, f(x) in descending powers of x, making sure that any  missing power is visible with a zero coefficient in front.
i.e.,  Write x5+4x4-10x2-x+6 like this: x5+4x4+0x3-10x2-x+6
note that the 3rd power of x is missing from the expression & we have to account for it with a zero coefficient.

2)  Now, drop down the coefficients (just the numbers) 1   4   0   -10   -1   6

3) Now, decide what you want to divide by, remember you will be dividing by linear expressions of the form x-b. So, a good place to start is to divide by x-1.  This uses the number 1 in the process.

 ~What's the process?

~Drop down the leading coefficient of 1 (far left). Multiply the leading coefficient (which is 1 in this case) by 1 (since you are dividing by x-1) (if you were dividing by x-2, you would use 2) (if you were dividing by x+1, you would use -1) (if you were dividing by x+3, you would  use -3) get it! We chose to start dividing by x-1, so we are using 1.

 ~After you multiply the leading coefficient by 1 (in your head), then add the  result to the next coefficient & write the result. Continue with this technique.  i.e., multiply that result by 1 (in your head) & add it to the next coefficient &  write the result. If you do it correctly, you will get the next row of numbers.
1    4    0    -10    -1    6
1    5    5    -5    -6    0   ←new row of numbers (using 1 in the process)

~The last number in the row is the REMAINDER. Notice that R=0, so we know that the division by x-1 was exact, so x-1 is a factor & x=1 is a root of the poly f(x)=0 since R=f(1)=0. Also, we get the quotient which is defined by the other numbers & is one degree less than the original expression. i.e., 1x4+5x3+5x2-5x-6.  We can now work with these coefficients to find more factors & more roots.

~We can leave those numbers up there & just change our reference row to
1    5    5    -5    -6

~What happens when we don't get a 0 at the end? Well, we still get the
remainder, f(b) but no factors or roots, so we try again with another
divisor, like x-2 (we would be using 2 in the process). We would continue
to try divisors of the form x-b until a 0 pops up. When that happens, then  
we can change our reference row to work on the quotient for the other
factors & roots.

~In some problems (not here) if we try 2 numbers & the remainder, f(b)
changes sign, we would know there is a root between them, since the
poly would have to cross the x axis & it is a continuous function (the
intermediate value theorem for continuous fcns). So, we can then try
a rational fraction in the process. (between those numbers). Yes, this
process works for rational fractions also. If you try a rational fraction
& get a fractional result (when multiplying, then adding) you can stop.
It won't work. Also, if the leading coefficient is A and the last coefficient
is B, all rational fractions that will work must have tops which are factors
of B & bottoms which are factors of A. Just a little time saving device.

~So, we have found one factor of the expression, namely, x-1.
Using Synthetic division on the new row, try 1 again (it's possible for an
expression to have repeated factors), So, here we go.

1    5    5    -5    -6
1    6    11    6    0 ←yes, another 0, so x-1 is a factor again.

~Change to the new row & try it again (it's possible for an expression to have
the same factor multiple number of times)

1    6    11    6    
1    7    18    24 ←no such luck,  x-1 is not a factor for the 3rd time

~Now, forget about that row (when there is no zero) & try x-2  (using 2)

1    6   11   6
1    8   27   60 ←we can see, no matter what positive number I use, I
                          will never get a zero here, so lets try some negatives.
                          lets test x+1 (using -1 in the process)
1    6   11   6
1    5    6   0 ←yes, so x+1 is a factor also. Now, the new quotient with
                         the coefficients 1,5, & 6 is quadratic, since we started
                         with the 5th power & found 3 factors (remember, every
                         time we divide, the quotient drops a degree)

~Since the 3rd quotient is quadratic, take it out of the problem & try to factor
this expression by conventional means.

x2+5x+6 → (x+2)(x+3)

~So, in summary, we have factored x5+4x4-10x2-x+6 into 5 linear factors
(x-1)(x-1)(x+1)(x+2)(x+3). And, if we set this poly =0 we find the 5 roots,
namely, 1,1,-1,-2,-3. These are the zeroes of the poly (x intercepts).

~In some cases, the poly might not have all linear factors. It could have a
factor like x2 +1 that can't be factored any further. Hopefully, these factors
can be found after finding the linear ones by Synthetic division (When the
last quotient is quadratic).

~Example in calculus:   Let's say you want the answer to the following limit:
                               lim     (x3+64)/(x+4) ,  it's that 0/0 again
                               x→ -4

~So, you'll need to fool with it. How do you factor x3+64? Well, if you forgot,
use Synthetic division to find them.

~Write x3+64    like  1    0    0     64   & try 4
                                 1    4    16    128 nope!, no 0, useless to try positives,
try -4                         1    -4    16    0  ←yes, a 0, so x+4 is a factor.
                                now, back to the limit

            lim      [(x+4)(x2-4x+16)]/(x+4)   =  lim (x2-4x+16) = 48,  nice!
           x→-4                                            x→-4

Notice that the 2nd factor was formulated by the other numbers, 1,-4,16 above.
That gives, x2-4x+16 for the other factor (which happens to be the quotient)

~Hope you enjoyed this lesson & it helps out on future problems.
Save it for reference.