Long Division, Synthetic Division, Remainder Theorem, and Rational Root (Zero) Theorem

Ok kiddies, let's learn Section 2.3.

This post is terribly long due to all the steps needed to complete example problems.  Don't be afraid.

Long Division of Polynomials

Say I want to divide   by

First, I have to set up the problem itself.   is the divisor and  is the dividend.

It's not that scary...well actually...I guess it is a little bit scary.

Now that you've set up the problem, completely ignore most of the problem.  Focus only on the leading divisor and the leading dividend.

Your mission now is to figure out how many times x goes into ...     (  x * ? = )

...

The answer is  in case you were wondering.

 

Long Division of Polynomials is very similar to Long Division that you learned when you were in Elementary school.  Your answer for the leading dividend goes above the second leading dividend, the answer for the second leading dividend goes above the third leading dividend, and so on.

Let's think about that 2.  Now that x has been multiplied by , the same needs to happen to the 2 from .  Use the distributive property.

Remember this: When subtracting this from the dividend, PUT IT IN PARENTHESES FIRST! Otherwise, things will get very, very confusing.

Now you can distribute and continue on with the problem.

The leading values will cancel out, and now you just repeat this basic process to solve the rest of the problem. 

(x multiplied by what equals , multiply the 2 by that answer, put it in parentheses, distribute, subtract, blah, blah, blah)

When you get to the very end and the last number does not equal zero, that is the remainder.  When writing the remainder, remember that the remainder is the numerator and the divisor is the denominator.

The correct way to write the final answer is 

BUT HOLD YOUR HORSES BATMAN! WHAT IF YOU HAVE A PROBLEM LIKE THIS?!

MISSING THE  VALUE? BLASPHEMOUS! 

Calm down champ.  Just do this.

Synthetic Division

This method is much, MUCH faster than Long Division.  Long Division is called Long Division for a reason.

We'll use the same problem as last time:

 divided by 

To set up a problem and use Synthetic Division, simply take the coefficients from the polynomial and lay them out like so.

See where I took these numbers from?  They're simply the coefficients from largest to smallest value.

Now before you try a billion numbers on the left side of the box, read the problem again.

divided by 

 

Don't even think about plugging any number in besides the one given to you!

This is the pattern to solve this problem.

Any numbers in the vertical pattern are added together.

Any numbers in the diagonal pattern are multiplied by the number outside.

See? 3 is simply dropped down (or it is the result of 3+0) and then it is multiplied by 2 to equal 6 in the next column.  6 will then be added by 2...

...and the pattern continues and continues.

Now that the problem has been solved, it needs to be written out correctly.

The rightmost bottom number is the remainder. 

(The remainder is written the same way as it is in Long Division)

All powers of x reduce by one because the remainder takes the place of  coefficient.

Much faster and simpler than Long Division, yes?

If     happens again, just remember to put a place holder for the missing x value, making the polynomial.  It is very important to list all coefficients when setting up the problem, even if some of them are zero.

Remainder Theorem

If a polynomial f(x) is divided by x-k, the remainder is r=f(k)

This sounds confusing and it is.

We'll use our beloved problem again.

  divided by 

What this is saying that if 

 and the remainder is 51,

 then (2)=51,

 and (2, 51) is a point on the graph of .

 This can be confirmed by substituting x=2 in the original function.

=51

Rational Root Theorem

Possible rational zeros          =           factors of constant term

                                                        factors of leading coefficient

Couldn't use the example I've used

throughout the entire post

because it doesn't have

any zeros :'(

The leading coefficient is 2 and the constant term is 3.

    factors of 3             =              

        factors of 2                               

= 

(1/1=1, 1/2=1/2, 3/1=3, 3/2=3/2)

From here, use synthetic division to test these eight possibilities.

Conveniently, positive 1 works.

That means that x=1 is a zero.

If you plug a zero you've found back into original function, you can find the other zeros without using Synthetic Division!

                                              

                                                       

                                                                           

Tada!

You made it to the end of the post.

Congratulations.