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Note: This is the alternative Synthetic Division page, for WWW browser which cannot display special symbols. In particular, I am using ^2 for "squared" and +- for "plus or minus" in this version. Please let me know if this is working for you or not, by sending me email. Return to the primary Synthetic Division page.
NOTE: The method shown below is not synthetic division. I will correct this article, eventually.
x+1 and 2x+3 are numbers. We don't know their values, because we don't know the value of x. But we can do arithmetic with them anyway. We can add them: (x+1)+(2x+3)=3x+4. I suppose you see how that is done. Here is a longer one:
(x^3+2x^2-x+4)+(x^2+x+2)=x^3+3x^2+6
The -x+x became zero, by the way. Subtraction is about the same. Be careful with the signs. Multiply these expressions:
(x+1)(2x+3)=2x^2+5x+3
How did we do that? Here is another arrangement, which may be clearer:
x+1
2x+3
------
3x+3
2x^2+2x
---------
2x^2+5x+3
It is a lot like normal multiplication. If x were 10, then this would be normal decimal multiplication. Anyway, you don't have to write it in columns, in order to multiply. But it may be easier for you. Let's multiply the two longer expressions. I will do it in columns:
x^3+2x^2- x+4
x^2+ x+2
--------------
2x^3+4x^2-2x+8
x^4+2x^3- x^2+4x
x^5+2x^4- x^3+4x^2
-----------------------
x^5+3x^4+3x^3+7x^2+2x+8
If we do this horizontally, without columns, we have to be careful that we multiply every combination of terms.
The normal method of dividing polynomials is called "synthetic division." Division comes in really handy when we try to factor polynomials. Let's try one which we already know comes out even: (2x^2+5x+3)/(x+1). This is the inverse of our first multiplication problem above. The easy way is to arrange things like long division.
------------
x+1 ) 2x^2+5x+3
The first term of our answer is 2x, because 2x(x+1)=2x^2+2x, which is approximately what we have under the divide sign. If x were 10, this would be the same as long division in base 10:
2x
-----------
x+1 ) 2x^2+5x+3
2x^2+2x
-------
3x+3
As you can see above, I subtracted, just as in long division. To continue, x+1 seems to go into 3x+3 3 times. So our answer is 2x+3, as we already knew from our multiplication problem:
2x+3
-----------
x+1 ) 2x^2+5x+3
2x^2+2x
------
3x+3
One thing we should remember is to save a place for the zeros, as in this problem:
---------------------
x+1 ) x^4+0x^3+0x^2+0x+1
That one comes out even, too. Change the sign and add when you subtract a negative term. How do we deal with remainders? Well if we have a remainder, we know that the division did not come out even. Sometimes, that is all we wanted to know, as when we factor. Otherwise, we end up with a fraction, as in 2x+1+3/(x-1).
Synthetic division is just like long division in almost every way.
Addendum (synthetic division):
Here is a larger division problem:
This diagram takes up a lot of space, and has a great deal of
redundancy. The following "shorthand" method (synthetic division) is fairly
popular:
2 | 1 -3 +0 +3 -1
| +2 -2 -4 -2
---------------------
1 -1 -2 -1 -3
As you can see, many of the numbers of our long division are there. The x-2 has become +2; we negate it so we can add later, instead of subtracting. And the answer is the bottom row, with the remainder at the right end. The way we do this is bring down the 1 and multiply it by 2 (the number that we are dividing by) to get the +2 in the next column. We add the -3 and +2 to get -1. We multiply that by 2 to get the -2 in the next column, etc. And we get our answer x^3-x^2-2x-1, with a remainder of -3.
This will work for polynomials which are of the forms x+a and x^n+cx^(n-1)+ . . . Most polynomials are not of those forms. But we often do need to divide by a binomial like px+q. Well, we could divide by p to get x+q/p. Similarly, we can divide bx^n+cx^(n-1)+ . . . by b, and then use synthetic division. We may end up with some fractions, which shouldn't complicate things too much. Synthetic division can be done with a computer, too.