Synthetic DivisionReference > Mathematics > Algebra > Polynomials
When you were in elementary school, you learned an algorithm for doing long division. You probably hated it, and wished you could use a calculator, but you still learned to do it. When you were in an algebra class, you probably learned that the same algorithm - with some slight tweaking - could be used to divide two polynomials. If one of the polynomials is a linear binomial with no leading coefficient, however, there is a short-cut method of long division you can use called synthetic division.
The simplest way to explain synthetic division is by example, so let's try some problems.
Example: x3 - 7x2 + 14x - 8 divided by x - 2.
Solution: First, we write out all the coefficients of the dividend in a row, with a box in front of them, and a line under them:
__| 1 -7 14 -8 ______________
Now take the opposite of the constant in the divisor. In this case, the constant is -2, so we want +2. Write this inside the box.
_2| 1 -7 14 -8 ______________
Now bring down the first coefficent beneath the line:
_2| 1 -7 14 -8 ______________ 1
Multiply that number by the number in the box and write it beneath the next coefficient:
_2| 1 -7 14 -8 _____2________ 1
Now add the two numbers in that column and write the result beneath the line:
_2| 1 -7 14 -8 _____2________ 1 -5
Multiply this number by the number in the box and write it in the next column:
_2| 1 -7 14 -8 _____2_-10____ 1 -5
Continue on in this way until the final column is filled:
_2| 1 -7 14 -8 _____2_-10___8 1 -5 4 0
Notice that the last number is a zero. This indicates that the division has no remainder. But what is the quotient? Easy! We simply take the numbers in the bottom row (except the zero at the end) and make them the coefficients of a new polynomial, with degree one less than the polynomial we started with:
1x2 -5x + 4. Of course, we don't need to write that leading coefficient of 1, so it's x2 -5x + 4.
Example: x5 + 32 divided by x + 2.
Solution: First, we note that there are several missing terms. These are terms with zero coefficients, and they need to be included in our setup:
__| 1 0 0 0 0 32 ______________________
Now we add in the opposite of the constant in our binomial divisor: -2
-2| 1 0 0 0 0 32 ______________________
And now we do the work:
-2| 1 0 0 0 0 32 ____-2___4__-8__16_-32 1 -2 4 -8 16 0
Again, we note that the remainder is zero, which means (x + 2) is a divisor of the original polynomial.
The quotient of this division is the fourth degree polynomial x4 - 2x3 + 4x2 - 8x + 16.
Example: 2x3 -9x2 +12x - 4 divided by 2x - 1.
Solution: This one is a little tricky, because we can only do synthetic division with a linear binomial with no leading coefficient, and this divisor has a leading coefficient of 2. So we can't use synthetic division. However, we can use synthetic division using the binomial (x - 1/2). Where did that come from? I took the divisor and divided it by two. It turns out that we often use synthetic division when trying to find roots, and if (2x - 1) is a factor, then so is (x - 1/2), so it works out well to do this. Here's our setup:
1/2| 2 -9 12 -4 _____________________
Now let's go for it!
1/2| 2 -9 12 -4 ________1____-4_____4 2 -8 8 0
Again, we see that this division comes out even, and the quotient is 2x2 - 8x + 8.
Thus, we could write 2x3 -9x2 +12x - 4 = (x - 1/2)(2x2 - 8x + 8). Since we actually wanted to divide by (2x - 1), we could multiply the first factor by two, and divide the second one by two:
2x3 -9x2 +12x - 4 = (2x - 1)(x2 - 4x + 4)