Division of Polynomials
By M Ransom
Polynomials are expressions involving x raised to a whole number power (exponent). Some examples are:
, a long division process must be used. Both methods will be demonstrated.
We will now look at some examples showing how these two procedures are done.
#1. Suppose we are to divide
.
The long division method is shown below:
.
The long division process is shown below using the standard process for division:
because the divisor involves a highest power of x which is 1 and the coefficient of x is 1. The problem is setup below. The first setup shows a 0 which is the coefficient of x2 because there is no x2 term in the problem. There must be a coefficient for each power of x in order for this method to work.
because the coefficient of x is not 1.
In this lesson we consider division of polynomials such as:
There are two ways to calculate a division of polynomials. One is long division and a second method is called synthetic division. Synthetic division can be used when the polynomial divisor such as x-2 has the highest power of x as 1 and the coefficient of x is also 1. Otherwise, such asand
.
We will now look at some examples showing how these two procedures are done.
#1. Suppose we are to divide
We use standard division notation:#2. Divide:.
Division of polynomials using this method requires us to divide by x, the first term in the divisor, x-2. We divide this into the first term ofgetting
and then follow the normal procedures for division:
After the first division, the problem looks like this:
- Step 1. Divide
- Step 2. Multiply
- Step 3. Subtract
- Step 4. Bring down….
We have divided x into x3 and the result is x2. Multiplying x2 by x-2 gives. Subtracting and bringing down the next term give us
.
We repeat the procedure dividing x into the first term of giving us –3x. Multiplying, subtracting, and bringing down the next term gives the result shown below.Following this procedure one more time gives the result below:This result means:Because we are dividing by x-2 we can do “synthetic division” to get this same result. We note that +2 is subtracted from x in the divisor and set up the problems using only the coefficients of
. The initial setup looks like this:
The process is somewhat different than traditional division.After the first step, our problem looks like:
- Step 1. Bring down (the first term is 1).
- Step 2. Multiply with the “divisor” 2 giving 2.
- Step 3. Add giving -3. The result is shown below
Then we go back to Step 2 and repeat the process: multiply -3 by the “divisor” 2 giving us -6. We add and get -4. The result is shown below on the right.
Continuing this process one more time gives us the result below.
Note that the remainder in this division is 0 and is the last number on the right of the bottom row. We use the bottom row to give coefficients of 1, -3, and -4 for a final answer which is the same result as with long division:
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#3. Synthetic division can be used to calculateThe results are shown step by step below.
- Step 1. Divide
- Step 2. Multiply
- Step 3. Subtract
- Step 4. Bring down
Step 1: Setup the problem:Step 2: Divide 3x into 3x3, place the result x2 above the division line and multiply this by the entire divisor 3x+2. That result is placed below 3x3+5x2 and is subtracted. The -7x is the term we bring down to prepare for the next division by 3x.Step 3: Divide 3x into -3x2. The result is shown below including multiplying, subtracting and bringing down the last term, 5.Step 4: Finish the problem by dividing 3x into -9x, multiplying the result -3 by the “divisor” 3x+2, and doing the final subtraction to get a remainder of 11. The result is shown below.The divisionis completed. The final answer involves a remainder of 11. Thus
. We cannot use synthetic division because the divisor is 3x+2 and the coefficient of x is not 1.
Step 1 the setup:#4. Long division is needed to divideStep 2: bring down the 2.Step 3: Multiply 8 x 4 = 32. Multiply 2 x 4 = 8 and add. Add 32 + 4 = 36.Step 4: The final step is to Multiply 36 x 4 = 144 and add to the – 7.This final result 137 is the remainder. Therefore,.
Step 1: setup the problem:Step 2: Divide 2x into 2x4 and get x3. Multiply x3 by the divisor 2x+3 and get 2x4+3x3. Write this under 2x4+x3 and subtract. Bring down the 0x2. The result is shown below.Step 3: Divide 2x into -2x3 and get -x2. Multiply -x2 by the divisor 2x+3 and get -2x3-3x. Write this under –2x3+x2 and subtract. Bring down the 0x. The result is shown below.Step 4: Divide 2x into 4x2 and get 2x. Multiply 2x by the divisor 2x+3 and get 4x2+6x. Write this under 4x2+0x and subtract. Bring down the -2. The result is shown below. There is one more step.Step 5: Divide 2x into -6x and get -3. Multiply -3 by 2x+3 and get -6x-9. Put the -6x-9 under the -6x-2 and subtract. The remainder is +7. The result is shown below.Thus.
1.
Use synthetic division:

2.
Use synthetic division:

3.
Use long division:
