3.1: Polynomial Division
 Page ID
 237
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Polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree, a generalized version of the familiar arithmetic technique called long division. It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones.
1. Review of Long Division
Example 1
Use long division to calculate
\(\dfrac{495}{12}\)
and write the steps for this process without using any numbers.
Solution
4  1  \(\frac{3}{12}\)  
12    4  9  5  
4  8  
1  5  
1  2  
3 
We see that we follow the steps:

Write it in long division form.

Determine what we need to multiply the quotient by to get the first term.

Place that number on top of the long division sign.

Multiply that number by the quotient and place the product below.

Subtract

Repeat the process until the degree of the difference is smaller than the degree of the quotient.

Write as sum of the top numbers + remainder/quotient.
\[\dfrac{P(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)}\]
Below is a nonsintactical version of a computer program:
while (degree of denominator < degree of remainder)
do
{
divide first term of remainder by first term of denominator and place above
quotient line;
multiply result by denominator and place product under the remainder;
subtract product from remainder for new remainder;
}
Write expression above the quotient line + remainder/denominator;
Exercises

\(\dfrac{(3x^2 + 5x + 7)}{(x + 1)}\)

\(\dfrac{(2x^4 + x  1)}{(x^2 + 3x + 1)}\)
2. Synthetic Division

Bring down the 2.

Multiply \((1)(2) = 2\) and place it under the 0.

Add \(0 + 2 = 2\) and place it in the third row.

Multiply \((1)(2) = 2\) and place it under the 1.

Add \(1 + (2) = 1\) and place it in the third row.

Multiply \((1)(1) = 1\) and place it under the 7.

Add \(7 + 1 = 8\) and place it in the third row.

Write \(2x^2 + 2x + 1 + \dfrac{8}{(x + 1)}\).
For the special case that the denominator is of the form \(x  r\), we can use a shorthand version of polynomial division called synthetic division. Here is a step by step method for synthetic division for \(\frac{P(x)}{(x  r)}\):
Step 1:
Drop all the x's filling in zeros where appropriate and set up the division
r  a b c d
and place a horizontal line leaving space between the numbers and the line.
Step 2:
Put the first coefficient under the line
r  a b c d
a
Step 3:
Multiply \(r\) by the number under the line and place the product below the second
coefficient.
r  a b c d
ra
a
Step 4:
Add the second column and place the sum below the line.
Step 5:
Repeat steps 3 and 4 until there are no more columns.
Step 6:
The last number is the remainder and the first numbers are the coefficients
of the polynomial \(Q(x)\).
Example 2
Use synthetic division to find
\[\dfrac{(2x^3 + x + 7)}{(x + 1)}.\]
Solution:
\[2x^2 + 2x + 1 + \dfrac{8}{(x + 1)}\]
_  _  _  _  _  
1    2  0  1  7 
2  2  1  
_  _  _  _  _  
2  2  1  8  
Steps:

Bring down the 2.

Multiply \((1)(2) = 2\) and place it under the 0.

Add \(0 + 2 = 2\) and place it in the third row.

Multiply \((1)(2) = 2\) and place it under the 1.

Add \(1 + (2) = 1\) and place it in the third row.

Multiply \((1)(1) = 1\) and place it under the 7.

Add \(7 + 1 = 8\) and place it in the third row.

Write \(2x^2 + 2x + 1 + \dfrac{8}{(x + 1)}\).
3. The Remainder Theorem
Theorem: The Remainder Theorem
For any polynomial \(P(x)\)
\[P(r) = \text{ the remainder of }\dfrac{P(x)}{(x  r)}.\]
In particular, if \(P(r) = 0\) then the remainder is also 0.
Proof:
\[\dfrac{P(x)}{(x  r)} = Q(x) + \dfrac{R}{(x  r)}\]
Multiply both sides by \(x  r\) to get
\[P(x) = Q(x)(x  r) + R\]
Plugging in \(r\), we have
\[P(r) = Q(r)(r  r) + R = R.\]
\(\box\)
Exercise
Verify that 2 is a root of
\[x^3  3x^2 + x + 2\]
using the remainder theorem.
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.