Download Remainder Theorem

Math 3
Name:_________________________10/11
Dividing Polynomials
First, let’s consider what happens when we divide numbers. Say we divide 13 by 5.
2  quotient
5 13
We could write this as
 10
13
3  remainder
 2 3
5
5
Another way of thinking about this example is:
13  2  5  3
Division of polynomials is similar. If we divide a polynomial f (x) by ( x  r ) we obtain a result of the form:
f ( x)  ( x  r )q(r )  R
where q (r ) is the quotient and R is the remainder.
Example 1: f ( x)  3x 2  5x  8 by ( x  2) .
3x  11
2
x  2 3x  5 x  8
3x 2  6 x
11x  8
11x  22
14
We can conclude that:
3x 2 5x  8  ( x  2)(3x  11)  14
where the quotient q( x)  3x  11 and the remainder R=14
We can also write our answer as:
14
3x 2  4 x  8)  ( x  2)  3x  11 
Dividing Polynomials
x2
Synthetic Division
This is another way of dividing a polynomial by a binomial.
Say we were dividing f ( x)  x 2  5 x  6 by ( x  1)
First, write the coefficients ONLY inside an upside-down division symbol:
Make sure you leave room inside, underneath the row of coefficients, to write another row of numbers later.
Since we are dividing by ( x  1) ,
we look at when this expression is equal to 0.
x  1  0 when x  1. Put the this number, x = 1, at the left:
Take the first number inside, representing the leading coefficient, and carry it down, unchanged, to below the
division symbol:
Multiply this carry-down value by the test zero, and carry the result up into the next column:
Add down the column:
Multiply the previous carry-down value by the test zero, and carry the new result up into the last column:
Add down the column:
This last carry-down value is the remainder.
Remainder Theorem
If a polynomial f(x) is divided by (x-r) and a remainder R is obtained, then f(r) = R.
Example 2: Use the remainder theorem to find the remainder for Example 1, which was
f ( x)  3x 2  5x  8 by ( x  2) .
Since we are dividing f ( x)  3x 2  5x  8 by ( x  2) , we let x = 2 because this is when x  2  0 .
Hence the remainder, R, is given by:
R  f (2)  3(2) 2  5(2)  8  14
This is the same remainder we achieved before.
Factor Theorem
If the remainder f(r) = R = 0, then (x-r) is a factor of f(x).
The Factor Theorem is powerful because it can be used to find the roots of polynomial equations.
Example 3: Is x  4 a factor of 3x 3  x 2  20 x  5 ?
For this question we need to find out if dividing 3x 3  x 2  20 x  5 by x  4 leaves a remainder. If the
remainder is 0, then x  4 is a factor of 3x 3  x 2  20 x  5 . We can find this easily by using the Remainder
Theorem.
Since x  4  0 when x  4 , we find what f (4) is:
f (4)  2(4) 3  (4) 2  20(4)  5  123
The remainder  0 , so x  4 is not a factor of 3x 3  x 2  20 x  5 .
Example 4: Is x  1 a factor of f ( x)  x 3  2 x 2  5x  6
In this case we need to test the remainder r = -1.
R  f (r )  f (1)  (1) 3  2(1) 2  5(1)  6  1  2  5  6  0
Therefore, since f (1)  0 , we conclude that x  1 is a factor of f ( x)  x 3  2 x 2  5x  6