1. Lesson 53
Factor Theorem & Remainder Theorem
Text: Chapter 4, section 3 & 4
 Flashback to algebraic long division:
Divide x3 4 x 2 5x 2 by x 2
Solution: Divisor and dividend
must be in descending
x2 2 x 1 order of powers and
x 2 x3 4 x2 5x 2 ALL powers must be
x3 2 x 2 represented.
0 2 x2 5x
.....2 x 2 4 x
..........0 x 2
...............x 2
..................0
To Check: (divisor)(quotient) + remainder = dividend
 Synthetic Division: can be used to simplify long division IF the divisor is
in the form x a . You then only need to use coefficients.
Example:
Divide 6 x3 19 x 2 16 x 4 by x 2
Solution:
Dividend
Divisor is 2 (solve x - 2=0) 2 6 -19 16 -4 coefficients
12 -14 4
To get the second line …
6 -7 2 0
bring down the first
coefficient. Multiply the
coefficient by divisor
and add. These are the coefficients
of the quotient. This is the
remainder
The quotient is 6 x 2 7 x 2

2. Example:
Divide: x 4 10 x 2 2 x 4 by x 3
Solution: Use Synthetic Division
 REMAINDER THEOREM – If p(x) is a polynomial then p(a) is equal to the
remainder when p(x) is divided by x – a.
Example:
Evaluate: f ( x) x4 10 x2 2 x 4 at f(-3)
Solution: f ( 3) ( 3)4 10( 3)2 2( 3) 4
81 90 6 4
1
This is equal to the remainder when we divided by x + 3.
Example:
Find the remainder when x 4 2 x 3 5 x 2 x 1
Solution: Use the remainder theorem

3.  FACTOR THEOREM: A polynomial P(x) has a factor x – a if and only if
P(a) = 0.
Example:
Determine whether x + 2 is a factor of f ( x) x3 6 x 4 . Determine all factors of
f(x).
Solution:
x + 2 is a factor if f(-2) = 0.
f ( 2) ( 2)3 6( 2) 4
f ( 2) 0
Therefore (x+2) is a factor.
Using synthetic division solve for the other factor.
-2 1 0 -6 -4
-2 4 4 f ( x) ( x 2)( x2 2 x 2)
1 -2 -2 0 ** ( x 2 2 x 2) does not factor
Example:
Factor completely: f ( x) x3 2 x 2 5 x 6
Solution:
Use the remainder theorem and factor theorem to find A factor. Then use
synthetic division to complete the factoring.
**Only try factors of 6 1, 2, 3, 6

4. Example:
Factor completely f ( x) 2 x3 3x 2 8x 3
Solution:
1 3
Possible “a” values are 1, 3, ,
2 2
(fractions are there because of the leading coefficient)
Assignment: Exercise 53 Q 1 to 17
Omit #5b & 6