1. Factor Theorem &
Rational Root
Theorem
Objective:
proves the Remainder Theorem, Factor Theorem and the Rational Root Theorem.
2. The Factor Theorem:
For a polynomial P(x), x – k is a factor iff P(k) =
0
iff
“if and only if”
It means that a theorem and its converse are true
3. If P(x) = x3
– 5x2
+ 2x + 8,
determine whether x – 4 is a factor.
4 1 -5 2 8
4 -4 -8
1 -1 -2 0
2 3 2
4 2 8
2 5
x x x x
x x
remainder is 0, therefore yes
other factor
4. Terminology:
Solutions (or roots) of polynomial equations
Zeros of polynomial functions
“k is a zero of the function f if f(k) = 0”
zeros of functions are the x values of the points
where the graph of the function crosses the x-axis
(x-intercepts where y = 0)
5. Ex 1: A polynomial function and
one of its zeros are given, find
the remaining zeros:
3 2
( ) 3 4 12; 2
P x x x x
2 1 3 -4 -12
2 10 12
1 5 6 0
2
5 6 0
2 3 0
2, 3
x x
x x
x
6. Ex 2: A polynomial function and
one of its zeros are given, find
the remaining zeros:
3
( ) 7 6; 3
P x x x
-3 1 0 -7 6
-3 9 -6
1 -3 2 0
2
3 2 0
1 2 0
1, 2
x x
x x
x
7. Rational Root Theorem:
Suppose that a polynomial equation with
integral coefficients has the root p/q , where
p and q are relatively prime integers. Then p
must be a factor of the constant term of the
polynomial and q must be a factor of the
coefficient of the highest degree term.
(useful when solving higher degree polynomial equations)
8. Solve using the Rational
Root Theorem:
4x2
+ 3x – 1 = 0 (any rational root must have a numerator
that is a factor of -1 and a denominator
that is a factor of 4)
factors of -1: ±1
factors of 4: ±1,2,4
possible rational roots: (now use synthetic division
to find rational roots)
1 1
1, ,
2 4
1 4 3 -1
4 7
4 7 6 no
-1 4 3 -1
-4 1
4 -1 0 !
yes
4 1 0
4 1
1
4
x
x
x
1
1,
4
x
(note: not all possible rational roots are zeros!)
9. Listing Possible Rational
Roots
When remembering how to find the list of all
possible rational roots of a polynomial,
remember the silly snake puts his tail over his
head (factors of the “tail of the polynomial”
over factors of the “head of the polynomial”).
11. Ex 3: Solve using the Rational Root
Theorem:3 2
2 13 10 0
x x x
1 1 2 -13 10
1 3 -10
1 3 -10 0 !
yes
2
3 10 0
5 2 0
5, 2
x x
x x
x
5,1, 2
x
1, 2, 5,10
possible rational roots:
12. Ex 4: Solve using the Rational Root
Theorem: 3 2
4 4 0
x x x
possible rational roots: 1, 2, 4
1 1 -4 -1 4
1 -3 -4
1 -3 -4 0 !
yes
2
3 4 0
4 1 0
1, 4
x x
x x
x
1,1, 4
x
13. Ex 5: Solve using the Rational
Root Theorem:
3 2
3 5 4 4 0
x x x
possible rational roots:
1 2 4
1, 2, 4, , ,
3 3 3
-1 3 -5 -4 4
-3 8
3 -8 -4
-4
0 !
yes
2
3 8 4 0
3 2 2 0
2
, 2
3
x x
x x
x
2
1, , 2
3
x
To find other roots can use synthetic division
using other possible roots on these coefficients.
(or factor and solve the quadratic equation)
2 3 -8 4 3 2 0
6 -4 3 2
3 -2 0
x
x
2
3
x
14. It’s your turn
x3
+ 4x2
– 15x – 18
x – 3
x3
– 11x2
+ 14x + 80
x – 8
x3
+ 4x2
– 15x – 18
x – 3
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