Slide 2:
If a polynomial P(x) has integer coefficients and P/Q is a rational
root, then …
p is a factor of the constant
q is a factor of the leading coefficient.
In other words: you can find all the POSSIBLE solutions by creating
all the fractions that follow the above rules.
For example:
Given 4x3 – 3x2 – 5x + 3, the possible solutions are: the factors of the constant are 3 , 1 The factors of the leading coefficient are 1, 2, 4 The possible solutions are:
Slide 3:
Find all the solutions to the following.
x3 + 2x2 – 5x – 6 = 0 P:
Q:
p/q: Use synthetic division to find one solution
(Guess and check):
-1 1 2 -5 -6
-1 -1 6
1 1 -6 0 Factor the bottom row to find the other solutions.
1x2 + 1x – 6 = 0
(x + 3)(x – 2) = 0
x + 3 = 0 x – 2 = 0
x = -3 x = 2 The solutions are -1, -3 and 2.