1. Lesson 2.6 Pre-Calc Part 2 When trying to ‘factor’ a quadratic into two binomials, we only ever concern ourselves with the factors of the ‘a’ --- leading coefficient, and ‘c’ – constant term. Example: 3x 2 – 11x – 4 = 0 (3x + 1)(x – 4) = 0 Solving for x  x = - 1/3. Or x = 4 (So we only concerned ourselves with the factors of ‘3’ and ‘4’ We call the possible factors of ‘c’  ‘p’ values We call the possible factors of ‘a’  ‘q’ values

2. This leads us into what is called the : Rational Roots theorem Let P(x) be a polynomial of degree ‘n’ with integral coefficients and a nonzero constant term. P(x) = a n x n + a n-1 x n-1 + …+ a 0 where a 0 /= 0 If one of the roots of the equation P(x) = 0 is x = ‘p/q’ where p and q are nonzero integers with no common factor other than 1, then ‘p’ must be a factor of a 0 and ‘q’ must be a factor of ‘a n ’ !

3. Example 3: a) According to the rational roots theorem what are the possible rational roots of : Px) = 3x 4 + 13x 3 + 15x 2 – 4 = 0 Note: If there are any ‘rational roots’, then they must be in the form of ‘p/q’. 1 st : List all possible ‘q’ values: + 1(+ 3) 2 nd : List all possible ‘p’ values: +1(+ 4) ; (+ 2)(+ 2) Therefore if there is a ‘rational root’ then it must come From this list of possible ‘p/q’ values: ‘p/q’  + (1/1, 1/3, 4/1, 4/3, 2/1, 2/3) or + 1,4,2,1/3,4/3,2/3  which means there are 12 possibilities!!!!

4. b)Determine whether any of the possible rational roots are ‘really’ roots. If so, then find all the roots??? Lets first evaluate x = 1, do you remember the quick and easy way to see if x = 1 is a root??? 2 nd : check x = - 1, do you remember the quick and easy way to see if x = - 1 is a root? Then check the other ‘possibilities’ using synthetic division: 3 13 15 0 - 4.. 6 38 106 212. 2) 3 19 53 106 208 Nope x ≠ 2, keep trying!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!