# How do you factor 6x^2+19x+3?

May 16, 2015

Use the new AC Method to factor.

$f \left(x\right) = 6 {x}^{2} + 19 x + 3 = \left(x - p\right) \left(x - q\right)$
Converted trinomial:$f ' \left(x\right) = {x}^{2} + 19 x + 18 = \left(x - p '\right) \left(x - q '\right) .$
Compose factor pairs of (a.c) = 18. Proceed (1, 18) = 19 = b. Then p' = 1 and q' = 18. We get: $p = \frac{p '}{a} = \frac{1}{6}$and $q = \frac{q '}{a} = \frac{18}{6} = 3$.
Factored form:$f \left(x\right) = \left(x + \frac{1}{6}\right) \left(x + 3\right) = \left(6 x + 1\right) \left(x + 3\right)$
Check: Develop$f \left(x\right) = 6 {x}^{2} + 18 x + x + 3 = 6 {x}^{2} + 19 x + 3$. Correct

Reminder about the Rule of Signs.
a. When a and c have same signs, compose factor pairs of c, or (a.c), with all positive numbers.
Example: f(x) = x^2 + 51x + 98 = (x - p)(x - q). Compose factor pairs of (a.c = 98). a and are both positive. Proceed: (1, 98)(2, 49). This last sum is 51 = b. Then p = 2 and q = 49.

b. When a and c have different signs, compose factor pairs of c, or (a.c), with all first numbers being negative.
Example. f(x) = x^2 - 11x - 42 = (x - p)(x - q). Compose factor pairs of (c = -42). Proceed: (-1, 42)(-2, 21)(-3, 14). This last sum is 11 = -b. Then the 2 real roots are -3 and 14.