# How do you factor 14x^2 - 41x + 15?

Aug 5, 2015

Factor $y = 14 {x}^{2} - 41 x + 15.$

Ans: (7x - 3)(2x - 5)

#### Explanation:

$y = 14 {x}^{2} - 41 x + 15 =$ 14(x - p)(x - q)
I use the new AC Method. (Google, Yahoo, Bing Search)
Converted trinomial:$y ' = {x}^{2} - 41 x + 210 =$(x - p')(x - q')
p' and q' have same sign (Rule of signs)
Factor pairs of 210 --> ...(5, 42)(6, 35). This sum is 41 = -b.
Then, p' = -6 and q' = -35.
Therefor, $p = \frac{p '}{a} = - \frac{6}{14} = - \frac{3}{7}$, and $q = - \frac{35}{14} = - \frac{5}{2}$
Factored form: y = `14(x - 3/7)(x - 5/2) = (7x - 3)(2x - 5)