How do you factor #7y^2+47y-14 #?
1 Answer
Aug 10, 2015
Factor: y = 7y^2 + 47y - 14.
Ans: (7x - 2)(x + 7)
Explanation:
I use the new AC Method: y = 7(x + p)(x + q).
Converted trinomial: y' = x^2 + 47y - 98. Find p' and q' knowing sum (+ 47) and product (- 98). p' and q' have opposite signs.
Factor pairs of (-98) --> (-2, 49). This sum is 47 = b.
Then p' = -2 and q' = 49.
Therefor: p = -2/7 and q = 49/7 = 7
y = 7(x - 2/7)(x + 7) = (7x - 2)(x + 7)
