How do you solve #35k^2 - 22k + 7 = 4# by factoring?

1 Answer
Aug 7, 2015

Answer:

Factor: #y = 35k^2 - 22k + 7 = 4#

Ans: (5x - 1)(7x - 3)

Explanation:

#y = 35k^2 - 22k + 3 = #35(x - p)(x - q).
I use the new AC Method to factor trinomials.
Converted trinomial #y' = k^2 - 22k + 105#. p' and q' have same sign.
Factor pairs of (105) --> ...(3, 35)(5, 21)(7, 15). This sum is 22 -= -b.
Then p' = -7 and q' = - 15.
Therefor, #p = (p')/a = -7/35 = -1/5# and # q = -15/35 = -3/7#

Factored form: #y = 35(x - 1/5)(x - 3/7) = (5x - 1)(7x - 3)#