How do you factor #6k ^ { 2} = 34+ 5#k?

2 Answers
Yiu A.
Nov 4, 2017

#k=17/6##or k=-2#

Explanation:

To solve this question, we can put all things to one side.

#6k^2=34+5k#

#6k^2-5k-34=0#

And we can use cross method to factorize the L.H.S. .

#(6k-17)(k+2)=0#

then, we get

#6k-17=0# #or k+2=0#

#k=17/6##or k=-2#

Here is the answer. Hope this can help you :)

Nghi N
Nov 4, 2017

(k + 2)(6k - 17)

Explanation:

Use the new AC Method to factor trinomials (Socratic Search).
#f(k) = 6k^2 - 5k - 34 =# 6(k + p)k + q)
Converted trinomial:
#f'(k) = k^2 - 5k - 204 =# (k + p')(k + q')
Proceed: find p' and q', then, divide them by a = 6 to get p and q.
Compose factor pairs of (ac = - 204) --> ...(6, - 34)(12, - 17). This last sum is -5. Therefor, p' = 12 and q' = - 17
Back to f(k), #p = (p')/a = 12/6 = 2#, and #q = (q')/a = - 17/6#
Factored form:
#f(k) = 6(k + 2)(k - 17/6) = (k + 2)(6k - 17)#

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