How do you factor by grouping #20c^2 - 17c - 10#?

Question

Savannah Anderson · Accepted Answer

Have you missed a leading #34c^3# term?

If you have, then we have a simple problem:

#34c^3+20c^2-17c-10 = (34c^3+20c^2)-(17c+10)#

#=2c^2(17c+10)-1(17c+10)#

#=(2c^2-1)(17c+10)#

If the quadratic as given is correct, it does not really group. It only has irrational factors:

#20c^2-17c-10#

#= 20(c-(17-sqrt(689))/40)(c-(17+sqrt(689))/40)#

Eva Abramson · Answer

There is another way. I use the new AC Method (Google , Yahoo Search) to factor trinomials. #f(x) = 20x2 - 17x - 10 = (x - p)(x - q)#
Converted trinomial: #f'(x) = x^2 - 17x + 200 = #(x - p')(x - q') with (a.c = 200). 
To find p' and q', compose factor pairs of 200. Proceed: (-4, 50)(-5, 40)(-8, 25). This last sum (25 - 8 = 17 = -b) . Then p' = 8 and q' = -25. 
We get: #p = (p')/a = 8/20 = 2/5#, and #q = (q')/a = -25/20 = -5/4#. Factored form: #f(x) = (x + 2/5)(x - 5/4) = (5x + 2)(4x - 5)# Check by developing.
#f(x) = 20x^2 - 25x + 8x - 10 = 20x^2 - 17x - 10.#  OK

Eva Adamson · Answer

undefined To factor by grouping 20c^2 - 17c - 10, you first multiply the leading coefficient (20) by the constant term (-10) to get -200. Then, find two numbers that multiply to give -200 and add to give the middle coefficient (-17). These numbers are -20 and 10. Next, split the middle term using these numbers: 20c^2 - 20c + 10c - 10. Factor by grouping: 20c(c - 1) + 10(c - 1). Now, factor out the common factor: (20c + 10)(c - 1). Finally, simplify: 10(2c + 1)(c - 1).