Question

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

Answer 1

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)`

Answer 2

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