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

May 16, 2015

Have you missed a leading $34 {c}^{3}$ term?

If you have, then we have a simple problem:

$34 {c}^{3} + 20 {c}^{2} - 17 c - 10 = \left(34 {c}^{3} + 20 {c}^{2}\right) - \left(17 c + 10\right)$

$= 2 {c}^{2} \left(17 c + 10\right) - 1 \left(17 c + 10\right)$

$= \left(2 {c}^{2} - 1\right) \left(17 c + 10\right)$

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

$20 {c}^{2} - 17 c - 10$

$= 20 \left(c - \frac{17 - \sqrt{689}}{40}\right) \left(c - \frac{17 + \sqrt{689}}{40}\right)$

May 16, 2015

There is another way. I use the new AC Method (Google , Yahoo Search) to factor trinomials.

$f \left(x\right) = 20 x 2 - 17 x - 10 = \left(x - p\right) \left(x - q\right)$
Converted trinomial: $f ' \left(x\right) = {x}^{2} - 17 x + 200 =$(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: $p = \frac{p '}{a} = \frac{8}{20} = \frac{2}{5}$, and $q = \frac{q '}{a} = - \frac{25}{20} = - \frac{5}{4}$.

Factored form: $f \left(x\right) = \left(x + \frac{2}{5}\right) \left(x - \frac{5}{4}\right) = \left(5 x + 2\right) \left(4 x - 5\right)$

Check by developing.
$f \left(x\right) = 20 {x}^{2} - 25 x + 8 x - 10 = 20 {x}^{2} - 17 x - 10.$ OK