How do you factor #40x^2 + 26x - 6#?

1 Answer
Mar 8, 2016

Factors are #40(x-(26+2sqrt409)/80)(x-(26-2sqrt409)/80)#

Explanation:

A simple way to factorize #ax^2+bx+c#, one needs to split middle term #bx# in two components, so that their product is #ac#. However, this is not possible in #40x^2+26x−6#, as product #ac# is #-240#, #26x# cannot be split into such set of integers.

Hence one ought to use quadratic formula. For this, let us find the roots of equation #40x^2+26x−6=0# using #(-b+-sqrt(b^2-4ac))/(2a)#

These are #(-26+-sqrt(26^2-4xx40xx(-6)))/(2xx40)# or

#(-26+-sqrt(676+960))/80# or #(-26+-sqrt(1636))/80# or

#(-26+-2sqrt409)/80#

Hence factors are #40(x-(26+2sqrt409)/80)(x-(26-2sqrt409)/80)#