How do you factor #3x^2 + 4x - 8#?

1 Answer
Jun 13, 2016

Answer:

Factors of #3x^2+4x-8# are #3(x+(2-sqrt28)/3)(x+(2+sqrt28)/3)#.

Explanation:

In #3x^2+4x-8#, the discriminant is #4^2-4*3*(-8)=(16+96)=-112#, is not the square of a rational number. Hence we cannot factorize it by splitting middle term.

Hence, the way is to find out zeros of quadratic trinomial #3x^2+4x-8#. Zeros of #ax^2+bx+c# are given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#.

So its zeros, which are two irratiional conjugate numbers are given by quadratic formula and are

#(-4+-sqrt112)/(2xx3)# or

#(-4+-2sqrt28)/6# or

#(-4+-2sqrt28)/6# i.e. #(-2-sqrt28)/3# and #(-2+sqrt28)/3#

Now, if #alpha# and #beta# are zeros of quadratic polynomial, then its factors are #(x-alpha)(x-beta)#. However as we have #3# as coefficient of #x^2#, we should multiply it by #3#

Hence factors of #3x^2+4x-8# are #3(x+(2-sqrt28)/3)(x+(2+sqrt28)/3)#.