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

Question

Eva Abramson · Accepted Answer

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

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)#.

Eva Abramson · Answer

undefined To factor the quadratic expression $3x^2 + 4x - 8$, you can use the quadratic formula or factor by grouping. Let's use the latter method. First, find two numbers that multiply to $3 \times (-8) = -24$ and add up to $4$. These numbers are $6$ and $-4$. Rewrite the middle term using these numbers: $3x^2 + 6x - 4x - 8$. Now, factor by grouping: $3x(x + 2) - 4(x + 2)$. Factor out the common factor $(x + 2)$: $(3x - 4)(x + 2)$. So, the factored form of $3x^2 + 4x - 8$ is $(3x - 4)(x + 2)$.