First things first, we need to create a common denominator. Let's factor everything first and see if anything divides out
(2(2x-1))/(3(x^2-2x-8)) + (3 xx 3)/(3(x^2-5x+4))2(2x−1)3(x2−2x−8)+3×33(x2−5x+4)
(2(2x-1))/(color(green)(3)color(green)((x-4))(x+2)) + (3 xx 3)/(color(green)(3)(x-1)color(green)((x-4)))2(2x−1)3(x−4)(x+2)+3×33(x−1)(x−4)
Nothing divides out, but it looks like the denominators are mostly the same, except that the right side doesn't have (x+2)(x+2), and the left side is missing (x-1)(x−1). Let's multiply both sides by the term they're missing (remember, multiply both the numerator and denominator)
(x-1)/(x-1) xx (4x+2)/(3(x-4)(x+2)) + (9)/(3(x-1)(x-4)(x+2)) xx (x+2)/(x+2)x−1x−1×4x+23(x−4)(x+2)+93(x−1)(x−4)(x+2)×x+2x+2
(4x^2-4x+2x-2)/(3(x-4)(x+2)(x-1)) + (9x+18)/(3(x-1)(x-4)(x+2))4x2−4x+2x−23(x−4)(x+2)(x−1)+9x+183(x−1)(x−4)(x+2)
Now that the denominators match, let's combine the numerators:
(4x^2-2x-2+9x+18)/(3(x-4)(x+2)(x-1))4x2−2x−2+9x+183(x−4)(x+2)(x−1)
That denominator could use some work...
3(x-4)(x+2)(x-1)3(x−4)(x+2)(x−1)
(3x-12)(x+2)(x-1)(3x−12)(x+2)(x−1)
(3x^2+6x-12x-24)(x-1)(3x2+6x−12x−24)(x−1)
(3x^2-6x-24)(x-1)(3x2−6x−24)(x−1)
3x^3-3x^2-6x^2+6x-24x+243x3−3x2−6x2+6x−24x+24
color(green)(3x^3-9x^2-18x+24)3x3−9x2−18x+24
Now we have
(4x^2+7x+16)/(color(green)(3x^3-9x^2-18x+24))4x2+7x+163x3−9x2−18x+24
I hope that helps!
