As posted, the expressions mean the following:
. . . . .\(\displaystyle \mbox{(a) }\, 5x^3\, -\, 17x^2\, -\, \dfrac{12x}{x^2}\, -\, 4x\)
. . . . .\(\displaystyle \mbox{(b) }\, 6x^2\, +\, 13x\, +\, \dfrac{5}{3x^2}\, +\, 26x\, +\, 35\)
However, from your subject line, I suspect that the grouping symbols and instructions were omitted, and that the intended exercises are:
. . . . .\(\displaystyle \mbox{(a') Simplify }\, \dfrac{5x^3\, -\, 17x^2\, -\, 12x}{x^2\, -\, 4x}\)
. . . . .\(\displaystyle \mbox{(b') Simplify }\, \dfrac{6x^2\, +\, 13x\, +\, 5}{3x^2\, +\, 26x\, +\, 35}\)
So you factored out the \(\displaystyle x\) from each of the numerator and denominator:
. . . . .\(\displaystyle \mbox{(a') }\, \dfrac{5x^3\, -\, 17x^2\, -\, 12x}{x^2\, -\, 4x}\, =\, \dfrac{x\, (5x^2\, -\, 17x\, -\, 12)}{x\, (x\, -\, 4)}\)
Making sure to note that \(\displaystyle x\) cannot equal 0, you cancelled to get:
. . . . .\(\displaystyle \mbox{(a') }\, \dfrac{5x^2\, -\, 17x\, -\, 12}{x\, -\, 4}\)
However, this is not the complete simplification. Now the quadratic in the numerator must be factored or, which ends up in the same place, the numerator must be divided by the denominator. Then cancel the common factor (noting the restriction on the domain) to get the final form.
Did your class not cover how to factor quadratics? To learn the usual method for this, try
here.
Once you have studied the lesson and learned the basic terms and techniques, apply them to the two quadratics in the second exercise. For the numerator, start by finding factors of (6)(+5) = +30 that add to +13. For the denominator, start by finding factors of (3)(+35) = +105 that add to +26.