$\text{1) }2x^3-x^2+5x+3\\ =2x^3-2x^2+x^2+6x-x+3\\ =\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\\ =2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\\ =\left(2x+1\right)\left(x^2-x+3\right)\\$

$\text{2) }\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+2x+5x+10\right)\left(x^2+3x+4x+12\right)-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\left(1\right)\\ \text{Đặt }x^2+7x+11=y\text{ }\text{ }\left(\text{*}\right)\\ Thay\text{ }\: \left(\text{*}\right)\text{ vào }\left(1\right)\\ \text{ }Ta\text{ }đư\text{ợc }:\\ \left(1\right)=\left(y-1\right)\left(y+1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\text{ }\text{ }\left(2\right)\\ Thay\text{ }\left(\text{*}\right)\text{ vào }\left(2\right)\\ \text{Ta lại được: }\left(2\right)=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+17\right)\\ =\left(x^2+6x+x+6\right)\left(x^2+7x+17\right)\\ =\left[\left(x^2+6x\right)+\left(x+6\right)\right]\left(x^2+7x+17\right)\\ =\left[x\left(x+6\right)+\left(x+6\right)\right]\left(x^2+7x+17\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+17\right)\\$

$\text{3) }27x^3-27x^2+18x-4\\ =27x^3-18x^2-9x^2+12x+6x-4\\ =\left(27x^3-18x^2+12x\right)-\left(9x^2-6x+4\right)\\ =3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\\ =\left(3x-1\right)\left(9x^2-6x+4\right)\\$