Write without absolute value: y=|x−3|+|x+2|−|x−5|, if x> 5

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

you can't solve x because it is 2 linear equation and you only have one equation