Given function : [tex]f(x)=5x^3-6x^2-59x+12.[/tex]
[tex]\mathrm{Factor\:}5x^3-6x^2-59x+12[/tex]
Splitting terms to factor it out.
[tex]5x^3-x^2-5x^2+x-60x+12[/tex]
Making it into groups
(5x^3-x^2) + (-5x^2+x) + (-60x+12)
Factoring out GCF of each group.
[tex]x^2(5x-1)-x(5x-1)-12(5x-1)[/tex]
[tex]=(5x-1)(x^2-x-12)[/tex]
Factoring x^2-x-12 into (x+3)(x-4)
[tex](5x-1)(x^2-x-12) = (5x-1)(x+3)(x-4)[/tex]
Applying zero product rule.
5x-1 =0
x+3=0 and
x-4= 0
On solving
x = 1/5 , x=-3 and x=4.
Therefore, zeros of the given function are [tex]x=-3,\:x=4,\:x=\frac{1}{5}[/tex].
