We have
[tex]f(x)=x^3-15x^2+50x[/tex].
To find zeros of this function we equate it with 0
[tex]x^3-15x^2+50x=0[/tex].
First we factor out x
[tex]x(x^2-15x+50)=0[/tex]
where we find that first zero is [tex]\boxed{x_1=0}[/tex].
Then we look at the expression in parentheses
[tex]x^2-15x+50=0[/tex]
using Viéts rule (factorisation)
[tex](x+a)(x+b)=x^2+x(a+b)+ab[/tex]
we can rewrite the equality
[tex](x-10)(x-5)=0[/tex].
If either of the terms is zero then the equality is true so we get two more zeros [tex]\boxed{x_2=10}[/tex] and [tex]\boxed{x_3=5}[/tex].
Hope this helps.
