Question 690722
Factoring {{{F(x)=x^3+x^2-42x}}} would help us find the zeros.
{{{F(x)=x^3+x^2-42x}}} --> {{{F(x)=x(x^2+x-42)}}} would be a first step (taking out the common factor {{{x}}}).
Factoring further is very easy if you have practiced factoring.
If not, here is how you would do it:
To factor {{{x^2+x-42=(x+a)(x+b)}}} <--> {{{x^2+x-42=x^2+(a+b)x+ab}}} 
we look for a pair of factors that multiply to give 42,
because we need to find numbers {{{a}}} and {{{b}}} such that {{{ab=-42}}} and {{{a+b=1}}}.
We look for the smaller factor, starting at 1 and going up until we get to the larger factor:
We can find four such pairs of factors:
{{{42=1*42}}}
{{{42=2*21}}}
{{{42=3*14}}}
({{{4}}} and {{{5}}} are not factors; they do not divide {{{42}}} evenly)
{{{42=6*7}}} ({{{42=7*6}}} is the same product and {{{7}}} is the larger factor here)
Now we consider that for the negative product {{{ab=-42}}}, one of those factors must be negative.
Since {{{42=6*7}}} and {{{(-6)+7=1}}}, {{{-6}}} and {{{7}}} are our numbers.
{{{x^2+x-42=(x+7)(x-6)}}} and
{{{F(x)=x(x+7)(x-6)}}} with zeros for x values that make
{{{x=0}}} <--> {{{highlight(x=0)}}},
{{{x+7=0}}} <--> {{{highlight(x=-7)}}}, and
{{{x-6=0}}} <--> {{{highlight(x=6)}}}.