Question 547294
{{{(2x-9)(3x^2+26x+48)=0}}}

The Principle of Zero Factors just says that if a product is zero, then at least one of the factors is zero.  This means that either 
{{{2x-9=0}}}
{{{2x=9}}}
{{{x=9/2}}}
OR
{{{(3x^2+26x+48)=0}}}

Using the quadratic formula we see that either 
{{{x=(-26+sqrt(26^2-4*3*48))/2*3}}}
{{{x=(-26+sqrt(676-576))/6}}}
{{{x=(-26+sqrt(100))/6}}}
{{{x=(-26+10)/6}}}
{{{x=(-16)/6}}}
{{{x=-8/3}}}
or
{{{x=(-26-sqrt(26^2-4*3*48))/2*3}}}
{{{x=(-26-sqrt(676-576))/6}}}
{{{x=(-26-sqrt(100))/6}}}
{{{x=(-26-10)/6}}}
{{{x=(-36)/6}}}
{{{x=-6}}}

Therefore there are three answers that would make the formula correct, x=-6, x=-8/3 and x=9/2


Hopefully this helps, let me know if you are still unsure:)