Question 815776
First of all we should rearrange the equation so the equation is equal to 0. 

So it would look like this: 
{{{27x^3+45x^2-3x-5=0}}} 

Since there are four terms using the 'grouping' method is probably our best choice to factor. One does this by first treating the equation as two separate parts. Like so: 

{{{27x^3+45x^x2}}} & {{{-3x-5}}} 

Then with each 'separate' part we factor out anything so: 

{{{27x^3+45x^2}}} factored out will be {{{(9x^2(3x+5))}}} 

{{{-3x-5}}} factored out will be {{{(-1(3x+5))}}} 

Then we cancel one of the like terms which is 3x+5 and combine the factored out elements so we are left with:

{{{(9x^2-1)(3x+5)}}} as factors. 

However, we are not done here as {{{9x^2-1}}} can be factored even more so it will be: 

{{{(3x+1)(3x-1)}}}

So our factors in the end would be: 

{{{(3x+1)(3x-1)(3x+5)}}}

So then we solve for x which I am sure you know how to do.