Factor

x³ + 5x + 3x² + 15

Factor the first two terms only by
taking out x

x(x² + 5) + 3x² + 15

Factor the last two terms only by
taking out 3

x(x² + 5) + 3(x² + 5)

Now you see that the term
x(x² + 5) and the term
3(x² + 5) have the common
factor (x² + 5)

x(x² + 5) + 3(x² + 5)

So factor out the red factor (x² + 5)
on the left and leave the black factors 
in parentheses on the right:

(x² + 5)(x + 3)

Edwin