The terms of #2x^2+10x+84#
have the obvious constant common factor #2#
and therefore can b factored as #2(x^2+5x+42)#
The determinant of #(x^2+5x+42)# is negative so there are no Real factors for this expression.
#color(white)("XXXX")#Side note
#color(white)("XXXX")#for an expression of the form #ax^2+bx+c#
#color(white)("XXXX")#the determinant is #Delta=b^2-4ac#
#color(white)("XXXX")#and it is a general rule that #Delta < 0#
#color(white)("XXXX")# implies no possible (Real) factors
However if Complex values are allowed then a factoring can be achieved using the quadratic formula.
#color(white)("XXXX")#Side note 2
#color(white)("XXXX")#The quadratic formula says that
#color(white)("XXXX")#an expression of the form #ax^2+bx+c#
#color(white)("XXXX")#can be factored with a term #(x-phi)#
#color(white)("XXXX")#where #phi = (-b+-sqrt(b^2-4ac))/(2a)#
Plugging in the obvious values for #a, b, c# from #1x^2+5x+42# gives the second answer as shown above.
