How do you factor the trinomial #2y^2+6y+24#?
1 Answer
Explanation:
This quadratic has negative discriminant, so we can tell that it has no factors wih Real coefficients, but we can factor it with Complex coefficients.
Note that the difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
I will use this with
Before that, complete the square then use the difference of squares identity, but first I will multiply by
#2(2y^2+6y+24)#
#=4y^2+12y+48#
#=(2y+3)^2-9+48#
#=(2y+3)^2+39#
#=(2y+3)^2+(sqrt(39))^2#
#=(2y+3)^2-(sqrt(39)i)^2#
#=((2y+3)-sqrt(39)i)((2y+3)+sqrt(39)i)#
#=(2y+3-sqrt(39)i)(2y+3+sqrt(39)i)#
So, dividing by
#2y^2+6y+24#
#=1/2 (2y+3-sqrt(39)i)(2y+3+sqrt(39)i)#
#=2(y+3/2-sqrt(39)/2i)(y+3/2+sqrt(39)/2i)#