How do you factor the trinomial #2y^2+6y+24#?

1 Answer
George C.
Apr 6, 2016

#2y^2+6y+24=2(y+3/2-sqrt(39)/2i)(y+3/2+sqrt(39)/2i)#

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 #a=(2y+3)# and #b=sqrt(39)i#.

Before that, complete the square then use the difference of squares identity, but first I will multiply by #2# to cut down on the arithmetic involving fractions. It will be necessary to divide by #2# at the end...

#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 #2# we find:

#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)#

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