How do you factor the expression #2 x^2 + 5 x +12#?

1 Answer
George C.
Aug 13, 2016

#2x^2+5x+12 = 1/8(4x+5-sqrt(71)i)(4x+5+sqrt(71)i)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this below with #a=(4x+5)# and #b=sqrt(71)i#.

#color(white)()#
Multiply by #8#, complete the square, then divide by #8#...

#8(2x^2+5x+12)#

#=16x^2+40x+96#

#=(4x+5)^2-25+96#

#=(4x+5)^2+71#

#=(4x+5)^2-(sqrt(71)i)^2#

#=((4x+5)-sqrt(71)i)((4x+5)+sqrt(71)i)#

#=(4x+5-sqrt(71)i)(4x+5+sqrt(71)i)#

So:

#2x^2+5x+12 = 1/8(4x+5-sqrt(71)i)(4x+5+sqrt(71)i)#

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