How do you factor the expression #x^2+2x+24#?

Question

2387 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-30T20:34:29

#x^2+2x+24=(x+1-sqrt(23)i)(x+1+sqrt(23)i)#

#x^2+2x+24# is in the form #ax^2+bx+c# with #a=1#, #b=2# and #c=24#. This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 2^2-(4*1*24) = 4 - 96 = -92#

Since this is negative, our quadratic has no factors with Real coefficients.

It does have zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#= (-b+-sqrt(Delta))/(2a)#

#= (-2+-sqrt(-92))/2#

#=-1+-sqrt(23)i#

and hence factors:

#x^2+2x+24 = (x+1-sqrt(23)i)(x+1+sqrt(23)i)#

#color(white)()#
Alternative Method

Use the difference of squares identity:

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

with #a=x+1# and #b=sqrt(23)i# as follows:

#x^2+2x+24#

#=x^2+2x+1+23#

#=(x+1)^2-(sqrt(23)i)^2#

#=((x+1)-sqrt(23)i)((x+1)+sqrt(23)i)#

#=(x+1-sqrt(23)i)(x+1+sqrt(23)i)#

#color(white)()#
Footnote

If the sign on the last term was #-# rather than #+#, then the factorisation would be a whole lot simpler:

#x^2+2x-24 = (x+6)(x-4)#

To find this, you might look for a pair of factors of #24# that differ by #2# and find that #6, 4# works.