How do you factor #3x^2 + 5#?

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Answer 1 · 2016-06-22T21:33:42

#3x^2+5=(sqrt(3)x-sqrt(5)i)(sqrt(3)x+sqrt(5)i)#

Note that if #x# is a Real number then #x^2 >= 0#.

Hence #3x^2+5 > 0#.

So this quadratic has no Real zeros and no linear factors with Real coefficients.

It can be factored with Complex numbers.

Use the difference of squares identity:

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

with #a=sqrt(3)x# and #b=sqrt(5)i# as follows:

#3x^2+5=(sqrt(3)x)^2-(sqrt(5)i)^2#

#=(sqrt(3)x-sqrt(5)i)(sqrt(3)x+sqrt(5)i)#