How do you factor #4x^2-5#?

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Answer 1 · 2017-06-13T10:29:08

It is not factorable.

In order for an expression to be factor able, there must be at least a common factor. Here, in this problem, there isn't any.

#4# and #5# don't have any common factors and not both terms have an #x# term. Therefore, this expression is not factorable.

Answer 2 · 2017-06-13T10:48:38

#4x^2-5=color(red)((2x+sqrt(5))(2x-sqrt(5)))#

Remember that:
#color(white)("XXX")color(cyan)a^2-color(magenta)b^2=(color(cyan)a+color(magenta)b)(color(cyan)a-color(magenta)b)#

#4x^2-5# can be written as #color(cyan)((2x)^2)-color(magenta)(sqrt(5)^2)#