How do you factor #4x^2-5x+6#?

1 Answer
Nov 7, 2016

Answer:

#4x^2-5x+6 = (2x-5/4-sqrt(71)/4i)(2x-5/4+sqrt(71)/4i)#

Explanation:

If the sign of the constant term was #-# rather than #+# then this would have a factorisation:

#4x^2-5x-6 = (x-2)(4x+3)#

For the example as given we find:

#4x^2-5x+6#

is in the form #ax^2+bx+c# with #a=4#, #b=-5# and #c=6#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-5)^2-4(4)(6) = 25 - 96 = -71#

Since #Delta < 0# this quadratic has no linear factors with Real coefficients.

We can still factor it, but we need Complex coefficients...

#4x^2-5x+6 = (2x)^2-2(5/4)(2x)+(5/4)^2-(5/4)^2+6#

#color(white)(4x^2-5x+6) = (2x-5/4)^2-25/16+96/16#

#color(white)(4x^2-5x+6) = (2x-5/4)^2+71/16#

#color(white)(4x^2-5x+6) = (2x-5/4)^2-(sqrt(71)/4i)^2#

#color(white)(4x^2-5x+6) = ((2x-5/4)-sqrt(71)/4i)((2x-5/4)+sqrt(71)/4i)#

#color(white)(4x^2-5x+6) = (2x-5/4-sqrt(71)/4i)(2x-5/4+sqrt(71)/4i)#