Question

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

Answer 1

`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)`