Question

For what values of `b` is `x^2+bx-2` factorable ?

Answer 1

For integer coefficients `b=+-1`.

For rational coefficients `b=(p^2-2)/p` for any non-zero rational number `p`.

For real coefficients any real value of `b` will work.

Explanation:

Given:

`x^2+bx-2`

Note that this is in the form `ax^2+bx+c` with `a=1` and `c=-2`

This quadratic has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = b^2-4(color(blue)(1))(color(blue)(-2)) = b^2+8`

`color(white)()`
Integer coefficients

The only integer values of `b` for which `Delta = b^2+8` is a perfect square are `b=+-1`, resulting in the factorable quadratics:

`x^2+x-2 = (x+2)(x-1)`

`x^2-x-2 = (x-2)(x+1)`

`color(white)()`
Rational coefficients

If we permit rational coefficients and rational values of `b` then there are some more possibilities.

For example, if `p` is any non-zero rational number then:

`(x+p)(x-2/p) = x^2+(p-2/p)x-2`

So `b=p-2/p=(p^2-2)/p` is a possibility for any non-zero rational number `p`.

`color(white)()`
Real coefficients

If we permit irrational coefficients, then we just require `Delta >= 0`, which is already true for any real value of `b`.

Then:

`x^2+bx-2 = (x+b/2)^2-(b^2/4+2)`

`color(white)(x^2+bx-2) = (x+b/2)^2-(sqrt(b^2/4+2))^2`

`color(white)(x^2+bx-2) = ((x+b/2)-sqrt(b^2/4+2))((x+b/2)+sqrt(b^2/4+2))`

`color(white)(x^2+bx-2) = (x+b/2-sqrt(b^2/4+2))(x+b/2+sqrt(b^2/4+2))`