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

##### 1 Answer

For integer coefficients

For rational coefficients

For real coefficients any real value of

#### Explanation:

Given:

#x^2+bx-2#

Note that this is in the form

This quadratic has discriminant

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

**Integer coefficients**

The only integer values of

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

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

**Rational coefficients**

If we permit rational coefficients and rational values of

For example, if

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

So

**Real coefficients**

If we permit irrational coefficients, then we just require

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))#