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))#