Is it possible to factor #y=x^2-4x-3 #? If so, what are the factors?

1 Answer
Feb 16, 2017

Answer:

Factors are #y=(x-2+sqrt7)(x-2-sqrt7)#

Explanation:

When you generally say - Ïs it possible to factor #y=ax^2+bx+c#, what is normally meant is to have rational factors of type #y=a(x+p)(x+q)#

This is possible if the discriminant #b^2-4ac# is square of a rational number.

Here in #y=x^2-4x-3#, we have #a=1#, #b=-4# and #c=-3# and hence #b^2-4ac=(-4)^2-4xx1xx(-3)=16+12=28#

as it is not the square of a rational number you cannot have rational factors. But it is still possible to have irrational factors.

As #y=x^2-4x-3=(x^2-2xx2xx x+2^2)-2^2-3#

= #(x-2)^2-7=(x-2)^2-(sqrt7)^2#

and as #a^2-b^2=(a+b)(a-b)#

factors are #y=(x-2+sqrt7)(x-2-sqrt7)#.