i don't get what i'm supposed to do
Find all integers k so that the trinomial can be factored. 3x^2+kx-1
Good hearing from you, Mathalina!
the left-most term is made up of 3x*x ...
the right-most term is made up of -1*1 ...
one mix is (3x-1)(x+1)= 3x^2 + 3x - x -1 = 3x^2 + 2x - 1 ==> k=2 ...
the other's (3x+1)(x-1)= 3x^2 - 3x + x -1 = 3x^2 - 2x - 1 ==> k= -2 ... Regards :)
ax2 + bx + c factors precisely when there are two integers u and v with u+v = b and uv = ac.
Thus for 3x2 + kx - 1 to factor, we want u+v = k and uv = -3.
Note that -3 has only two factorizations, either -1*3 or -3*1.
If u = -1 and v = 3 then k = u + v = -1 + 3 = 2.
If u = -3 and k = 1, then k = -3 + 1 = -2.
So the only possible values of k are k = ±2.
I'm not really sure what you're *supposed* to do, but here's how I would go about this:
3 (the x^2 coefficient) can only be factored as 3 and 1 or (-3) and (-1). (-1) (the constant) factors as (-1) and 1. Those are the integral possibilities.
So the possible binomials are (3x + 1) (x - 1), (3x - 1) (x + 1), (-3x + 1) (-x - 1), and (-3x - 1) (x + 1).
k for these four cases would turn out to be (-2), 2, 2, and (-2).
So the possible values for k are (-2) and 2.
(Note: It really shouldn't be necessary to try both negative and positive coefficients of x in the binomials.)