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

# 3 Answers

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 :)

# Comments

Brad, just fix the free coefficient in on the bottom line. It's (-1)

ax^{2} + bx + c factors precisely when there are two integers u and v with u+v = b and uv = ac.

Thus for 3x^{2} + 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.)

## Comments

N.D. -- thanks for the edit, ma'am ...

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