Can someone show me how to even do this problem.

Find all positive values for K for which each of the following can be factored.

x^2 + x - k

News flash: It can be factored with any term (although the roots may not be real roots).

Now if you mean factored with real values for the factors a and b, ...
IT has to add to one, multiply to k.

a+b=1
ab=k
k=0 is a solution, but zero is not positive nor negative.
k=2 a=2, b=-1
k=6 (a=+3, b=-2) is a solution
k=20 a=5, b=-4 is a solution
Now, finding all positive values for k will be a great task, because this goes on ad infinitum. Example...
k=8.4525 a= 3.45 b=-2.45

you got me here i don't understand. So how do i determine what to substitue/.

To determine the positive values for K that allow the expression x^2 + x - k to be factored, you can follow these steps:

1. First, you need to understand the requirements for factoring this quadratic expression. The expression can be factored if there exist two real numbers, let's call them "a" and "b," such that when you add "a" and "b," you get 1 (the coefficient of the x term), and when you multiply "a" and "b," you get -k (the constant term).

2. Write down the two conditions that need to be satisfied.
a + b = 1
a * b = -k

3. Start substituting different values for "k" and solve the equations to find corresponding values for "a" and "b."

For example, let's say k = 2:
Substitute this value of k into the equations:
a + b = 1 (1)
a * b = -2 (2)

You can solve this system of equations in various ways, such as substitution or elimination. Let's use substitution:

From equation (1), solve for a in terms of b:
a = 1 - b

Substitute this value of a in equation (2):
(1 - b) * b = -2

Simplify the equation:
b - b^2 = -2
Rearrange:
b^2 - b - 2 = 0

Factor the quadratic equation:
(b - 2)(b + 1) = 0

Solve for b:
b = 2 or b = -1

Substitute these values back into equation (1) to find the corresponding values for a:
For b = 2: a = 1 - 2 = -1
For b = -1: a = 1 - (-1) = 2

So, when k = 2, the values for a and b are -1 and 2, respectively.

4. Repeat this process for different values of k to obtain the corresponding values for a and b.

Continuing from the previous example:
k = 6:
(b - 3)(b + 2) = 0
b = 3 or b = -2
For b = 3: a = 1 - 3 = -2
For b = -2: a = 1 - (-2) = 3

k = 20:
(b - 5)(b + 4) = 0
b = 5 or b = -4
For b = 5: a = 1 - 5 = -4
For b = -4: a = 1 - (-4) = 5

By substituting different values for k and solving the corresponding equations, you can find different values for a and b.

5. Note that the set of positive values for k is infinite because you can always find different combinations of a and b that satisfy the conditions. So, there is no specific "list" of positive values for k, but you can continue substituting different positive numbers into the equations to identify many more solutions.