Question

How do you find all values of k so that the polynomial `x^2-5x+k` can be factored with integers?

Answer 1

`k={0,4,6}`

Explanation:

Considering `k ge 0` and solving for `x` we have

`x=(5pmsqrt(25-4k))/2`

Here for `k=0` we have as roots `x=0,x=-5`

If `25-4k = m^2` then this condition requires

`5^2-m^2=4k` or

`(5-m)(5+m)=4k` so we have the possibilities

`{(5-m=f_i),(5+m=(4k)/f_i):}` with `f_i={1,2,4}`

so we have for

`f_2=2->m=3,k=4`
`f_3=4->m=1,k=6`

analogously with

`{(5+m=f_i),(5-m=(4k)/f_i):}` with `f_i={1,2,4}`

we obtain

`m=-3,k=4` and
`m=-1,k=6`

so finally `k={0,4,6}` and

for
`k=0->x(x-5)`
`k=4->(x-1)(x-4)`
`k=6->(x-2)(x-3)`

Answer 2

`k in {-5a-a^2 | AAa in ZZ}`

Explanation:

Let `a` and `b` be integers so that we have the factoring:
`color(white)("XXX")x^2color(red)(-5)x+color(magenta)k=color(blue)((x+a)(x+b))`

Since
`color(white)("XXX")(x+a)(x+b) = x^2+color(red)(""(a+b))x+color(magenta)(ab)`
we are looking for pairs of integers `a, b` such that
[1]`color(white)("XXX")a+b=-5` and
[2]`color(white)("XXX")ab=k`

From [1] we have
`color(white)("XXX")b=-5-a`
Substituting this back into [2], we get
`color(white)("XXX")k=a(-5-a) = -5a-a^2=-(a^2+5a)`

That is for any integer `a`, setting `k=-(a^2+5a)`
`color(white)("XXX")x^2-5x+k =x^2-5x+(-a^2-5a)`
can be factored as `(x+a)(x+(-a-5))`
with integer values for `a` and `(-a-5)`