How do you find all values of k so that the polynomial #x^2-5x+k# can be factored with integers?
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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)#
#k in {-5a-a^2 | AAa in ZZ}#
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)#
