How do you find all values of k so that the polynomial #x^2+kx+14# can be factored with integers?

2 Answers
Dec 5, 2016

#k in {-15,-9,9,15}#

Explanation:

In order to factorise this quadratic with integers we would require factors of #14# that add up to #k#. We can just list these:

# {: ("factor1", "factor2", "sum", "factored quadratic") , (14,1,15,(x+14)(x+1)), (7,2,9,(x+7)(x+2)), (-14,-1,-15,(x-14)(x-1)), (-7,-2,-9,(x-7)(x-2)) :} #

So the values are #k in {-15,-9,9,15}#

Dec 6, 2016

#k in {-15,-9,9,15}#

Explanation:

Solving #x^2+k x+14=0# we have

#x=1/2(-kpmsqrt(k^2-56))#. We need

#k^2-50= m^2# where #m, k# are integers

so

#k^2-m^2=56#

Here #56 = 1 cdot 2^3 cdot 7#

so the possible factors are

#f = {1,2,4,7,8,14,28,56}#.

The solutions for #m,k# are the integer solutions for

#{(k+m=f_i),(k-m=56/f_i),(-k+m=f_i),(-k-m=56/f_i):}#

The positive solutions kor #k# are

#((k,m),(9,-5),(15,-13),(9,5),(15,13))#

Finally, the feasible values for #k# are #{-15,-9,9,15}#

Note that

#x = 1/2(-kpmm)# and #-kpm m# is an even number so is divisible by #2#.

The factorizations are

#(x+2)(x+7)# and
#(x+1)(x+14)#
#(x-2)(x-7)# and
#(x-1)(x-14)#