What are values of #k# so that the trinomial #x^2+kx-35# can be factored using integers?

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Answer 1 · 2018-03-27T16:46:32

#k = +-2# or #k = +-34#

Given:

#x^2+kx-35#

Any integer factorisation must take the form:

#(x+m)(x+n) = x^2+(m+n)x+mn#

So:

#mn = -35#

#m + n = k#

The possible values of #m, n# are the factors of #-35#, namely:

#+-1, +-5, +-7, +-35#

So possible values of #k# are:

#1-35 = -34#

#5-7 = -2#

#7-5 = 2#

#35-1 = 34#