How do you find all values of k so that the polynomial #x^2+kx-19# can be factored with integers?

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Answer 1 · 2016-12-23T03:11:14

#k=+-20#

When we factor a trinomial of form

#mx^2+nx+p#

we end up with a general form:

#(ax+b)(cx+d)#

where:

#ac=m#
#bd=p#
#ad+bc=n#

So let's now move to the statement in question:

#x^2+kx-19#

And so we have:

#m=1, n=k, p=19#

We're asked to show all values of #k# where the factors, #a, b, c, d# are integers.

From this, we know that:

since #m=1, a=c=+-1 #
since #p=19, bd=19#, so we can have either #b=+-1, d=+-19#, with the signs moving in concert (so if b is positive, d is positive). And so #b+d=+-20#

and so this means that for:

#ad+bc=n#

We can have:

#1(1)+1(19)=20, (x+1)(x+19)#
#1(-1)+1(-19)=-20, (x-1)(x-19)#
#(-1)(1)+(-1)(19)=-20, (-x+1)(-x+19)#
#(-1)(-1)+(-1)(-19)=20, (-x-1)(-x-19)#

And so #k=+-20#