How do you find all values of k so that the polynomial x^2+kx-19x2+kx19 can be factored with integers?

1 Answer

k=+-20k=±20

Explanation:

When we factor a trinomial of form

mx^2+nx+pmx2+nx+p

we end up with a general form:

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

where:

ac=mac=m
bd=pbd=p
ad+bc=nad+bc=n

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

x^2+kx-19x2+kx19

And so we have:

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

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

From this, we know that:

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

and so this means that for:

ad+bc=nad+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,(x1)(x19)
(-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,(x1)(x19)

And so k=+-20k=±20