Question 20121
Try this: For {{{x^2 + kx + 12}}}

What are the integral factors of +12?

{{{+12 = (1)(12)}}}
{{{+12 = (2)(6)}}}
{{{+12 = (3)(4)}}}
{{{+12 = (-1)(-12)}}}
{{{+12 = (-2)(-6)}}}
{{{+12 = (-3)(-4)}}}

Now find the sum of each pair of factors:

1 + 12 = 13
2 + 6 = 8
3 + 4 = 7
(-1)+(-12) = -13
(-2)+(-6) = -8
(-3)+(-4) = -7

k can be: 13, 8, 7, -13, -8, or -7

Check:
{{{(x+1)(x+12) = x^2+13x+12}}}
{{{(x+2)(x+6) = x^2+8x+12}}}
{{{(x+3)(x+4) = x^2+7x+12}}}
{{{(x-1)(x-12) = x^2-13x+12}}}
{{{(x-2)(x-6) = x^2-8x+12}}}
{{{(x-3)(x-4) = x^2-7x+12}}}

You can use the same procedure for {{{r^2+kr+20}}}