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

$k = \pm 20$

#### Explanation:

When we factor a trinomial of form

$m {x}^{2} + n x + p$

we end up with a general form:

$\left(a x + b\right) \left(c x + d\right)$

where:

$a c = m$
$b d = p$
$a d + b c = n$

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

${x}^{2} + k x - 19$

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

and so this means that for:

$a d + b c = n$

We can have:

$1 \left(1\right) + 1 \left(19\right) = 20 , \left(x + 1\right) \left(x + 19\right)$
$1 \left(- 1\right) + 1 \left(- 19\right) = - 20 , \left(x - 1\right) \left(x - 19\right)$
$\left(- 1\right) \left(1\right) + \left(- 1\right) \left(19\right) = - 20 , \left(- x + 1\right) \left(- x + 19\right)$
$\left(- 1\right) \left(- 1\right) + \left(- 1\right) \left(- 19\right) = 20 , \left(- x - 1\right) \left(- x - 19\right)$

And so $k = \pm 20$