# For what values of b is x^2+bx-2 factorable ?

Apr 26, 2017

For integer coefficients $b = \pm 1$.

For rational coefficients $b = \frac{{p}^{2} - 2}{p}$ for any non-zero rational number $p$.

For real coefficients any real value of $b$ will work.

#### Explanation:

Given:

${x}^{2} + b x - 2$

Note that this is in the form $a {x}^{2} + b x + c$ with $a = 1$ and $c = - 2$

This quadratic has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {b}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{- 2}\right) = {b}^{2} + 8$

$\textcolor{w h i t e}{}$
Integer coefficients

The only integer values of $b$ for which $\Delta = {b}^{2} + 8$ is a perfect square are $b = \pm 1$, resulting in the factorable quadratics:

${x}^{2} + x - 2 = \left(x + 2\right) \left(x - 1\right)$

${x}^{2} - x - 2 = \left(x - 2\right) \left(x + 1\right)$

$\textcolor{w h i t e}{}$
Rational coefficients

If we permit rational coefficients and rational values of $b$ then there are some more possibilities.

For example, if $p$ is any non-zero rational number then:

$\left(x + p\right) \left(x - \frac{2}{p}\right) = {x}^{2} + \left(p - \frac{2}{p}\right) x - 2$

So $b = p - \frac{2}{p} = \frac{{p}^{2} - 2}{p}$ is a possibility for any non-zero rational number $p$.

$\textcolor{w h i t e}{}$
Real coefficients

If we permit irrational coefficients, then we just require $\Delta \ge 0$, which is already true for any real value of $b$.

Then:

${x}^{2} + b x - 2 = {\left(x + \frac{b}{2}\right)}^{2} - \left({b}^{2} / 4 + 2\right)$

$\textcolor{w h i t e}{{x}^{2} + b x - 2} = {\left(x + \frac{b}{2}\right)}^{2} - {\left(\sqrt{{b}^{2} / 4 + 2}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + b x - 2} = \left(\left(x + \frac{b}{2}\right) - \sqrt{{b}^{2} / 4 + 2}\right) \left(\left(x + \frac{b}{2}\right) + \sqrt{{b}^{2} / 4 + 2}\right)$

$\textcolor{w h i t e}{{x}^{2} + b x - 2} = \left(x + \frac{b}{2} - \sqrt{{b}^{2} / 4 + 2}\right) \left(x + \frac{b}{2} + \sqrt{{b}^{2} / 4 + 2}\right)$