I am wondering if there are any easy ways or variety of ways to show how to factorize quadratics, especially if the variable is given in its inverse form. Or if the highest power coefficient is not 1.
for example, lets consider this: $1 - 0.4x^{-1} - 0.6x^{-2}$
we can get, $(1 - x^{-1})(1 + 0.6x^{-1})$
my question is really about teaching this. what methods might you suggest and how might you break this down for learners?
On
Factorize $1 - 0.4x^{-1} - 0.6x^{-2}$
Personally, if a lucky guess doesn't present itself in less than a minute, I always resort to brute force, which is reasonably quick.
First, I will re-express the function as
$$\frac{1}{x^2} \times f(x) ~: ~f(x) = x^2 - 0.4x - 0.6.$$
Then, I will apply brute force against $~f(x).~$ Then, I will re-apply the factor $~\dfrac{1}{x}~$ to each of the resulting factors.
$$f(x) = 0 \iff x = \frac{1}{2} \left[ ~0.4 \pm \sqrt{0.16 + (4 \times 0.6)} ~\right] \iff $$
$$x = \frac{0.4 \pm \sqrt{2.56}}{2} = 0.2 \pm 0.8 \iff $$
$$x = \frac{2 \pm 8}{10}.$$
Therefore,
$$f(x) ~\text{factors to}~ \left( ~x - \frac{2 + 8}{10} ~\right) \times \left( ~x - \frac{2 - 8}{10} ~\right) $$
$$= ~(x-1) \times \left(x + \frac{3}{5}\right) \implies $$
$$\frac{1}{x^2} ~f(x) ~\text{factors to}~ \left( ~1 - \frac{1}{x} ~\right) \times \left( ~1 + \frac{3}{5x} ~\right). $$
For me, the easiest method is to assume x^-1=t ; (x^-1)^2=t^2 (Squaring both sides)
and then 1-0.4t-0.6t^2 and then solving it. OR else the other way can be using the sum of roots and multiplication of roots
like:
alpha+beta=-b/a and alpha*beta= c/a
or graphical method