This binomial is part of a bigger problem that I need to solve, however, I am little stuck on how to factor it. $(t-1)(t-1)(t+1)$ does not work.
$\endgroup$3 Answers
$\begingroup$$$t^4-2=\left(t^2-\sqrt 2\right)\left(t^2+\sqrt 2\right)=\left(t-\sqrt[4]{2}\right)\left(t+\sqrt[4]{2}\right)\left(t^2+\sqrt 2\right).$$
$\endgroup$ $\begingroup$To find the roots of $t^4 -2$, let $$t^4 = 2$$
$$t^2 = \pm \sqrt{2}$$
Assuming $t$ is real, this gives
$$t = \pm \sqrt[4]{2}$$
This means you can write $$t^4 -2 = \left(t-\sqrt[4]{2}\right)\left(t+\sqrt[4]{2}\right) \left(t^2 + \sqrt{2}\right)$$
The two real roots correspond to the first two factors above. The other factor gives two complex roots that aren't real.
$\endgroup$ 3 $\begingroup$$$t^4-2=(t^2)^2-(\sqrt2)^2=(t^2-\sqrt2)(t^2+\sqrt2)=(t-\sqrt[4]{2})(t+\sqrt[4]{2})(t^2+\sqrt2)$$
$\endgroup$
