Expresión del cuadrado perfecto
Expresemos el cuadrado perfecto del trinomio cuadrático
$$\left(y^{4} + y^{2}\right) - 10$$
Para eso usemos la fórmula
$$a y^{4} + b y^{2} + c = a \left(m + y^{2}\right)^{2} + n$$
donde
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
En nuestro caso
$$a = 1$$
$$b = 1$$
$$c = -10$$
Entonces
$$m = \frac{1}{2}$$
$$n = - \frac{41}{4}$$
Pues,
$$\left(y^{2} + \frac{1}{2}\right)^{2} - \frac{41}{4}$$
/ ____________\ / ____________\ / ______________\ / ______________\
| / ____ | | / ____ | | / ____ | | / ____ |
| / 1 \/ 41 | | / 1 \/ 41 | | / 1 \/ 41 | | / 1 \/ 41 |
|x + I* / - + ------ |*|x - I* / - + ------ |*|x + / - - + ------ |*|x - / - - + ------ |
\ \/ 2 2 / \ \/ 2 2 / \ \/ 2 2 / \ \/ 2 2 /
$$\left(x - i \sqrt{\frac{1}{2} + \frac{\sqrt{41}}{2}}\right) \left(x + i \sqrt{\frac{1}{2} + \frac{\sqrt{41}}{2}}\right) \left(x + \sqrt{- \frac{1}{2} + \frac{\sqrt{41}}{2}}\right) \left(x - \sqrt{- \frac{1}{2} + \frac{\sqrt{41}}{2}}\right)$$
(((x + i*sqrt(1/2 + sqrt(41)/2))*(x - i*sqrt(1/2 + sqrt(41)/2)))*(x + sqrt(-1/2 + sqrt(41)/2)))*(x - sqrt(-1/2 + sqrt(41)/2))