Sr Examen
Descomponer -b^4-2*b^2-6 al cuadrado
Solución
Factorización
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$$\left(b + \left(\sqrt[4]{6} \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)} - \sqrt[4]{6} i \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)}\right)\right) \left(b + \left(\sqrt[4]{6} \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)} + \sqrt[4]{6} i \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)}\right)\right) \left(b + \left(- \sqrt[4]{6} \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)} + \sqrt[4]{6} i \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)}\right)\right) \left(b + \left(- \sqrt[4]{6} \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)} - \sqrt[4]{6} i \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{5} \right)}}{2} \right)}\right)\right)$$
(((b + 6^(1/4)*sin(atan(sqrt(5))/2) + i*6^(1/4)*cos(atan(sqrt(5))/2))*(b + 6^(1/4)*sin(atan(sqrt(5))/2) - i*6^(1/4)*cos(atan(sqrt(5))/2)))*(b - 6^(1/4)*sin(atan(sqrt(5))/2) + i*6^(1/4)*cos(atan(sqrt(5))/2)))*(b - 6^(1/4)*sin(atan(sqrt(5))/2) - i*6^(1/4)*cos(atan(sqrt(5))/2))
Expresión del cuadrado perfecto
Expresemos el cuadrado perfecto del trinomio cuadrático
$$\left(- b^{4} - 2 b^{2}\right) - 6$$
Para eso usemos la fórmula
$$a b^{4} + b^{3} + c = a \left(b^{2} + m\right)^{2} + n$$
donde
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
En nuestro caso
$$a = -1$$
$$b = -2$$
$$c = -6$$
Entonces
$$m = 1$$
$$n = -5$$
Pues,
$$- \left(b^{2} + 1\right)^{2} - 5$$
$$\left(- b^{4} - 2 b^{2}\right) - 6$$
Para eso usemos la fórmula
$$a b^{4} + b^{3} + c = a \left(b^{2} + m\right)^{2} + n$$
donde
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
En nuestro caso
$$a = -1$$
$$b = -2$$
$$c = -6$$
Entonces
$$m = 1$$
$$n = -5$$
Pues,
$$- \left(b^{2} + 1\right)^{2} - 5$$
Unión de expresiones racionales
[src]
2 / 2\ -6 + b *\-2 - b /
$$b^{2} \left(- b^{2} - 2\right) - 6$$
-6 + b^2*(-2 - b^2)