Factorización compleja .

Question

¿Cuál sería la factorización del siguiente término?

$$15x^{2}-24y^{2}-6z^{2}+2xy+24yz-zx$$

Traté de convertirlo en un cuadrado perfecto como $$(a+b+c){2}$$ pero sin ningún propósito. ¿Qué debo hacer?

Answer 1

Pista: escríbalo como una cuadrática en $x$ (por ejemplo) y resolverlo para las raíces que luego permiten la factorización:

$$ 15x^{2}+(2y-z)x-24y^{2}-6z^{2}+24yz $$

Obsérvese que la constante (en $x$ ) es:

$$ -24y^{2}-6z^{2}+24yz = -6 (2 y - z)^2 $$

Entonces el discriminante resulta ser convenientemente un cuadrado:

$$ \Delta_x = (2y-z)^2 + 4\cdot15\cdot 6 (2y-z)^2 = 361 (2 y - z)^2 $$

Answer 2

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 15 } & 1 & 0 \\ - \frac{ 1 }{ 30 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 30 & 0 & 0 \\ 0 & - \frac{ 722 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 15 } & - \frac{ 1 }{ 30 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 30 & 2 & - 1 \\ 2 & - 48 & 24 \\ - 1 & 24 & - 12 \\ \end{array} \right) $$

$$ 15 \left(x + \frac{y}{15} - \frac{z}{30} \right)^2 - \frac{361}{15} \left( y - \frac{z}{2} \right)^2 = 15 x^2 - 24 y^2 - 6 z^2 + 24 yz -zx + 2xy \; \; , $$ $$ 15^2 \left(x + \frac{y}{15} - \frac{z}{30} \right)^2 - 19^2 \left( y - \frac{z}{2} \right)^2 = 15 \left( 15 x^2 - 24 y^2 - 6 z^2 + 24 yz -zx + 2xy \right) \; \; , $$ $$ \left(15x + y - \frac{z}{2} \right)^2 - \left(19 y - \frac{19z}{2} \right)^2 = 15 \left( 15 x^2 - 24 y^2 - 6 z^2 + 24 yz -zx + 2xy \right) \; \; , $$ $$ \left( 15x + 20 y - 10 z \right) \left( 15x -18 y +9 z \right) = 15 \left( 15 x^2 - 24 y^2 - 6 z^2 + 24 yz -zx + 2xy \right) \; \; , $$ $$ \left( 3x + 4 y - 2 z \right) \left( 5x -6 y +3 z \right) = 15 x^2 - 24 y^2 - 6 z^2 + 24 yz -zx + 2xy \; \; . $$

Me gusta. Si obtenemos la matriz diagonal $D$ con entradas diagonales $A,-B,0,$ pero $AB$ no es un cuadrado racional, entonces obtenemos una factorización donde algunos coeficientes implican raíces cuadradas.

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 30 & 2 & - 1 \\ 2 & - 48 & 24 \\ - 1 & 24 & - 12 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 15 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 15 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 15 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 30 & 0 & - 1 \\ 0 & - \frac{ 722 }{ 15 } & \frac{ 361 }{ 15 } \\ - 1 & \frac{ 361 }{ 15 } & - 12 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & \frac{ 1 }{ 30 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 15 } & \frac{ 1 }{ 30 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 15 } & - \frac{ 1 }{ 30 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 30 & 0 & 0 \\ 0 & - \frac{ 722 }{ 15 } & \frac{ 361 }{ 15 } \\ 0 & \frac{ 361 }{ 15 } & - \frac{ 361 }{ 30 } \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 15 } & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 15 } & - \frac{ 1 }{ 30 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 30 & 0 & 0 \\ 0 & - \frac{ 722 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 15 } & 1 & 0 \\ 0 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 30 & 2 & - 1 \\ 2 & - 48 & 24 \\ - 1 & 24 & - 12 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 15 } & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 30 & 0 & 0 \\ 0 & - \frac{ 722 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 15 } & 1 & 0 \\ - \frac{ 1 }{ 30 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 30 & 0 & 0 \\ 0 & - \frac{ 722 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 15 } & - \frac{ 1 }{ 30 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 30 & 2 & - 1 \\ 2 & - 48 & 24 \\ - 1 & 24 & - 12 \\ \end{array} \right) $$

Answer 3

También podemos hacer algo si la forma binaria es positiva, sólo requerirá números complejos. En este caso, supongo que el extra $15$ se absorbería dividiendo cada uno de los dos factores por $\sqrt{15}$

$$ f(x,y,z) = 15 x^2 + 20 y^2 + 31 z^2 + 48 yz + 34 zx + 32 xy $$

$$ f = 15 \left( x + \frac{16y}{15} + \frac{17 z}{15} \right)^2 + \frac{44}{15} (y + 2 z)^2 \; \; , $$ $$ 15 f = (15 x + 16 y + 17 z)^2 + 44 (y+2z)^2 $$ $$ 15 f = \left(15 x + (16 + 2i \sqrt {11}) y + (17 + 4i \sqrt {11}) z \right) \; \cdot \; \left(15 x + (16 - 2i \sqrt {11}) y + (17 - 4i \sqrt {11}) z \right) $$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 16 }{ 15 } & 1 & 0 \\ \frac{ 17 }{ 15 } & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 15 & 0 & 0 \\ 0 & \frac{ 44 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 16 }{ 15 } & \frac{ 17 }{ 15 } \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 15 & 16 & 17 \\ 16 & 20 & 24 \\ 17 & 24 & 31 \\ \end{array} \right) $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 15 & 16 & 17 \\ 16 & 20 & 24 \\ 17 & 24 & 31 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 16 }{ 15 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 16 }{ 15 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 16 }{ 15 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 15 & 0 & 17 \\ 0 & \frac{ 44 }{ 15 } & \frac{ 88 }{ 15 } \\ 17 & \frac{ 88 }{ 15 } & 31 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 17 }{ 15 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 16 }{ 15 } & - \frac{ 17 }{ 15 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 16 }{ 15 } & \frac{ 17 }{ 15 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 15 & 0 & 0 \\ 0 & \frac{ 44 }{ 15 } & \frac{ 88 }{ 15 } \\ 0 & \frac{ 88 }{ 15 } & \frac{ 176 }{ 15 } \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 16 }{ 15 } & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 16 }{ 15 } & \frac{ 17 }{ 15 } \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 15 & 0 & 0 \\ 0 & \frac{ 44 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 16 }{ 15 } & 1 & 0 \\ 1 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 15 & 16 & 17 \\ 16 & 20 & 24 \\ 17 & 24 & 31 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 16 }{ 15 } & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 15 & 0 & 0 \\ 0 & \frac{ 44 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 16 }{ 15 } & 1 & 0 \\ \frac{ 17 }{ 15 } & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 15 & 0 & 0 \\ 0 & \frac{ 44 }{ 15 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 16 }{ 15 } & \frac{ 17 }{ 15 } \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 15 & 16 & 17 \\ 16 & 20 & 24 \\ 17 & 24 & 31 \\ \end{array} \right) $$

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