Квадрат үшмүшені көбейткіштерге жіктеу

Теңдеуді шешеміз: $6 x^2-x-2=0$ $${x_1} = \frac{2}{3},\quad {x_2} = – \frac{1}{2}$$ $$ = 3\left( {x – \frac{2}{3}} \right) \cdot 2\left( {x + \frac{1}{2}} \right) = (3x – 2)(2x + 1)$$

$${x^3} – x – 2x – 2 = \left( {{x^3} – x} \right) – (2x + 2) = x\left( {{x^2} – 1} \right) – 2(x + 1) = $$ $$ = x(x – 1)(x + 1) – 2(x + 1) = (x + 1)\left( {{x^2} – x – 2} \right) = $$ $$ = \left\| \begin{array}{l}{x^2} – x – 2 = 0\\{x_1} = 2,\quad {x_2} = – 1\end{array} \right\| = (x + 1)(x + 1)(x – 2) = {(x + 1)^2}(x – 2)$$

$$4 – 7x – 2{x^2} = – \left( {2{x^2} + 7x – 4} \right) = \left\| {\begin{array}{*{20}{l}}{2{x^2} + 7x – 4 = 0}\\{{x_1} = – 4;{x_2} = \frac{1}{2}}\end{array}} \right\| = $$ $$ = – 2(x + 4)\left( {x – \frac{1}{2}} \right) = (x + 4)(1 – 2x)$$

$9 x^2-30 x y+24 y^2=0$ теңдеуін $x$-ке қатысты шешеміз: $$a = 9,\quad b = – 30y,\quad c = 24{y^2}$$ $$D = {b^2} – 4ac = {( – 30y)^2} – 4 \cdot 9 \cdot 24{y^2} = 900{y^2} – 864{y^2} = 36{y^2}$$ $${x_1} = \frac{{30y + 6y}}{{18}} = 2y,\quad {x_2} = \frac{{30y – 6y}}{{18}} = \frac{4}{3}y$$ $$9{x^2} – 30xy + 24{y^2} = 9(x – 2y)\left( {x – \frac{4}{3}y} \right) = $$ $$ = (x – 2y) \cdot 9\left( {x – \frac{4}{3}y} \right) = (x – 2y)(9x – 12y)$$

$${a^4} + {a^2} – 2 = \left| {\begin{array}{*{20}{l}}{{a^2} = x}\\{{x^2} + x – 2 = 0}\\{{x_1} = – 2,\quad {x_2} = 1}\end{array}} \right| = (x + 2)(x – 1) = $$ $$ = \left( {{a^2} + 2} \right)\left( {{a^2} – 1} \right) = \left( {{a^2} + 2} \right)(a – 1)(a + 1)$$