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

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