Rešiti sledeću jednačinu:

$2{x}^{5}+{x}^{4}-19{x}^{3}+19{x}^{2}-x-2=0$

$2{x}^{5}+{x}^{4}-19{x}^{3}+19{x}^{2}-x-2=0$

${2}{x}^{5}+{1}{x}^{4}-{19}{x}^{3}+{19}{x}^{2}{-}{1}x{-}{2}=0$
$\begin{array}{ccccccc}& {2}& {1}& -19& {19}& -1& -2\\ --& --& --& --& --& --& --\\ {1}& {2}& {3}& -16& {3}& {2}& 0\end{array}$

${{x}}_{1}{=}{1}$

$\left(x{-}{1}\right)\left({2}{x}^{4}+{3}{x}^{3}{-}{16}{x}^{2}+{3}x+{2}\right)=0$
$2{x}^{4}+3{x}^{3}-16{x}^{2}+3x+2=0$

Podelite čitavu jednačinu sa ${x}^{2}$.

$\frac{2{x}^{{4}}}{{{x}}^{2}}+\frac{3{x}^{{3}}}{{{x}}^{2}}-\frac{16{{x}}^{2}}{{{x}}^{2}}+\frac{3{x}}{{{x}}^{2}}+\frac{2}{{x}^{2}}=0$
${2}{{x}}^{2}+{3}{x}-16+\frac{3}{x}+\frac{2}{{x}^{2}}=0$
${2}{{x}}^{2}+\frac{2}{{x}^{2}}+{3}{x}+\frac{3}{x}-16=0$
${2}\left({x}^{2}{+}\frac{1}{{x}^{2}}\right)+{3}\left(x{+}\frac{1}{x}\right)-16=0$
$t={x}{+}\frac{1}{x}$
${t}^{2}={\left(x+\frac{1}{x}\right)}^{2}={\left(\frac{{x}^{2}+1}{x}\right)}^{2}=\frac{{x}^{4}+2{x}^{2}+1}{{x}^{2}}=\frac{{x}^{4}}{{x}^{2}}+\frac{2{x}^{2}}{{x}^{2}}+\frac{1}{{x}^{2}}={x}^{2}+2+\frac{1}{{x}^{2}}$
${t}^{2}-2={{x}}^{2}{-}\frac{1}{{x}^{2}}$
$2\left({{t}}^{2}{-}{2}\right)+3{t}-16=0$
$2{t}^{2}-4+3t-16=0$
$2{t}^{2}+3t-20=0$
${t}_{1,2}=\frac{-3±\sqrt{{3}^{3}-4·2·\left(-20\right)}}{2·2}=\frac{-3±\sqrt{9+160}}{4}=\frac{-3±13}{4}$
$\frac{5}{2}·2x=x·2x+\frac{1}{x}·2x$
$5x=2{x}^{2}+2$
$2{x}^{2}-5x+2=0$
${x}_{2,3}=\frac{5±\sqrt{{5}^{2}-4·2·2}}{2·2}=\frac{5±\sqrt{25-16}}{4}=\frac{5±\sqrt{9}}{4}=\frac{5±3}{4}$
${x}_{2}=\frac{5+3}{4}=\frac{8}{4};{x}_{3}=\frac{5-3}{4}=\frac{2}{4}$

${{t}}_{2}{=}{-4}{}{⇒}-4=x+\frac{1}{x}$
$-4=x+\frac{1}{x}\overline{)·x}$
$-4·x=x·x+\frac{1}{{x}}·{x}$
$-4x={x}^{2}+1$
${x}^{2}+4x+1=0$
${x}_{4,5}=\frac{-4±\sqrt{{4}^{2}-4·1·1}}{2·1}=\frac{-4±\sqrt{16-4}}{2}=\frac{4±\sqrt{12}}{2}=\frac{4±\sqrt{4·3}}{2}=\frac{4±\sqrt{4}·\sqrt{3}}{2}=\frac{4±2\sqrt{3}}{2}=\frac{2\left(2±\sqrt{3}\right)}{2}=2±\sqrt{3}$

${x}_{1}=1;{x}_{2}=2;{x}_{3}=-\frac{1}{2};{x}_{4}=-2+\sqrt{3};{x}_{5}=-2-\sqrt{3}$

${x}_{1}=1;{x}_{2}=2;{x}_{3}=-\frac{1}{2};{x}_{4}=-2+\sqrt{3};{x}_{5}=-2-\sqrt{3}$