Exercise:
Let $p\left( x \right)=-125{{x}^{4}}+50{{x}^{2}}-x-4\in \mathbb{R}\left[ x \right]$
Write $p\left( x \right)$ as product of two factors to be determined then deduce its real roots .
Solution: we have $p\left( x \right)=-125{{x}^{4}}+50{{x}^{2}}-x-4=-\left( 125{{x}^{4}}-50{{x}^{2}}+x+4 \right)$
$=-\left[ 5\left( 25{{x}^{4}}-10{{x}^{2}} \right)+x+4 \right]=-5\left( {{\left( 5{{x}^{2}} \right)}^{2}}-2\times 5{{x}^{2}}+1-1 \right)-x-4$
$\Rightarrow p\left( x \right)=-5{{\left( 5{{x}^{2}}-1 \right)}^{2}}+5-x-4=-5{{\left( 1-5{{x}^{2}} \right)}^{2}}-x+1$
Now Take $t=1-5{{x}^{2}}\Rightarrow t-1=-5{{x}^{2}}$
$\Rightarrow p\left( x \right)=-5{{t}^{2}}-x+t-t+1=-5{{t}^{2}}-\left( x-t \right)-\left( t-1 \right)$
$=-5{{t}^{2}}-\left( x-t \right)+5{{x}^{2}}=5\left( {{x}^{2}}-{{t}^{2}} \right)-\left( x-t \right)=5\left( x-t \right)\left( x+t \right)-\left( x-t \right)$
\(\therefore p\left( x \right)=\left( x-t \right)\left( 5\left( x+t \right)-1 \right)=\left( x+5{{x}^{2}}-1 \right)\left( 5x+4-25{{x}^{2}} \right)\)
To obtain the roots it’s more a simply just use $p\left( x \right)=0$
$\Rightarrow 5{{x}^{2}}+x-1=0\,\,\,\,or\,\,\,\,-25{{x}^{2}}+5x+4=0$ so each equation have two zeros
$5{{x}^{2}}+x-1=0\Rightarrow {{x}^{2}}+\frac{2}{2}\frac{1}{5}x-\frac{1}{5}=0\Rightarrow {{x}^{2}}+\frac{x}{5}+\frac{1}{100}-\frac{1}{100}-\frac{1}{5}=0$
$\Rightarrow {{\left( x+\frac{1}{10} \right)}^{2}}=\frac{1}{100}+\frac{1}{5}=\frac{1+20}{100}=\frac{21}{100}\Rightarrow x+\frac{1}{10}=\pm \frac{\sqrt{21}}{10}\Rightarrow x=\frac{-1\pm \sqrt{21}}{10}$
Also in the second equation we have two zeros
$-25{{x}^{2}}+5x+4=0\Rightarrow {{x}^{2}}-\frac{5}{25}x-\frac{4}{25}=0\Rightarrow {{x}^{2}}-\frac{x}{5}-\frac{4}{25}=0\Rightarrow {{x}^{2}}-\frac{x}{5}+\frac{1}{100}-\frac{1}{100}-\frac{4}{25}=0$
$\Rightarrow {{\left( x-\frac{1}{10} \right)}^{2}}=\frac{1}{100}+\frac{4}{25}=\frac{1+16}{100}=\frac{17}{100}$ $\Rightarrow x-\frac{1}{10}=\pm \frac{\sqrt{17}}{10}\Rightarrow x=\frac{1\pm \sqrt{17}}{10}$
No comments:
Post a Comment