$\text{a)}W\left(x\right)=x^2-x+\frac{1}{4}=x^2-2\cdot x\cdot \frac{1}{2}+{\left(\frac{1}{2}\right)}^2={\left(x-\frac{1}{2}\right)}^2$   

 

$\text{b)}W\left(x\right)=8x^3+27={\left(2x\right)}^3+3^3=\left(2x+3\right)\left({\left(2x\right)}^2-2x\cdot 3+3^2\right)=$  

$=\left(2x+3\right)\underset{P\left(x\right)}{\underbrace{\left(4x^2-6x+9\right)}}$  

Wyznaczamy pierwiastki wielomianu P. 

$4x^2-6x+9=0$  

$\Delta ={\left(-6\right)}^2-4\cdot 4\cdot 9=36-144=-108<0$  

Wielomian P nie ma pierwiastków. 

 

$\text{c)}W\left(x\right)=8x^3+12x^2+6x+1={\left(2x\right)}^2+3\cdot {\left(2x\right)}^2\cdot 1+3\cdot 2x\cdot 1+1^3={\left(2x+1\right)}^3$  

 

$\text{d)}W\left(x\right)=125x^3-8={\left(5x\right)}^3-2^3=\left(5x-2\right)\left({\left(5x\right)}^2+5x\cdot 2+2^3\right)=$  

$=\left(5x-2\right)\underset{P\left(x\right)}{\underbrace{\left(25x^2+10x+8\right)}}$  

Wyznaczamy pierwiastki wielomianu P. 

$25x^2+10x+8=0$  

$\Delta ={10}^2-4\cdot 25\cdot 8=100-800=-700<0$  

Wielomian P nie ma pierwiastków. 

 

$\text{e)}W\left(x\right)=81x^4-16={\left(9x^2\right)}^2-4^2=\left(9x^2+4\right)\left(9x^2-4\right)=$  

$=\left(9x^2+4\right)\left({\left(3x\right)}^2-2^2\right)=\left(9x^2+4\right)\left(3x+2\right)\left(3x-2\right)$   

 

$\text{f)}W\left(x\right)={\left(x-1\right)}^3+8={\left(x-1\right)}^3+2^3=\left(x-1+2\right)\left({\left(x-1\right)}^2-\left(x-1\right)\cdot 2+2^2\right)=$   

$=\left(x+1\right)\left(x^2-2x+1-2x+2+4\right)=\left(x+1\right)\underset{P\left(x\right)}{\underbrace{\left(x^2-4x+7\right)}}$     

Wyznaczamy pierwiastki wielomianu P. 

$\Delta ={\left(-4\right)}^2-4\cdot 1\cdot 7=16-28=-12<0$  

Wielomian P nie ma pierwiastków.