$\text{a)}$  

$\left(x-2\right)\left(x^2-6x+10\right)=2\left(x-2\right)\mid -2\left(x-2\right)$  

$\left(x-2\right)\left(x^2-6x+10\right)-2\left(x-2\right)=0$  

$\left(x-2\right)\left(x^2-6x+10-2\right)=0$         

$\left(x-2\right)\left(x^2-6x+8\right)=0$  

$\Delta ={\left(-6\right)}^2-4\cdot 1\cdot 8=36-32=4$  

$x_1=\frac{6-2}{2}=\frac{4}{2}=2$  

$x_2=\frac{6+2}{2}=\frac{8}{2}=4$  

 

$\left(x-2\right)\left(x-2\right)\left(x-4\right)=0$  

${\left(x-2\right)}^2\left(x-4\right)=0$  

$x-2=0\vee x-4=0$  

$x=2\vee x=4$  

Odp. 2 jest pierwiastkiem dwukrotnym, a 4 jest pierwiastkiem jednokrotnym.

---

$\text{b)}$  

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

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

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

$\Delta ={\left(-2\right)}^2-4\cdot 1\cdot \left(-3\right)=4+12=16$  

$x_1=\frac{2-4}{2}=\frac{-2}{2}=-1$  

$x_2=\frac{2+4}{2}=\frac{6}{2}=3$  

 

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

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

$x+1=0\vee x-3=0$  

$x=-1\vee x=3$  

Odp. -1 jest pierwiastkiem trzykrotnym, a 3 jest pierwiastkiem jednokrotnym.

---

$\text{c)}$  

$\left(x+2\right)\left(4x^3+8x^2\right)=x^2+4x+4$  

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

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

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

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

$x+2=0\vee 2x-1=0\vee 2x+1=0$  

$x=-2\vee x=\frac{1}{2}\vee x=-\frac{1}{2}$  

Odp. -2 jest pierwiastkiem dwukrotnym, a  $\frac{1}{2}\text{i}-\frac{1}{2}$   jest pierwiastkiem jednokrotnym.

---

$\text{d)}$  

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

${\left(x-1\right)}^2{\left(x+2\right)}^2\cdot \left[\left(x+2\right)+\left(x-1\right)\right]=0$  

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

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

$x-1=0\vee x+2=0\vee 2x+1=0$  

$x=1\vee x=-2\vee x=-\frac{1}{2}$  

Odp. 1 i -2 jest pierwiastkiem dwukrotnym, a $-\frac{1}{2}$   jest pierwiastkiem jednokrotnym.