$a)$  

$a^2-5a=\underline{a}\cdot a-5\underline{a}=a\left(a-5\right)$    

 

$b)$  

$2b^2-4b+6=\underline{2}{b}^2-\underline{2}\cdot 2\cdot b+\underline{2}\cdot 3=2\left(b^2+2b+3\right)$ $2b^2-4b+6=\underline{2}{b}^2-\underline{2}\cdot 2\cdot b+\underline{2}\cdot 3=2\left(b^2+2b+3\right)$    

 

$c)$  

$3x^3+6x^2-12=\underline{3}{x}^3+\underline{3}\cdot 2\cdot x^2-\underline{3}\cdot 4=3\left(x^3+2x^2-4\right)$   

 

$d)$  

$3a^{2}b-2a{b}^2+ab=3\cdot \underline{a}\cdot a\cdot \underline{b}-2\cdot \underline{a}\cdot \underline{b}\cdot b+\underline{a}\cdot \underline{b}=ab\left(3a-2b+1\right)$  

 

$e)$  

$\frac{1}{2}ab-\frac{1}{4}a{b}^2+\frac{1}{2}a=\underline{\frac{1}{2}}\cdot \underline{a}\cdot b-\underline{\frac{1}{2}}\cdot \frac{1}{2}\cdot \underline{a}\cdot b^2+\underline{\frac{1}{2}}\cdot \underline{a}=\frac{1}{2}a\left(b-\frac{1}{2}{b}^2+1\right)$   

 

$f)$  

$10x^2+\frac{5}{x}=2\cdot \underline{5}\cdot x^2+\underline{5}\cdot \frac{1}{x}=5\left(2x^2+\frac{1}{x}\right)$  

 

$g)$  

$5x^3-25x^2-55x=\underline{5}\cdot x\cdot x\cdot \underline{x}-\underline{5}\cdot 5\cdot \underline{x}\cdot x-\underline{5}\cdot 11\cdot \underline{x}=5x\left(x^2-5x-11\right)$   

 

$h)$  

$-\frac{28}{x}+\frac{7}{x^2}+\frac{21}{x^3}=-\underline{\frac{7}{x}}\cdot 4+\underline{\frac{7}{x}}\cdot \frac{1}{x}+\underline{\frac{7}{x}}\cdot \frac{3}{x^2}=\frac{7}{x}\left(-4+\frac{1}{x}+\frac{3}{x^2}\right)=-\frac{7}{x}\left(4-\frac{1}{x}-\frac{3}{x^2}\right)$