From Förberedande kurs i matematik 1

First, we multiply the second bracket by \displaystyle x from the first bracket,

\displaystyle \left( x+2 \right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...

Then, do the same for \displaystyle 2 from the first bracket:

\displaystyle \left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5

Now, collect together \displaystyle x^{3}-, \displaystyle x^{2}-, \displaystyle x- and the constant terms:

\displaystyle 3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10

The coefficient in front of \displaystyle x^{2} is \displaystyle 5 and the coefficient in front of x is \displaystyle 3.