$x^3-8x^2+30x-20=0$ has roots $a$, $b$, $c$. Find the equation with roots $a+2$, $b+2$, $c+2$ I have the equation $$x^3-8x^2+30x-20=0$$ let's call the roots $a,b,c$
It's easy to find the equation with roots $(a+2)(b+2)(c+2)$
these are the steps in my books

so the equation is $$ aX^3+ bX^2 + cX + D = 0 $$
$$ x^3 -14x^2 +74x -120 = 0$$
my problem is, I can't understand why divide by $-2$ not $2$
how do I do this division with long division? like what am I dividing by if the symbols were here?!

Alternatively: for the roots $a,b,c$ of $x^3-8x^2+30x-20=0$, by Vieta's we have:
$$a+b+c=8\\
ab+bc+ca=30\\
abc=20$$
Now we want the roots $a+2,b+2,c+2$ of $X^3+kX^2+lX+m=0$. Hence:
$$a+2+b+2+c+2=a+b+c+6=14=-k\\
(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2)=\\
(ab+bc+ca)+4(a+b+c)+12=74=l\\
(a+2)(b+2)(c+2)=\\
abc+2(ab+bc+ca)+4(a+b+c)+8=120=-m \Rightarrow \\
x^3-14x^2+74x-120=0.$$
